Answer: (i) Since \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), we have \(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}\). Multiplying by \(\mathbf{A}\) again gives

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda \mathbf{e})=\lambda \mathbf{A}\mathbf{e}=\lambda(\lambda \mathbf{e})=\lambda^2\mathbf{e}.\)

Hence \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\) with eigenvalue \(\lambda^2\).

(ii) First find the eigenvalues of \(\mathbf{A}\). Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries \(n, 2n, 3n\). Therefore the eigenvalues of \(\mathbf{A}+n\mathbf{I}\) are

\(n+n=2n, \quad 2n+n=3n, \quad 3n+n=4n.\)

By part (i), the eigenvalues of \(\mathbf{B}=(\mathbf{A}+n\mathbf{I})^2\) are the squares of these, so they are \(4n^2, 9n^2, 16n^2\).

Now use the fact that for an upper triangular matrix, an eigenvector for the first diagonal entry can be chosen from the first basis vector, and similarly for the others here. Compute eigenvectors of \(\mathbf{A}+n\mathbf{I}\):

• For eigenvalue \(2n\), solve \((\mathbf{A}+n\mathbf{I}-2n\mathbf{I})\mathbf{x}=\mathbf{0}\), i.e. \((\mathbf{A}-n\mathbf{I})\mathbf{x}=\mathbf{0}\). This gives \(x_2=x_3=0\), so an eigenvector is \((1,0,0)^T\).
• For eigenvalue \(3n\), solve \((\mathbf{A}+n\mathbf{I}-3n\mathbf{I})\mathbf{x}=\mathbf{0}\), i.e. \((\mathbf{A}-2n\mathbf{I})\mathbf{x}=\mathbf{0}\). This gives \(x_2=0\) and \((n-2n)x_1+3x_3=0\), so \(-n x_1+3x_3=0\). A suitable eigenvector is \((3,0,n)^T\).
• For eigenvalue \(4n\), solve \((\mathbf{A}+n\mathbf{I}-4n\mathbf{I})\mathbf{x}=\mathbf{0}\), i.e. \((\mathbf{A}-3n\mathbf{I})\mathbf{x}=\mathbf{0}\). This gives \(x_3=0\), \(2n-3n=-n\) so \(x_2=0\), and hence an eigenvector is \((0,0,1)^T\).

Therefore the corresponding eigenvectors of \(\mathbf{B}\) are the same, and we may take

\(\mathbf{P}=\begin{pmatrix}1&3&0\\0&0&0\\0&n&1\end{pmatrix}\)

but this is singular, so we choose a better set of independent eigenvectors. A simpler choice is

\(\mathbf{v}_1=(1,0,0)^T, \quad \mathbf{v}_2=(3,0,n)^T, \quad \mathbf{v}_3=(0,1,0)^T,\)

where \(\mathbf{v}_3\) is an eigenvector corresponding to \(4n\) after re-checking the third eigenspace. In fact, because \(\mathbf{A}+n\mathbf{I}\) is upper triangular with distinct diagonal entries, it is diagonalisable, and the eigenvectors can be arranged so that

\(\mathbf{P}=\begin{pmatrix}1&3&0\\0&0&1\\0&n&0\end{pmatrix}, \qquad \mathbf{D}=\begin{pmatrix}4n^2&0&0\\0&9n^2&0\\0&0&16n^2\end{pmatrix}.\)

Then \(\mathbf{B}=\mathbf{PDP}^{-1}\).

(i) If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), then

\(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}.\)

Apply \(\mathbf{A}\) once more:

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda(\lambda\mathbf{e})=\lambda^2\mathbf{e}.\)

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\), with corresponding eigenvalue \(\lambda^2\).

(ii) Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries \(n,2n,3n\). Hence the eigenvalues of \(\mathbf{A}+n\mathbf{I}\) are

\(2n,3n,4n.\)

By part (i), squaring the matrix squares its eigenvalues, so \(\mathbf{B}=(\mathbf{A}+n\mathbf{I})^2\) has eigenvalues

\(4n^2,9n^2,16n^2.\)

We now find eigenvectors of \(\mathbf{A}+n\mathbf{I}\), which are also eigenvectors of \(\mathbf{B}\).

For \(\lambda=2n\):

\(\mathbf{A}+n\mathbf{I}-2n\mathbf{I}=\begin{pmatrix}-n&1&3\\0&n&0\\0&0&2n\end{pmatrix}.\)

Solving \((\mathbf{A}+n\mathbf{I}-2n\mathbf{I})\mathbf{x}=\mathbf{0}\) gives \(x_2=0\), \(x_3=0\), and \(x_1\) free. So we may take

\(\mathbf{v}_1=\begin{pmatrix}1\\0\\0\end{pmatrix}.\)

For \(\lambda=3n\):

\(\mathbf{A}+n\mathbf{I}-3n\mathbf{I}=\begin{pmatrix}-2n&1&3\\0&0&0\\0&0&n\end{pmatrix}.\)

Solving gives \(x_3=0\) and \(-2n x_1+x_2=0\), so \(x_2=2n x_1\). A suitable eigenvector is

\(\mathbf{v}_2=\begin{pmatrix}1\\2n\\0\end{pmatrix}.\)

For \(\lambda=4n\):

\(\mathbf{A}+n\mathbf{I}-4n\mathbf{I}=\begin{pmatrix}-3n&1&3\\0&-n&0\\0&0&0\end{pmatrix}.\)

Solving gives \(x_2=0\), then \(-3n x_1+3x_3=0\), so \(x_3=nx_1\). A suitable eigenvector is

\(\mathbf{v}_3=\begin{pmatrix}1\\0\ \end{pmatrix}.\)

Take these as the columns of \(\mathbf{P}\):

\(\mathbf{P}=\begin{pmatrix}1&1&1\\0&2n&0\\0&0&n\end{pmatrix}.\)

Then \(\mathbf{P}\) is non-singular because its determinant is \(2n^2 eq 0\) since \(n eq 0\).

The corresponding diagonal matrix is

\(\mathbf{D}=\begin{pmatrix}4n^2&0&0\\0&9n^2&0\\0&0&16n^2\end{pmatrix}.\)

Therefore

\(\mathbf{B}=\mathbf{PDP}^{-1}\),

with

\(\mathbf{P}=\begin{pmatrix}1&1&1\\0&2n&0\\0&0&n\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}4n^2&0&0\\0&9n^2&0\\0&0&16n^2\end{pmatrix}.\)

Answer: (i) \(A=\dfrac{1}{b+1}\begin{pmatrix}1+2b&1&1\\-b&3b+2&-1\\2b&-2b&b+3\end{pmatrix}\).

(ii) \(A^{-1}\begin{pmatrix}0\\2\\-2\end{pmatrix}=\dfrac{2}{3}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\\frac23\\-\frac23\end{pmatrix}\).

(iii) \(n=2\) and \(b=1\).

(i) Put the three eigenvectors as the columns of a matrix \(P\), in the same order as their eigenvalues:

\(P=\begin{pmatrix}1&-1&0\\1&0&1\\0&b&-1\end{pmatrix}\), and \(D=\begin{pmatrix}2&0&0\\0&1&0\\0&0&3\end{pmatrix}\).

Then \(A=PDP^{-1}\).

Since \(b\) is positive, \(b+1\neq0\), so \(P\) is invertible. Its inverse is

\(P^{-1}=\dfrac{1}{b+1}\begin{pmatrix}b&1&1\\-1&1&1\\-b&b&-1\end{pmatrix}\).

Now

\(PD=\begin{pmatrix}2&-1&0\\2&0&3\\0&b&-3\end{pmatrix}\).

Therefore

\(A=PDP^{-1}=\dfrac{1}{b+1}\begin{pmatrix}2&-1&0\\2&0&3\\0&b&-3\end{pmatrix}\begin{pmatrix}b&1&1\\-1&1&1\\-b&b&-1\end{pmatrix}\).

Multiplying gives

\(A=\dfrac{1}{b+1}\begin{pmatrix}1+2b&1&1\\-b&3b+2&-1\\2b&-2b&b+3\end{pmatrix}\).

(ii) Notice that

\(\begin{pmatrix}0\\2\\-2\end{pmatrix}=2\begin{pmatrix}0\\1\\-1\end{pmatrix}\).

The vector \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) is an eigenvector of \(A\) with eigenvalue \(3\). Hence applying \(A^{-1}\) to it divides it by \(3\):

\(A^{-1}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\dfrac13\begin{pmatrix}0\\1\\-1\end{pmatrix}\).

So

\(A^{-1}\begin{pmatrix}0\\2\\-2\end{pmatrix}=\dfrac23\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\\frac23\\-\frac23\end{pmatrix}\).

(iii) Since \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is an eigenvector with eigenvalue \(2\),

\(A^n\begin{pmatrix}1\\1\\0\end{pmatrix}=2^n\begin{pmatrix}1\\1\\0\end{pmatrix}\).

But this is given as \(\begin{pmatrix}4\\4\\0\end{pmatrix}\), so \(2^n=4\). Therefore \(n=2\).

The vector \(\begin{pmatrix}-1\\0\\b\end{pmatrix}\) is an eigenvector with eigenvalue \(1\), so

\(A^n\begin{pmatrix}-1\\0\\b\end{pmatrix}=1^n\begin{pmatrix}-1\\0\\b\end{pmatrix}=\begin{pmatrix}-1\\0\\b\end{pmatrix}\).

This is given as \(\begin{pmatrix}-1\\0\\b^{-1}\end{pmatrix}\), so \(b=b^{-1}\). Since \(b\) is positive, \(b^2=1\) gives \(b=1\).

Answer: (i) Since \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), multiplying by \(\mathbf{A}\) again gives

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\).

Multiplying once more by \(\mathbf{A}\),

\(\mathbf{A}^3\mathbf{e}=\mathbf{A}(\lambda^2\mathbf{e})=\lambda^2\mathbf{A}\mathbf{e}=\lambda^3\mathbf{e}\).

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^3\) with corresponding eigenvalue \(\lambda^3\).

(ii) First find \(\mathbf{A}^3\). Since \(\mathbf{A}\) is triangular, its eigenvalues are its diagonal entries \(2\) and \(3\), so \(\mathbf{A}^3+\mathbf{I}\) has eigenvalues \(2^3+1=9\) and \(3^3+1=28\).

Now compute \(\mathbf{A}^3\) directly:

\(\mathbf{A}^2=\begin{pmatrix}2&0\\-1&3\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\),

so

\(\mathbf{A}^3=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}8&0\\-19&27\end{pmatrix}.\)

Hence

\(\mathbf{A}^3+\mathbf{I}=\begin{pmatrix}9&0\\-19&28\end{pmatrix}.\)

To diagonalise this matrix, find its eigenvectors.

For eigenvalue \(9\):

\(\begin{pmatrix}0&0\\-19&19\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\), so \(-19x+19y=0\), hence \(x=y\). Take \(\mathbf{v}_1=\begin{pmatrix}1\\1\end{pmatrix}.\)

For eigenvalue \(28\):

\(\begin{pmatrix}-19&0\\-19&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\begin{pmatrix}0\\0\end{pmatrix}\), so \(x=0\). Take \(\mathbf{v}_2=\begin{pmatrix}0\\1\end{pmatrix}.\)

Using these eigenvectors as columns of \(\mathbf{P}\),

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix}, \qquad \mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)

Then \(\mathbf{A}^3+\mathbf{I}=\mathbf{PDP}^{-1}\).

(i) Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\).

Applying \(\mathbf{A}\) once more gives

\(\mathbf{A}^3\mathbf{e}=\mathbf{A}(\mathbf{A}^2\mathbf{e})=\mathbf{A}(\lambda^2\mathbf{e})=\lambda^2\mathbf{A}\mathbf{e}=\lambda^3\mathbf{e}\).

Therefore \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^3\), with corresponding eigenvalue \(\lambda^3\).

(ii) We consider \(\mathbf{B}=\mathbf{A}^3+\mathbf{I}\), where

\(\mathbf{A}=\begin{pmatrix}2&0\\-1&3\end{pmatrix}.\)

First calculate \(\mathbf{A}^3\):

\(\mathbf{A}^2=\begin{pmatrix}2&0\\-1&3\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\),

so

\(\mathbf{A}^3=\mathbf{A}^2\mathbf{A}=\begin{pmatrix}4&0\\-5&9\end{pmatrix}\begin{pmatrix}2&0\\-1&3\end{pmatrix}=\begin{pmatrix}8&0\\-19&27\end{pmatrix}.\)

Hence

\(\mathbf{B}=\mathbf{A}^3+\mathbf{I}=\begin{pmatrix}9&0\\-19&28\end{pmatrix}.\)

This is triangular, so its eigenvalues are the diagonal entries \(9\) and \(28\).

For \(\lambda=9\), solve \((\mathbf{B}-9\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\begin{pmatrix}0&0\\-19&19\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{0}\),

so \(-19x+19y=0\), hence \(x=y\). A suitable eigenvector is \(\begin{pmatrix}1\\1\end{pmatrix}\).

For \(\lambda=28\), solve \((\mathbf{B}-28\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\begin{pmatrix}-19&0\\-19&0\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=\mathbf{0}\),

so \(x=0\). A suitable eigenvector is \(\begin{pmatrix}0\\1\end{pmatrix}\).

Take these eigenvectors as the columns of \(\mathbf{P}\):

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix}.\)

Then the corresponding diagonal matrix is

\(\mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)

Therefore \(\mathbf{A}^3+\mathbf{I}=\mathbf{PDP}^{-1}\), with

\(\mathbf{P}=\begin{pmatrix}1&0\\1&1\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}9&0\\0&28\end{pmatrix}.\)

Answer: (i) Since \(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}\) and \(\mathbf{B}\mathbf{e}=\mu \mathbf{e}\), we have

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{A}\mathbf{e}+\mathbf{B}\mathbf{e}=\lambda \mathbf{e}+\mu \mathbf{e}=(\lambda+\mu)\mathbf{e}.\)

So \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\) with eigenvector \(\mathbf{e}\).

(ii) Let the eigenvectors be \(\mathbf{v}_1=\begin{pmatrix}1\\2\\1\end{pmatrix}\), \(\mathbf{v}_2=\begin{pmatrix}1\\4\\-1\end{pmatrix}\), and \(\mathbf{v}_3=\begin{pmatrix}0\\1\\0\end{pmatrix}\).

For \(\mathbf{v}_1\):

\(\mathbf{A}\mathbf{v}_1=\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}3\\7\\3\end{pmatrix}=3\begin{pmatrix}1\\2\\1\end{pmatrix}.\)

So the eigenvalue is \(3\).

For \(\mathbf{v}_2\):

\(\mathbf{A}\mathbf{v}_2=\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}1\\4\\-1\end{pmatrix}=\begin{pmatrix}1\\6\\-1\end{pmatrix}=1\begin{pmatrix}1\\4\\-1\end{pmatrix}\) is not correct if read directly, so compute carefully:

First component: \(2(1)+0(4)+1(-1)=1\).

Second component: \(-1(1)+2(4)+3(-1)=-1+8-3=4\).

Third component: \(1(1)+0(4)+2(-1)=-1\).

Hence \(\mathbf{A}\mathbf{v}_2=\begin{pmatrix}1\\4\\-1\end{pmatrix}=1\mathbf{v}_2\), so the eigenvalue is \(1\).

For \(\mathbf{v}_3\):

\(\mathbf{A}\mathbf{v}_3=\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\2\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}.\)

So the eigenvalue is \(2\).

Therefore the eigenvalues of \(\mathbf{A}\) are \(3,1,2\), corresponding respectively to \(\begin{pmatrix}1\\2\\1\end{pmatrix}, \begin{pmatrix}1\\4\\-1\end{pmatrix}, \begin{pmatrix}0\\1\\0\end{pmatrix}\).

(iii) Using part (i), the eigenvalues of \(\mathbf{A}+\mathbf{B}\) are the sums of the corresponding eigenvalues of \(\mathbf{A}\) and \(\mathbf{B}\):

• for \(\begin{pmatrix}1\\2\\1\end{pmatrix}\): \(3+4=7\),
• for \(\begin{pmatrix}1\\4\\-1\end{pmatrix}\): \(1+5=6\),
• for \(\begin{pmatrix}0\\1\\0\end{pmatrix}\): \(2+1=3\).

So \(\mathbf{A}+\mathbf{B}\) is diagonalizable with eigenvectors given by these three vectors. Take

\(\mathbf{P}=\begin{pmatrix}1&1&0\\2&4&1\\1&-1&0\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}7&0&0\\0&6&0\\0&0&3\end{pmatrix}.\)

Then \(\mathbf{A}+\mathbf{B}=\mathbf{PDP}^{-1}\), so

\((\mathbf{A}+\mathbf{B})^3=\mathbf{P D}^3\mathbf{P}^{-1}\)

with

\(\mathbf{D}^3=\begin{pmatrix}343&0&0\\0&216&0\\0&0&27\end{pmatrix}.\)

Hence

\((\mathbf{A}+\mathbf{B})^3=\mathbf{P}\begin{pmatrix}343&0&0\\0&216&0\\0&0&27\end{pmatrix}\mathbf{P}^{-1},\)

where \(\mathbf{P}=\begin{pmatrix}1&1&0\\2&4&1\\1&-1&0\end{pmatrix}\).

(i) Let \(\mathbf{e}\) be a common eigenvector of \(\mathbf{A}\) and \(\mathbf{B}\), with \(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}\) and \(\mathbf{B}\mathbf{e}=\mu \mathbf{e}\). Then

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{A}\mathbf{e}+\mathbf{B}\mathbf{e}=\lambda \mathbf{e}+\mu \mathbf{e}=(\lambda+\mu)\mathbf{e}.\)

Therefore \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\), with the same eigenvector \(\mathbf{e}\).

(ii) For each given eigenvector, compute \(\mathbf{A}\mathbf{v}\) and compare with \(\mathbf{v}\).

For \(\mathbf{v}_1=\begin{pmatrix}1\\2\\1\end{pmatrix}\):

\(\mathbf{A}\mathbf{v}_1=\begin{pmatrix}2&0&1\\-1&2&3\\1&0&2\end{pmatrix}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}2+1\\-1+4+3\\1+2\end{pmatrix}=\begin{pmatrix}3\\6\\3\end{pmatrix}=3\begin{pmatrix}1\\2\\1\end{pmatrix}.\)

So the eigenvalue is \(3\).

For \(\mathbf{v}_2=\begin{pmatrix}1\\4\\-1\end{pmatrix}\):

\(\mathbf{A}\mathbf{v}_2=\begin{pmatrix}2(1)+0(4)+1(-1)\\-1(1)+2(4)+3(-1)\\1(1)+0(4)+2(-1)\end{pmatrix}=\begin{pmatrix}1\\4\\-1\end{pmatrix}=1\begin{pmatrix}1\\4\\-1\end{pmatrix}.\)

So the eigenvalue is \(1\).

For \(\mathbf{v}_3=\begin{pmatrix}0\\1\\0\end{pmatrix}\):

\(\mathbf{A}\mathbf{v}_3=\begin{pmatrix}0\\2\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix}.\)

So the eigenvalue is \(2\).

Thus the eigenpairs for \(\mathbf{A}\) are:

• \(\begin{pmatrix}1\\2\\1\end{pmatrix}\) with eigenvalue \(3\),
• \(\begin{pmatrix}1\\4\\-1\end{pmatrix}\) with eigenvalue \(1\),
• \(\begin{pmatrix}0\\1\\0\end{pmatrix}\) with eigenvalue \(2\).

(iii) The corresponding eigenvalues of \(\mathbf{B}\) are given as \(4,5,1\) for the same three eigenvectors. By part (i), the eigenvalues of \(\mathbf{A}+\mathbf{B}\) are therefore

\(3+4=7,\quad 1+5=6,\quad 2+1=3.\)

Take the eigenvectors as the columns of \(\mathbf{P}\):

\(\mathbf{P}=\begin{pmatrix}1&1&0\\2&4&1\\1&-1&0\end{pmatrix}.\)

Then the diagonal matrix of eigenvalues is

\(\mathbf{D}=\begin{pmatrix}7&0&0\\0&6&0\\0&0&3\end{pmatrix}.\)

Since the columns of \(\mathbf{P}\) are eigenvectors, \(\mathbf{A}+\mathbf{B}=\mathbf{PDP}^{-1}\), so

\((\mathbf{A}+\mathbf{B})^3=\mathbf{P D}^3\mathbf{P}^{-1}.\)

Now

\(\mathbf{D}^3=\begin{pmatrix}343&0&0\\0&216&0\\0&0&27\end{pmatrix}.\)

Hence a suitable answer is

\(\mathbf{P}=\begin{pmatrix}1&1&0\\2&4&1\\1&-1&0\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}343&0&0\\0&216&0\\0&0&27\end{pmatrix},\)

so that \((\mathbf{A}+\mathbf{B})^3=\mathbf{PDP}^{-1}.\)

Answer: (a) \(\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector because \(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-2\\-2\\-2\end{pmatrix}=-2\begin{pmatrix}1\\1\\1\end{pmatrix}\). The corresponding eigenvalue is \(-2\).

(b) The characteristic equation is \(\lambda^3-19\lambda-30=0\). Since \(\lambda=-2\) is a root, \(\lambda^3-19\lambda-30=(\lambda+2)(\lambda^2-2\lambda-15)=(\lambda+2)(\lambda-5)(\lambda+3)\). The other eigenvalues are \(5\) and \(-3\).

(c) \(\mathbf{A}^{-1}=\frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)

(a) Multiply \(\mathbf{A}\) by \(\begin{pmatrix}1\\1\\1\end{pmatrix}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-11+1+8\\0-2+0\\-16+1+13\end{pmatrix}=\begin{pmatrix}-2\\-2\\-2\end{pmatrix}.\)

Since this equals \(-2\begin{pmatrix}1\\1\\1\end{pmatrix}\), the vector is an eigenvector and the eigenvalue is \(\lambda=-2\).

(b) The characteristic equation is found from \(\det(\mathbf{A}-\lambda\mathbf{I})=0\):

\(\det\begin{pmatrix}-11-\lambda&1&8\\0&-2-\lambda&0\\-16&1&13-\lambda\end{pmatrix}=0.\)

Expanding along the second row,

\((-2-\lambda)\det\begin{pmatrix}-11-\lambda&8\\-16&13-\lambda\end{pmatrix}=0.\)

So

\((-2-\lambda)\big(( -11-\lambda)(13-\lambda)+128\big)=0.\)

Now

\(( -11-\lambda)(13-\lambda)+128=\lambda^2-2\lambda-143+128=\lambda^2-2\lambda-15.\)

Hence

\((-2-\lambda)(\lambda^2-2\lambda-15)=0.\)

This is equivalent to

\(\lambda^3-19\lambda-30=0.\)

Using the root \(\lambda=-2\),

\(\lambda^3-19\lambda-30=(\lambda+2)(\lambda^2-2\lambda-15)=(\lambda+2)(\lambda-5)(\lambda+3).\)

So the other eigenvalues are \(5\) and \(-3\).

(c) By Cayley-Hamilton, \(\mathbf{A}\) satisfies its characteristic equation:

\(\mathbf{A}^3-19\mathbf{A}-30\mathbf{I}=\mathbf{0}.\)

Multiply by \(\mathbf{A}^{-1}\):

\(\mathbf{A}^2-19\mathbf{I}-30\mathbf{A}^{-1}=\mathbf{0}.\)

So

\(30\mathbf{A}^{-1}=\mathbf{A}^2-19\mathbf{I}.\)

Now calculate

\(\mathbf{A}^2=\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}\begin{pmatrix}-11&1&8\\0&-2&0\\-16&1&13\end{pmatrix}=\begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix}.\)

Therefore

\(\mathbf{A}^2-19\mathbf{I}=\begin{pmatrix}-7&-5&16\\0&4&0\\-32&-5&41\end{pmatrix}-\begin{pmatrix}19&0&0\\0&19&0\\0&0&19\end{pmatrix}=\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)

Hence

\(\mathbf{A}^{-1}=\frac{1}{30}\begin{pmatrix}-26&-5&16\\0&-15&0\\-32&-5&22\end{pmatrix}.\)

Answer: (a) The eigenvalues of \(\mathbf{P}\) are \(1, 2, -1\).

(b) \(\mathbf{P}^{-1}=\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right)\).

(c) \(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a} \\ 0 & 2 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right)\).

(a) Since \(\mathbf{P}\) is upper triangular, its eigenvalues are the diagonal entries.

Hence the eigenvalues are \(1, 2, -1\).

(b) The characteristic polynomial is

\((\lambda-1)(\lambda-2)(\lambda+1)=\lambda^3-2\lambda^2-\lambda+2.\)

So \(\mathbf{P}\) satisfies

\(\mathbf{P}^3-2\mathbf{P}^2-\mathbf{P}+2\mathbf{I}=\mathbf{0}.\)

Since \(\mathbf{P}\) is invertible, multiply by \(\mathbf{P}^{-1}\):

\(\mathbf{P}^2-2\mathbf{P}-\mathbf{I}+2\mathbf{P}^{-1}=\mathbf{0},\)

so

\(2\mathbf{P}^{-1}=\mathbf{I}+2\mathbf{P}-\mathbf{P}^2.\)

Now

\(\mathbf{P}^2=\left(\begin{array}{ccc}1 & -3 & -1 \\ 0 & 4 & 1 \\ 0 & 0 & 1\end{array}\right).\)

Also,

\(\mathbf{I}+2\mathbf{P}=\left(\begin{array}{ccc}3 & -2 & 2 \\ 0 & 5 & 2 \\ 0 & 0 & -1\end{array}\right).\)

Therefore

\(2\mathbf{P}^{-1}=\left(\begin{array}{ccc}2 & 1 & 3 \\ 0 & 1 & 1 \\ 0 & 0 & -2\end{array}\right),\)

hence

\(\mathbf{P}^{-1}=\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right).\)

(c) Form the matrix of eigenvectors

\(\mathbf{P}=\left(\begin{array}{ccc}1 & -1 & 1 \\ 0 & 2 & 1 \\ 0 & 0 & -1\end{array}\right).\)

Then

\(\mathbf{A}=\mathbf{P}\left(\begin{array}{ccc}a & 0 & 0 \\ 0 & \frac{1}{2} & 0 \\ 0 & 0 & 2\end{array}\right)\mathbf{P}^{-1},\)

so

\(\mathbf{A}^{-1}=\mathbf{P}\left(\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right)\mathbf{P}^{-1}.\)

First,

\(\mathbf{P}\left(\begin{array}{ccc}\frac{1}{a} & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & \frac{1}{2}\end{array}\right)=\left(\begin{array}{ccc}\frac{1}{a} & -2 & \frac{1}{2} \\ 0 & 4 & \frac{1}{2} \\ 0 & 0 & -\frac{1}{2}\end{array}\right).\)

Then

\(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & -2 & \frac{1}{2} \\ 0 & 4 & \frac{1}{2} \\ 0 & 0 & -\frac{1}{2}\end{array}\right)\left(\begin{array}{ccc}1 & \frac{1}{2} & \frac{3}{2} \\ 0 & \frac{1}{2} & \frac{1}{2} \\ 0 & 0 & -1\end{array}\right).\)

Multiplying gives

\(\mathbf{A}^{-1}=\left(\begin{array}{ccc}\frac{1}{a} & \frac{1-2a}{2a} & \frac{3-3a}{2a} \\ 0 & 2 & \frac{3}{2} \\ 0 & 0 & \frac{1}{2}\end{array}\right).\)

Answer: (a) The eigenvalues of \(\mathbf{A}\) are \(1\), \(2\) and \(-3\).

(b) \(\mathbf{A}^{-1}=p\mathbf{A}^2+q\mathbf{I}\) with

\(p=-\frac{1}{6}\), \(q=\frac{7}{6}\).

So \(\mathbf{A}^{-1}=-\frac{1}{6}\mathbf{A}^2+\frac{7}{6}\mathbf{I}\).

(a) The eigenvalues are found from the characteristic equation

\(\det(\mathbf{A}-\lambda \mathbf{I})=0\).

Thus

\(\left|\begin{array}{ccc}6-\lambda & -9 & 5 \\ 5 & -8-\lambda & 5 \\ 1 & -1 & 2-\lambda\end{array}\right|=0\).

Expanding gives

\(\lambda^3-7\lambda+6=0\).

Factorising,

\(\lambda^3-7\lambda+6=(\lambda-1)(\lambda-2)(\lambda+3)\).

Hence the eigenvalues are

\(\lambda=1,\;2,\;-3\).

(b) Using the characteristic equation from part (a), and applying it to \(\mathbf{A}\),

\(\mathbf{A}^3-7\mathbf{A}+6\mathbf{I}=\mathbf{0}\).

Since none of the eigenvalues is \(0\), the matrix \(\mathbf{A}\) is invertible. Multiplying through by \(\mathbf{A}^{-1}\) gives

\(\mathbf{A}^2-7\mathbf{I}+6\mathbf{A}^{-1}=\mathbf{0}\).

So

\(6\mathbf{A}^{-1}=-\mathbf{A}^2+7\mathbf{I}\),

and therefore

\(\mathbf{A}^{-1}=-\frac{1}{6}\mathbf{A}^2+\frac{7}{6}\mathbf{I}\).

Thus \(p=-\frac{1}{6}\) and \(q=\frac{7}{6}\).

Answer: (i) The eigenvector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\) corresponds to the eigenvalue \(2\).

(ii) The negative eigenvalue is \(-2\), and a corresponding eigenvector is \(\begin{pmatrix}3\\-4\\0\end{pmatrix}\).

(iii) An eigenvalue of \(A+A^6\) is \(66\), with corresponding eigenvector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\).

Since \(A\) is upper triangular, its eigenvalues are the diagonal entries \(2\), \(-2\) and \(1\).

(i) For the eigenvector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\), we have

\(A\begin{pmatrix}1\\0\\0\end{pmatrix}=\begin{pmatrix}2\\0\\0\end{pmatrix}=2\begin{pmatrix}1\\0\\0\end{pmatrix}.\)

So the corresponding eigenvalue is \(2\).

(ii) The negative eigenvalue is \(-2\). To find a corresponding eigenvector, solve \((A+2I)\mathbf{x}=\mathbf{0}\):

\(A+2I=\begin{pmatrix}4&3&1\\0&0&1\\0&0&3\end{pmatrix}.\)

Let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then

\(\begin{cases}4x+3y+z=0,\\ z=0.\\\end{cases}\)

From \(z=0\), the first equation becomes \(4x+3y=0\), so we may take \(x=3\), \(y=-4\). Hence a corresponding eigenvector is

\(\begin{pmatrix}3\\-4\\0\end{pmatrix}.\)

(iii) If \(\mathbf{v}\) is an eigenvector of \(A\) with eigenvalue \(\lambda\), then

\(A\mathbf{v}=\lambda \mathbf{v}\)

and therefore

\(A^6\mathbf{v}=\lambda^6 \mathbf{v}.\)

So \(\mathbf{v}\) is also an eigenvector of \(A+A^6\), with eigenvalue \(\lambda+\lambda^6\).

Using the eigenvalue \(2\) from part (i) and eigenvector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\), we get

\(\left(A+A^6\right)\begin{pmatrix}1\\0\\0\end{pmatrix}=(2+2^6)\begin{pmatrix}1\\0\\0\end{pmatrix}=66\begin{pmatrix}1\\0\\0\end{pmatrix}.\)

So an eigenvalue of \(A+A^6\) is \(66\), with corresponding eigenvector \(\begin{pmatrix}1\\0\\0\end{pmatrix}\).

Answer: (a) A suitable set of eigenvectors is

• for eigenvalue \(1\): \(\mathbf{i}+2\mathbf{j}+2\mathbf{k}\)
• for eigenvalue \(-1\): \(\mathbf{i}+\mathbf{k}\)
• for eigenvalue \(-2\): \(\mathbf{j}+\mathbf{k}\)

(b) The eigenvalues of \(\mathbf{B}=\mathbf{A}-2\mathbf{I}\) are \(-1\), \(-3\) and \(-4\). A corresponding set of eigenvectors is the same as for \(\mathbf{A}\): \(\mathbf{i}+2\mathbf{j}+2\mathbf{k}\), \(\mathbf{i}+\mathbf{k}\) and \(\mathbf{j}+\mathbf{k}\).

(a) For each eigenvalue \(\lambda\), we find a non-zero vector \(\mathbf{x}\) such that \(\mathbf{A}\mathbf{x}=\lambda \mathbf{x}\). Solving the three systems gives the following eigenvectors.

For \(\lambda=1\), one eigenvector is \(\begin{pmatrix}1\\2\\2\end{pmatrix}\), i.e. \(\mathbf{i}+2\mathbf{j}+2\mathbf{k}\).

For \(\lambda=-1\), one eigenvector is \(\begin{pmatrix}1\\0\\1\end{pmatrix}\), i.e. \(\mathbf{i}+\mathbf{k}\).

For \(\lambda=-2\), one eigenvector is \(\begin{pmatrix}0\\1\\1\end{pmatrix}\), i.e. \(\mathbf{j}+\mathbf{k}\).

So a set of corresponding eigenvectors is \(\mathbf{i}+2\mathbf{j}+2\mathbf{k}\), \(\mathbf{i}+\mathbf{k}\), and \(\mathbf{j}+\mathbf{k}\).

(b) If \(\mathbf{A}\mathbf{x}=\lambda \mathbf{x}\), then

\(\mathbf{B}\mathbf{x}=(\mathbf{A}-2\mathbf{I})\mathbf{x}=\mathbf{A}\mathbf{x}-2\mathbf{x}=(\lambda-2)\mathbf{x}.\)

So each eigenvalue of \(\mathbf{B}\) is 2 less than the corresponding eigenvalue of \(\mathbf{A}\).

Hence the eigenvalues of \(\mathbf{B}\) are \(1-2=-1\), \(-1-2=-3\), and \(-2-2=-4\).

The eigenvectors are unchanged, so a corresponding set is

• for \(-1\): \(\mathbf{i}+2\mathbf{j}+2\mathbf{k}\)
• for \(-3\): \(\mathbf{i}+\mathbf{k}\)
• for \(-4\): \(\mathbf{j}+\mathbf{k}\)

Answer: (i) The corresponding eigenvalue is \(-2\).

(ii) A corresponding eigenvector is \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).

(iii) The third eigenvalue is \(5\), and a corresponding eigenvector is \(\begin{pmatrix}1\\1\\1\end{pmatrix}\).

(iv) One suitable choice is \(\mathbf{P}=\begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix}\) and \(\mathbf{D}=\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&5\end{pmatrix}\). Hence \(\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}\), so

\(\mathbf{A}^{n}=\begin{pmatrix}5^{n}+(-1)^{n}-(-2)^{n} & 2(-2)^{n}+(-1)^{n+1}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-2)^{n} & 2(-2)^{n}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-1)^{n} & (-1)^{n}-5^{n} & 5^{n}\end{pmatrix}.\)

(i) Since \(\begin{pmatrix}1\\1\\0\end{pmatrix}\) is an eigenvector, we compute \(\mathbf{A}\begin{pmatrix}1\\1\\0\end{pmatrix}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}6-8\\7-9\\6-6\end{pmatrix}=\begin{pmatrix}-2\\-2\\0\end{pmatrix}= -2\begin{pmatrix}1\\1\\0\end{pmatrix}.\)

So the corresponding eigenvalue is \(\lambda=-2\).

(ii) To find an eigenvector for \(\lambda=-1\), solve \((\mathbf{A}+\mathbf{I})\mathbf{x}=\mathbf{0}\).

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}7&-8&7\\7&-8&7\\6&-6&6\end{pmatrix}.\)

Let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then the equations are

\(7x-8y+7z=0\) and \(6x-6y+6z=0\), so from the second equation \(x-y+z=0\), i.e. \(x=y-z\).

Substitute into the first:

\(7(y-z)-8y+7z=0 \Rightarrow -y=0 \Rightarrow y=0.\)

Then \(x=-z\). Taking \(z=-1\) gives \(x=1\), so one eigenvector is

\(\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)

(iii) The determinant is the product of the eigenvalues. First calculate \(\det\mathbf{A}\):

Expand along the first row:

\(\det\mathbf{A}=6\begin{vmatrix}-9&7\\-6&5\end{vmatrix}-(-8)\begin{vmatrix}7&7\\6&5\end{vmatrix}+7\begin{vmatrix}7&-9\\6&-6\end{vmatrix}.\)

So

\(\det\mathbf{A}=6(({-9})(5)-7({-6})) + 8(7\cdot 5-7\cdot 6) + 7(7({-6})-({-9})6).\)

\(=6(-45+42)+8(35-42)+7(-42+54)=6(-3)+8(-7)+7(12).\)

\(=-18-56+84=10.\)

Thus the product of the three eigenvalues is \(10\). Since two eigenvalues are \(-2\) and \(-1\), the third eigenvalue \(\lambda_3\) satisfies

\(({-2})(-1)\lambda_3=10 \Rightarrow 2\lambda_3=10 \Rightarrow \lambda_3=5.\)

For \(\lambda=5\), solve \((\mathbf{A}-5\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{A}-5\mathbf{I}=\begin{pmatrix}1&-8&7\\7&-14&7\\6&-6&0\end{pmatrix}.\)

From the third row, \(6x-6y=0\), so \(x=y\). Then the first row gives

\(x-8y+7z=0.\)

With \(x=y\), this becomes \(-7y+7z=0\), so \(y=z\). Hence \(x=y=z\), and a corresponding eigenvector is

\(\begin{pmatrix}1\\1\\1\end{pmatrix}.\)

(iv) Use the eigenvectors as the columns of \(\mathbf{P}\), in the same order as the eigenvalues in \(\mathbf{D}\):

\(\mathbf{P}=\begin{pmatrix}1&1&1\\1&0&1\\0&-1&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}-2&0&0\\0&-1&0\\0&0&5\end{pmatrix}.\)

Then \(\mathbf{P}^{-1}\mathbf{A}\mathbf{P}=\mathbf{D}\), so \(\mathbf{A}=\mathbf{P}\mathbf{D}\mathbf{P}^{-1}\), and therefore

\(\mathbf{A}^n=\mathbf{P}\mathbf{D}^n\mathbf{P}^{-1}.\)

Now \(\mathbf{D}^n=\begin{pmatrix}(-2)^n&0&0\\0&(-1)^n&0\\0&0&5^n\end{pmatrix}.\)

Also \(\det\mathbf{P}=-1\), so

\(\mathbf{P}^{-1}=\begin{pmatrix}-1&2&-1\\1&-1&0\\1&-1&1\end{pmatrix}.\)

Multiplying out gives

\(\mathbf{A}^{n}=\begin{pmatrix}5^{n}+(-1)^{n}-(-2)^{n} & 2(-2)^{n}+(-1)^{n+1}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-2)^{n} & 2(-2)^{n}-5^{n} & 5^{n}-(-2)^{n}\\5^{n}-(-1)^{n} & (-1)^{n}-5^{n} & 5^{n}\end{pmatrix}.\)

Answer:

Suppose \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{A}\) is non-singular and \(\mathbf{e}\ne\mathbf{0}\).

Multiplying by \(\mathbf{A}^{-1}\) gives

\(\mathbf{e}=\lambda\mathbf{A}^{-1}\mathbf{e}.\)

Since \(\lambda\ne0\), divide by \(\lambda\):

\(\mathbf{A}^{-1}\mathbf{e}=\lambda^{-1}\mathbf{e}.\)

Therefore \(\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}^{-1}\), with the same eigenvector \(\mathbf{e}\).

Also,

\(\mathbf{A}\mathbf{e}+\mathbf{A}^{-1}\mathbf{e}=\lambda\mathbf{e}+\lambda^{-1}\mathbf{e},\)

so

\(\left(\mathbf{A}+\mathbf{A}^{-1}\right)\mathbf{e}=\left(\lambda+\lambda^{-1}\right)\mathbf{e}.\)

Hence \(\lambda+\lambda^{-1}\) is an eigenvalue of \(\mathbf{A}+\mathbf{A}^{-1}\).

For the eigenvalue \(1\), solve \(\left(\mathbf{A}-\mathbf{I}\right)\mathbf{x}=\mathbf{0}\):

\(\begin{pmatrix}1&0&1\\-1&1&3\\1&0&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\)

The first row gives \(x+z=0\), so \(z=-x\). The second row gives \(-x+y+3z=0\), so \(y=4x\). Taking \(x=-1\) gives the eigenvector

\(\begin{pmatrix}-1\\-4\\1\end{pmatrix}.\)

Now check the two given eigenvectors:

\(\mathbf{A}\begin{pmatrix}0\\1\\0\end{pmatrix}=\begin{pmatrix}0\\2\\0\end{pmatrix}=2\begin{pmatrix}0\\1\\0\end{pmatrix},\)

so the corresponding eigenvalue is \(2\).

Also,

\(\mathbf{A}\begin{pmatrix}1\\2\\1\end{pmatrix}=\begin{pmatrix}3\\6\\3\end{pmatrix}=3\begin{pmatrix}1\\2\\1\end{pmatrix},\)

so the corresponding eigenvalue is \(3\).

The eigenvalues of \(\mathbf{A}+\mathbf{A}^{-1}\) are therefore \(2\), \(\frac{5}{2}\) and \(\frac{10}{3}\). Hence the eigenvalues of \(\left(\mathbf{A}+\mathbf{A}^{-1}\right)^3\) are

\(2^3=8,\qquad \left(\frac{5}{2}\right)^3=\frac{125}{8},\qquad \left(\frac{10}{3}\right)^3=\frac{1000}{27}.\)

Using the corresponding eigenvectors as columns, one valid choice is

\(\mathbf{P}=\begin{pmatrix}-1&0&1\\-4&1&2\\1&0&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}8&0&0\\0&\frac{125}{8}&0\\0&0&\frac{1000}{27}\end{pmatrix}.\)

Therefore \(\left(\mathbf{A}+\mathbf{A}^{-1}\right)^3=\mathbf{PDP}^{-1}\).

Answer: The corresponding eigenvalue is \(-1\).

The other eigenvalues are \(-3\) and \(-4\). Corresponding eigenvectors may be taken as \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\) for \(-3\), and \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\) for \(-4\).

Let \(\mathbf{v}=\begin{pmatrix}0\\1\\-1\end{pmatrix}\). Then

\(\mathbf{M}\mathbf{v}=\begin{pmatrix}-2 & 2 & 2\\2 & 1 & 2\\-3 & -6 & -7\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\-1\\1\end{pmatrix}= -1\begin{pmatrix}0\\1\\-1\end{pmatrix}.\)

So the corresponding eigenvalue is \(-1\).

Now let the other two eigenvalues be \(\lambda_2\) and \(\lambda_3\). The trace of \(\mathbf{M}\) is

\(-2+1-7=-8.\)

Hence

\(\lambda_1+\lambda_2+\lambda_3=-8.\)

Since \(\lambda_1=-1\),

\(\lambda_2+\lambda_3=-7.\)

Also the determinant is the product of the eigenvalues. Compute

\(\det\mathbf{M}=\begin{vmatrix}-2 & 2 & 2\\2 & 1 & 2\\-3 & -6 & -7\end{vmatrix}.\)

Expanding along the first row,

\(\det\mathbf{M}=-2\begin{vmatrix}1 & 2\\-6 & -7\end{vmatrix}-2\begin{vmatrix}2 & 2\\-3 & -7\end{vmatrix}+2\begin{vmatrix}2 & 1\\-3 & -6\end{vmatrix}.\)

So

\(\det\mathbf{M}=-2( -7+12 )-2( -14+6 )+2( -12+3 )=-2(5)-2(-8)+2(-9)=-10+16-18=-12.\)

Therefore

\(\lambda_1\lambda_2\lambda_3=-12.\)

With \(\lambda_1=-1\), this gives

\(\lambda_2\lambda_3=12.\)

So \(\lambda_2\) and \(\lambda_3\) satisfy

\(t^2-(\lambda_2+\lambda_3)t+\lambda_2\lambda_3=0\),

that is

\(t^2+7t+12=0.\)

Hence

\((t+3)(t+4)=0,\)

so the other eigenvalues are \(-3\) and \(-4\).

For \(\lambda=-3\), solve \((\mathbf{M}+3\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+3\mathbf{I}=\begin{pmatrix}1 & 2 & 2\\2 & 4 & 2\\-3 & -6 & -4\end{pmatrix}.\)

From the first equation, \(x+2y+2z=0\). Taking \(z=0\) and \(x=2\) gives \(y=-1\), so an eigenvector is

\(\begin{pmatrix}2\\-1\\0\end{pmatrix}.\)

For \(\lambda=-4\), solve \((\mathbf{M}+4\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+4\mathbf{I}=\begin{pmatrix}2 & 2 & 2\\2 & 5 & 2\\-3 & -6 & -3\end{pmatrix}.\)

Taking \(y=0\) and \(x=1\) gives \(z=-1\), so an eigenvector is

\(\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)

Answer:

For \(\lambda=1\), solve \((A-I)\mathbf{x}=0\):

\(A-I=\begin{pmatrix}0&0&0\\10&-8&10\\7&-5&7\end{pmatrix}.\)

If \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\), the two non-zero rows give

\(10x-8y+10z=0,\qquad 7x-5y+7z=0.\)

These equations give \(y=0\) and \(x=-z\). A convenient eigenvector is therefore

\(\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)

For \(\lambda=3\),

\(A-3I=\begin{pmatrix}-2&0&0\\10&-10&10\\7&-5&5\end{pmatrix}.\)

The first row gives \(x=0\), and the remaining rows give \(y=z\). Hence a corresponding eigenvector is

\(\begin{pmatrix}0\\1\\1\end{pmatrix}.\)

Now check the given eigenvector:

\(A\begin{pmatrix}0\\2\\1\end{pmatrix}=\begin{pmatrix}0\\-4\\-2\end{pmatrix}=-2\begin{pmatrix}0\\2\\1\end{pmatrix}.\)

So its eigenvalue is \(-2\).

To diagonalise, place the eigenvectors as the columns of \(P\) in the same order as the eigenvalues in \(D\):

\(P=\begin{pmatrix}0&1&0\\2&0&1\\1&-1&1\end{pmatrix},\qquad D=\begin{pmatrix}-2&0&0\\0&1&0\\0&0&3\end{pmatrix}.\)

Then

\(P^{-1}=\begin{pmatrix}-1&1&-1\\1&0&0\\2&-1&2\end{pmatrix},\)

and these matrices satisfy

\(P^{-1}AP=D.\)

Answer: \(\alpha=-1\).

The other eigenvalues are \(\lambda_1=6\) and \(\lambda_2=3\).

Corresponding eigenvectors are \(\begin{pmatrix}1\\2\\-2\end{pmatrix}\) for \(\lambda=-9\), \(\mathbf e_1=\begin{pmatrix}-2\\2\\1\end{pmatrix}\) for \(\lambda_1=6\), and \(\mathbf e_2=\begin{pmatrix}2\\1\\2\end{pmatrix}\) for \(\lambda_2=3\).

\(\mathbf M\mathbf x=6a\mathbf e_1+3b\mathbf e_2\), so \(p=6a\) and \(q=3b\).

Since \(-9\) is an eigenvalue, \(\det(\mathbf M+9\mathbf I)=0\).

Thus

\(\begin{vmatrix}12&-4&2\\-4&\alpha+9&6\\2&6&7\end{vmatrix}=0\).

Expanding gives \(80\alpha+80=0\), so \(\alpha=-1\).

With \(\alpha=-1\),

\(\mathbf M=\begin{pmatrix}3&-4&2\\-4&-1&6\\2&6&-2\end{pmatrix}\).

The characteristic equation is

\(\det(\mathbf M-\lambda\mathbf I)=0\),

which simplifies to

\(\lambda^3-63\lambda+162=0\).

Since \(-9\) is already a root,

\(\lambda^3-63\lambda+162=(\lambda+9)(\lambda-6)(\lambda-3)\).

Therefore the other two eigenvalues are \(\lambda_1=6\) and \(\lambda_2=3\).

Solving \((\mathbf M-\lambda\mathbf I)\mathbf e=\mathbf0\) for each eigenvalue gives the following corresponding eigenvectors:

for \(\lambda=-9\), \(\mathbf e=\begin{pmatrix}1\\2\\-2\end{pmatrix}\);

for \(\lambda_1=6\), \(\mathbf e_1=\begin{pmatrix}-2\\2\\1\end{pmatrix}\);

for \(\lambda_2=3\), \(\mathbf e_2=\begin{pmatrix}2\\1\\2\end{pmatrix}\).

Now \(\mathbf x=a\mathbf e_1+b\mathbf e_2\). Since \(\mathbf e_1\) and \(\mathbf e_2\) are eigenvectors corresponding to \(6\) and \(3\), respectively,

\(\mathbf M\mathbf e_1=6\mathbf e_1\), and \(\mathbf M\mathbf e_2=3\mathbf e_2\).

Hence

\(\mathbf M\mathbf x=\mathbf M(a\mathbf e_1+b\mathbf e_2)=a\mathbf M\mathbf e_1+b\mathbf M\mathbf e_2=6a\mathbf e_1+3b\mathbf e_2\).

Therefore \(p=6a\) and \(q=3b\).

Answer: The matrix is \(\mathbf{A}=\begin{pmatrix}-1&2&-2\\0&1&1\\0&0&2\end{pmatrix}\).

Let the eigenvectors be \(v_1=\begin{pmatrix}1\\0\\0\end{pmatrix}\), \(v_2=\begin{pmatrix}1\\1\\0\end{pmatrix}\), and \(v_3=\begin{pmatrix}0\\1\\1\end{pmatrix}\), corresponding to eigenvalues \(-1\), \(1\), and \(2\) respectively.

Since \(A v = \lambda v\) for each eigenpair, we can write the columns of \(A\) in terms of these eigenvectors by forming the matrix \(P\) whose columns are the eigenvectors and the diagonal matrix \(D\) of eigenvalues:

\(P=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}, \qquad D=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}.\)

Then \(A=PD P^{-1}\). First, \(P\) is upper triangular with determinant \(1\), so it is invertible. Its inverse is

\(P^{-1}=\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}.\)

Now calculate \(PD\):

\(PD=\begin{pmatrix}1&1&0\\0&1&1\\0&0&1\end{pmatrix}\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}=\begin{pmatrix}-1&1&0\\0&1&2\\0&0&2\end{pmatrix}.\)

So

\(A=(PD)P^{-1}=\begin{pmatrix}-1&1&0\\0&1&2\\0&0&2\end{pmatrix}\begin{pmatrix}1&-1&1\\0&1&-1\\0&0&1\end{pmatrix}=\begin{pmatrix}-1&2&-2\\0&1&1\\0&0&2\end{pmatrix}.\)

Therefore \(\mathbf{A}=\begin{pmatrix}-1&2&-2\\0&1&1\\0&0&2\end{pmatrix}\).

Answer: (i) \(\mathbf{ABe}=\lambda\mu\mathbf e\).

(ii) The eigenvalues of \(\mathbf A\) are \(-1,2,3\), with corresponding eigenvectors \(\begin{pmatrix}1\\-1\\0\end{pmatrix}\), \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) and \(\begin{pmatrix}1\\0\\1\end{pmatrix}\).

(iii) The eigenvalues of \(\mathbf{AB}\) are \(3,-4,12\), with the same corresponding eigenvectors in that order.

(i) Since \(\mathbf{Ae}=\lambda\mathbf e\) and \(\mathbf{Be}=\mu\mathbf e\),

\(\mathbf{ABe}=\mathbf A(\mathbf{Be})=\mathbf A(\mu\mathbf e)=\mu\mathbf{Ae}=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

(ii) The characteristic equation is

\((\lambda+1)(\lambda^2-5\lambda+6)=0=(\lambda+1)(\lambda-2)(\lambda-3).\)

So \(\lambda=-1,2,3\). Solving \((\mathbf A-\lambda\mathbf I)\mathbf x=\mathbf0\) gives the corresponding eigenvectors

\(\lambda=-1:\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \lambda=2:\begin{pmatrix}1\\-1\\1\end{pmatrix},\qquad \lambda=3:\begin{pmatrix}1\\0\\1\end{pmatrix}.\)

(iii) For the first eigenvector,

\(\mathbf B\begin{pmatrix}1\\-1\\0\end{pmatrix}=\begin{pmatrix}-3\\3\\0\end{pmatrix}=-3\begin{pmatrix}1\\-1\\0\end{pmatrix},\)

so the corresponding eigenvalue of \(\mathbf B\) is \(-3\). Similarly, the other two eigenvalues of \(\mathbf B\) are \(-2\) and \(4\). By part (i), the eigenvalues of \(\mathbf{AB}\) are the products of the matching eigenvalues:

\((-1)(-3)=3,\qquad 2(-2)=-4,\qquad 3(4)=12.\)

The corresponding eigenvectors are

\(\begin{pmatrix}1\\-1\\0\end{pmatrix},\qquad \begin{pmatrix}1\\-1\\1\end{pmatrix},\qquad \begin{pmatrix}1\\0\\1\end{pmatrix}.\)

Answer: The eigenvalues of \(\mathbf{A}\) are \(2\), \(1\) and \(-1\).

For \(\mathbf{B}\), the vector \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector with eigenvalue \(3\).

Hence an eigenvector of \(\mathbf{A B}\) is \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\), with corresponding eigenvalue \(6\).

First, for \(\mathbf{A}\) and \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\),

\(\mathbf{A e}=\begin{pmatrix}4 & -5 & 3\\3 & -4 & 3\\1 & -1 & 2\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}4-5+3\\3-4+3\\1-1+2\end{pmatrix}=\begin{pmatrix}2\\2\\2\end{pmatrix}=2\mathbf{e}.\)

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\), with eigenvalue \(2\).

To find the other eigenvalues, use the characteristic equation:

\(\det(\mathbf{A}-\lambda\mathbf{I})=0.\)

This gives

\(\lambda^3-2\lambda^2-\lambda+2=0.\)

Factorising,

\(\lambda^3-2\lambda^2-\lambda+2=(\lambda-2)(\lambda^2-1)=(\lambda-2)(\lambda-1)(\lambda+1).\)

Hence the other two eigenvalues are \(1\) and \(-1\).

Now for \(\mathbf{B}\),

\(\mathbf{B e}=\begin{pmatrix}-1 & 4 & 0\\-1 & 3 & 1\\1 & -1 & 3\end{pmatrix}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1+4+0\\-1+3+1\\1-1+3\end{pmatrix}=\begin{pmatrix}3\\3\\3\end{pmatrix}=3\mathbf{e}.\)

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{B}\), with eigenvalue \(3\).

Therefore

\(\mathbf{A B e}=\mathbf{A}(\mathbf{B e})=\mathbf{A}(3\mathbf{e})=3\mathbf{A e}=3(2\mathbf{e})=6\mathbf{e}.\)

So \(\mathbf{e}=\begin{pmatrix}1\\1\\1\end{pmatrix}\) is an eigenvector of \(\mathbf{A B}\), with corresponding eigenvalue \(6\).

Answer: Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

\(\mathbf{B}(\mathbf{M}\mathbf{e})=\mathbf{MAM}^{-1}(\mathbf{M}\mathbf{e})=\mathbf{MAe}=\mathbf{M}(\lambda\mathbf{e})=\lambda(\mathbf{M}\mathbf{e}).\)

Since \(\mathbf{M}\) is non-singular, \(\mathbf{M}\mathbf{e}\neq \mathbf{0}\). Therefore \(\mathbf{M}\mathbf{e}\) is an eigenvector of \(\mathbf{B}\) with corresponding eigenvalue \(\lambda\).

The eigenvalues of \(\mathbf{A}\) are \(-1,1,2\).

Corresponding eigenvectors for \(\mathbf{A}\) may be taken as

• for \(\lambda=-1\): \(\left(\begin{array}{c}1\\0\\0\end{array}\right)\)
• for \(\lambda=1\): \(\left(\begin{array}{c}1\\1\\0\end{array}\right)\)
• for \(\lambda=2\): \(\left(\begin{array}{c}9\\4\\1\end{array}\right)\)

The eigenvalues of \(\mathbf{B}\) are also \(-1,1,2\).

Using \(\mathbf{M}\mathbf{e}\):

• for \(\lambda=-1\): \(\mathbf{M}\left(\begin{array}{c}1\\0\\0\end{array}\right)=\left(\begin{array}{c}1\\0\\0\end{array}\right)\)
• for \(\lambda=1\): \(\mathbf{M}\left(\begin{array}{c}1\\1\\0\end{array}\right)=\left(\begin{array}{c}1\\1\\0\end{array}\right)\)
• for \(\lambda=2\): \(\mathbf{M}\left(\begin{array}{c}9\\4\\1\end{array}\right)=\left(\begin{array}{c}10\\4\\1\end{array}\right)\)

So corresponding eigenvectors of \(\mathbf{B}\) may be taken as \(\left(\begin{array}{c}1\\0\\0\end{array}\right)\), \(\left(\begin{array}{c}1\\1\\0\end{array}\right)\), \(\left(\begin{array}{c}10\\4\\1\end{array}\right)\).

Let \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{e}\neq \mathbf{0}\). Since

\(\mathbf{B}=\mathbf{MAM}^{-1},\)

we have

\(\mathbf{B}(\mathbf{M}\mathbf{e})=\mathbf{MAM}^{-1}(\mathbf{M}\mathbf{e})=\mathbf{MAe}=\mathbf{M}(\lambda\mathbf{e})=\lambda(\mathbf{M}\mathbf{e}).\)

Also, \(\mathbf{M}\) is non-singular, so \(\mathbf{M}\mathbf{e}\neq \mathbf{0}\). Hence \(\mathbf{M}\mathbf{e}\) is an eigenvector of \(\mathbf{B}\) with eigenvalue \(\lambda\).

Now

\(\mathbf{A}=\begin{pmatrix}-1&2&1\\0&1&4\\0&0&2\end{pmatrix}.\)

Since \(\mathbf{A}\) is upper triangular, its eigenvalues are the diagonal entries:

\(\lambda=-1,\ 1,\ 2.\)

Eigenvectors of \(\mathbf{A}\)

For \(\lambda=-1\):

\(\mathbf{A}+\mathbf{I}=\begin{pmatrix}0&2&1\\0&2&4\\0&0&3\end{pmatrix}.\)

So \(z=0\), then \(y=0\), and \(x\) is free. Take \(x=1\):

\(\mathbf{e}_{-1}=\left(\begin{array}{c}1\\0\\0\end{array}\right).\)

For \(\lambda=1\):

\(\mathbf{A}-\mathbf{I}=\begin{pmatrix}-2&2&1\\0&0&4\\0&0&1\end{pmatrix}.\)

So \(z=0\), and \(-2x+2y=0\Rightarrow x=y\). Take \(x=y=1\):

\(\mathbf{e}_{1}=\left(\begin{array}{c}1\\1\\0\end{array}\right).\)

For \(\lambda=2\):

\(\mathbf{A}-2\mathbf{I}=\begin{pmatrix}-3&2&1\\0&-1&4\\0&0&0\end{pmatrix}.\)

Let \(z=t\). Then from the second row, \(-y+4t=0\Rightarrow y=4t\). From the first row, \(-3x+2y+z=0\), so

\(-3x+8t+t=0\Rightarrow 3x=9t\Rightarrow x=3t.\)

Taking \(t=1\), one eigenvector is

\(\mathbf{e}_{2}=\left(\begin{array}{c}3\\4\\1\end{array}\right).\)

So a suitable set of eigenvectors for \(\mathbf{A}\) is

\(\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad \left(\begin{array}{c}1\\1\\0\end{array}\right),\quad \left(\begin{array}{c}3\\4\\1\end{array}\right).\)

Now

\(\mathbf{M}=\begin{pmatrix}1&0&1\\0&1&0\\0&0&1\end{pmatrix}.\)

The eigenvalues of \(\mathbf{B}=\mathbf{MAM}^{-1}\) are the same as those of \(\mathbf{A}\), namely

\(-1,\ 1,\ 2.\)

Corresponding eigenvectors are \(\mathbf{M}\mathbf{e}\):

For \(\lambda=-1\):

\(\mathbf{M}\left(\begin{array}{c}1\\0\\0\end{array}\right)=\left(\begin{array}{c}1\\0\\0\end{array}\right).\)

For \(\lambda=1\):

\(\mathbf{M}\left(\begin{array}{c}1\\1\\0\end{array}\right)=\left(\begin{array}{c}1\\1\\0\end{array}\right).\)

For \(\lambda=2\):

\(\mathbf{M}\left(\begin{array}{c}3\\4\\1\end{array}\right)=\left(\begin{array}{c}4\\4\\1\end{array}\right).\)

Therefore corresponding eigenvectors of \(\mathbf{B}\) may be taken as

\(\left(\begin{array}{c}1\\0\\0\end{array}\right),\quad \left(\begin{array}{c}1\\1\\0\end{array}\right),\quad \left(\begin{array}{c}4\\4\\1\end{array}\right).\)

Answer: For an eigenvector \(\mathbf{e}\) of \(\mathbf{A}\) with eigenvalue \(\lambda\), we have \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\). Then

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\).

So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\) with corresponding eigenvalue \(\lambda^2\).

The eigenvalues of \(\mathbf{B}\) are \(-2\), \(1\), and \(4\).

Hence the eigenvalues of \(\mathbf{B}^2\) are \(4\), \(1\), and \(16\), and the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are obtained by applying \(x^4+2x^2+3\) to each eigenvalue of \(\mathbf{B}\):

For \(x=-2\): \((-2)^4+2(-2)^2+3=16+8+3=27\).

For \(x=1\): \(1^4+2(1)^2+3=1+2+3=6\).

For \(x=4\): \(4^4+2(4)^2+3=256+32+3=291\).

Therefore the eigenvalues are \(27\), \(6\), and \(291\).

Let \(\mathbf{e}\neq \mathbf{0}\) be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\). Then

\(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\).

Multiplying both sides by \(\mathbf{A}\) gives

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda(\lambda\mathbf{e})=\lambda^2\mathbf{e}.\)

Hence \(\mathbf{e}\) is also an eigenvector of \(\mathbf{A}^2\), with corresponding eigenvalue \(\lambda^2\).

Now consider

\(\mathbf{B}=\begin{pmatrix}1&3&0\\2&0&2\\1&1&2\end{pmatrix}.\)

The characteristic equation is

\(\det(\mathbf{B}-\lambda\mathbf{I})=0.\)

Expanding, this gives

\((1-\lambda)(\lambda-4)(\lambda+2)=0.\)

So the eigenvalues of \(\mathbf{B}\) are

\(\lambda=-2,\ 1,\ 4.\)

To find the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\), use the fact that if \(\lambda\) is an eigenvalue of \(\mathbf{B}\), then \(\mathbf{B}^2\) has eigenvalue \(\lambda^2\), and more generally a polynomial in \(\mathbf{B}\) has the corresponding polynomial in \(\lambda\) as an eigenvalue.

So the required eigenvalues are

\(\lambda^4+2\lambda^2+3\)

for \(\lambda=-2,1,4\).

• For \(\lambda=-2\): \((-2)^4+2(-2)^2+3=16+8+3=27\).
• For \(\lambda=1\): \(1^4+2(1)^2+3=1+2+3=6\).
• For \(\lambda=4\): \(4^4+2(4)^2+3=256+32+3=291\).

Therefore the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are \(27\), \(6\), and \(291\).

Answer: (a) If \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), then multiplying by \(\mathbf{A}\) gives \(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda^2\mathbf{e}\). So \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\) with eigenvalue \(\lambda^2\).

The eigenvalues of \(\mathbf{B}\) are \(-2\), \(1\) and \(4\).

The eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are \(6\), \(27\) and \(291\).

(a) Suppose \(\mathbf{A}\mathbf{e}=\lambda\mathbf{e}\), where \(\mathbf{e}\neq \mathbf{0}\). Then

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda\mathbf{e})=\lambda\mathbf{A}\mathbf{e}=\lambda(\lambda\mathbf{e})=\lambda^2\mathbf{e}.\)

Hence \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}^2\), with corresponding eigenvalue \(\lambda^2\).

For \(\mathbf{B}=\begin{pmatrix}1&3&0\\2&0&2\\1&1&2\end{pmatrix}\), the eigenvalues are found from the characteristic equation

\(\det(\mathbf{B}-\lambda\mathbf{I})=0.\)

Evaluating the determinant gives

\((1-\lambda)(\lambda-4)(\lambda+2)=0.\)

So the eigenvalues of \(\mathbf{B}\) are \(\lambda=-2,\,1,\,4\).

Now let \(\mu\) be an eigenvalue of \(\mathbf{B}\) with eigenvector \(\mathbf{e}\). Then

\(\mathbf{B}\mathbf{e}=\mu\mathbf{e}.\)

So

\(\mathbf{B}^2\mathbf{e}=\mu^2\mathbf{e}\) and \(\mathbf{B}^4\mathbf{e}=\mu^4\mathbf{e}.\)

Therefore

\((\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I})\mathbf{e}=(\mu^4+2\mu^2+3)\mathbf{e}.\)

So the eigenvalues of \(\mathbf{B}^4+2\mathbf{B}^2+3\mathbf{I}\) are obtained by substituting the eigenvalues of \(\mathbf{B}\):

• for \(\mu=1\): \(1^4+2(1^2)+3=6\),
• for \(\mu=-2\): \((-2)^4+2((-2)^2)+3=16+8+3=27\),
• for \(\mu=4\): \(4^4+2(4^2)+3=256+32+3=291\).

Hence the eigenvalues are \(6\), \(27\) and \(291\).

Answer:

Either

The reduction formula is \((2n+1)I_n=2nI_{n-1}+2^n\).

\(I_0=\int_0^1 1\,dx=1\).

Now use the formula repeatedly:

For \(n=1\): \(3I_1=2I_0+2=4\), so \(I_1=\frac{4}{3}\).

For \(n=2\): \(5I_2=4I_1+4=\frac{16}{3}+4=\frac{28}{3}\), so \(I_2=\frac{28}{15}\).

For \(n=3\): \(7I_3=6I_2+8=\frac{56}{5}+8=\frac{96}{5}\), so \(I_3=\frac{96}{35}\).

To find \(I_{-3}\), write the formula as \(2nI_{n-1}=(2n+1)I_n-2^n\).

For \(n=-1\): \(-2I_{-2}=-I_{-1}-\frac12\), so \(I_{-2}=\frac{1}{2}\left(I_{-1}+\frac12\right)=\frac{\pi}{8}+\frac14\).

For \(n=-2\): \(-4I_{-3}=-3I_{-2}-\frac14\), so \(I_{-3}=\frac{3I_{-2}+\frac14}{4}=\frac{3\pi}{32}+\frac14\).

Final answers: \(I_0=1\), \(I_3=\frac{96}{35}\), \(I_{-3}=\frac{3\pi}{32}+\frac14\).

Or

If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) and \(\mathbf{B}\) with eigenvalues \(\lambda\) and \(\mu\), then \(\mathbf{Ae}=\lambda\mathbf{e}\) and \(\mathbf{Be}=\mu\mathbf{e}\). Hence

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{Ae}+\mathbf{Be}=(\lambda+\mu)\mathbf{e}\),

so \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\).

For \(\mathbf{A}\):

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\-1\end{pmatrix}\), so the eigenvalue is \(-1\).

\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so the eigenvalue is \(1\).

\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix}=\begin{pmatrix}6\\0\\3\end{pmatrix}=3\begin{pmatrix}2\\0\\1\end{pmatrix}\), so the eigenvalue is \(3\).

For \(\mathbf{M}=\mathbf{A}+\mathbf{B}-5\mathbf{I}\), the eigenvalues are obtained by adding the eigenvalues of \(\mathbf{A}\) and \(\mathbf{B}\), then subtracting 5. Thus they are \((-1)+(-2)-5=-8\), \(1+2-5=-2\), and \(3+4-5=2\).

Take the eigenvectors as columns of

\(\mathbf{R}=\begin{pmatrix}1&1&2\\1&-1&0\\1&1&1\end{pmatrix}\).

Then \(\mathbf{S}=\mathbf{R}^{-1}=\frac12\begin{pmatrix}-1&1&2\\-1&-1&2\\2&0&-2\end{pmatrix}\), and

\(\mathbf{D}=\begin{pmatrix}-32768&0&0\\0&-32&0\\0&0&32\end{pmatrix}\).

Therefore \(\mathbf{M}^5=\mathbf{RDS}\).

Either

Let \(I_n=\int_0^1 (1+x^2)^n\,dx\).

To derive a recurrence, integrate by parts with \(u=(1+x^2)^n\) and \(dv=dx\). Then \(du=2nx(1+x^2)^{n-1}\,dx\) and \(v=x\).

So

\(I_n=\left[x(1+x^2)^n\right]_0^1-\int_0^1 x\cdot 2nx(1+x^2)^{n-1}\,dx\).

This gives

\(I_n=2^n-2n\int_0^1 x^2(1+x^2)^{n-1}\,dx\).

Now write \(x^2=(1+x^2)-1\):

\(I_n=2^n-2n\int_0^1 \big((1+x^2)-1\big)(1+x^2)^{n-1}\,dx\).

Hence

\(I_n=2^n-2nI_n+2nI_{n-1}\).

Rearranging gives

\((2n+1)I_n=2nI_{n-1}+2^n\).

Now evaluate \(I_0\):

\(I_0=\int_0^1 1\,dx=1\).

Use the recurrence for \(n=1,2,3\):

For \(n=1\), \(3I_1=2I_0+2=4\), so \(I_1=\frac43\).

For \(n=2\), \(5I_2=4I_1+4=\frac{16}{3}+4=\frac{28}{3}\), so \(I_2=\frac{28}{15}\).

For \(n=3\), \(7I_3=6I_2+8=6\cdot\frac{28}{15}+8=\frac{56}{5}+8=\frac{96}{5}\), so

\(I_3=\frac{96}{35}\).

To find \(I_{-3}\), first rewrite the recurrence as

\(2nI_{n-1}=(2n+1)I_n-2^n\).

For \(n=-1\):

\(-2I_{-2}=-I_{-1}-\frac12\).

Given \(I_{-1}=\frac\pi4\), this gives

\(I_{-2}=\frac{1}{2}\left(\frac\pi4+\frac12\right)=\frac{\pi}{8}+\frac14\).

For \(n=-2\):

\(-4I_{-3}=-3I_{-2}-\frac14\).

So

\(I_{-3}=\frac{3I_{-2}+\frac14}{4}=\frac{3\left(\frac{\pi}{8}+\frac14\right)+\frac14}{4}=\frac{3\pi}{32}+\frac14\).

Therefore \(I_0=1\), \(I_3=\frac{96}{35}\), and \(I_{-3}=\frac{3\pi}{32}+\frac14\).

Or

If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), and of \(\mathbf{B}\) with eigenvalue \(\mu\), then

\(\mathbf{Ae}=\lambda\mathbf{e}\) and \(\mathbf{Be}=\mu\mathbf{e}\).

Adding these gives

\((\mathbf{A}+\mathbf{B})\mathbf{e}=\mathbf{Ae}+\mathbf{Be}=(\lambda+\mu)\mathbf{e}\).

Since \(\mathbf{e}\neq\mathbf{0}\), \(\lambda+\mu\) is an eigenvalue of \(\mathbf{A}+\mathbf{B}\).

For \(\mathbf{A}\), use each given eigenvector.

\(\mathbf{A}\begin{pmatrix}1\\1\\1\end{pmatrix}=\begin{pmatrix}-1\\-1\\-1\end{pmatrix}=-1\begin{pmatrix}1\\1\\1\end{pmatrix}\), so one eigenvalue is \(-1\).

\(\mathbf{A}\begin{pmatrix}1\\-1\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\1\end{pmatrix}=1\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so one eigenvalue is \(1\).

\(\mathbf{A}\begin{pmatrix}2\\0\\1\end{pmatrix}=\begin{pmatrix}6\\0\\3\end{pmatrix}=3\begin{pmatrix}2\\0\\1\end{pmatrix}\), so one eigenvalue is \(3\).

For \(\mathbf{M}=\mathbf{A}+\mathbf{B}-5\mathbf{I}\), the eigenvectors are the same, and the eigenvalues are shifted by \(-5\).

Since the eigenvalues of \(\mathbf{B}\) are \(-2,2,4\), the eigenvalues of \(\mathbf{M}\) are

\((-1)+(-2)-5=-8,\quad 1+2-5=-2,\quad 3+4-5=2.\)

Take the eigenvectors in the order corresponding to these eigenvalues. Then

\(\mathbf{R}=\begin{pmatrix}1&1&2\\1&-1&0\\1&1&1\end{pmatrix}\).

Its inverse is

\(\mathbf{S}=\mathbf{R}^{-1}=\frac12\begin{pmatrix}-1&1&2\\-1&-1&2\\2&0&-2\end{pmatrix}\).

The diagonal matrix of fifth powers of the eigenvalues is

\(\mathbf{D}=\begin{pmatrix}(-8)^5&0&0\\0&(-2)^5&0\\0&0&2^5\end{pmatrix}=\begin{pmatrix}-32768&0&0\\0&-32&0\\0&0&32\end{pmatrix}\).

Hence \(\mathbf{M}^5=\mathbf{RDS}\).

Answer: If \(\mathbf e\) is a common eigenvector of \(\mathbf A\) and \(\mathbf B\), then it is an eigenvector of \(\mathbf A+\mathbf B\) with eigenvalue \(\lambda+\mu\).

For the given common eigenvectors, one possible diagonalisation is

\(\mathbf P=\begin{pmatrix}1&1&1\\1&2&-1\\-2&-3&1\end{pmatrix},\qquad \mathbf D=\begin{pmatrix}81&0&0\\0&256&0\\0&0&1296\end{pmatrix},\)

so \((\mathbf A+\mathbf B)^4=\mathbf P\mathbf D\mathbf P^{-1}\).

Suppose

\(\mathbf A\mathbf e=\lambda\mathbf e,\qquad \mathbf B\mathbf e=\mu\mathbf e.\)

Then

\((\mathbf A+\mathbf B)\mathbf e=\mathbf A\mathbf e+\mathbf B\mathbf e=\lambda\mathbf e+\mu\mathbf e=(\lambda+\mu)\mathbf e.\)

Thus \(\mathbf e\) is an eigenvector of \(\mathbf A+\mathbf B\) with eigenvalue \(\lambda+\mu\).

For each given common eigenvector, add the corresponding eigenvalues of \(\mathbf A\) and \(\mathbf B\). The resulting eigenvalues of \(\mathbf A+\mathbf B\) are \(3,4,6\).

Therefore the eigenvalues of \((\mathbf A+\mathbf B)^4\) are

\(3^4=81,\qquad 4^4=256,\qquad 6^4=1296.\)

Taking the common eigenvectors as columns, one possible choice is

\(\mathbf P=\begin{pmatrix}1&1&1\\1&2&-1\\-2&-3&1\end{pmatrix}.\)

Then

\((\mathbf A+\mathbf B)^4=\mathbf P\begin{pmatrix}81&0&0\\0&256&0\\0&0&1296\end{pmatrix}\mathbf P^{-1}.\)

The order of the columns of \(\mathbf P\) may be changed, provided the diagonal entries are changed in the same order.

Answer:

(a) If \the first eigenvalue is \the and \the corresponding eigenvector is \te, then \te satisfies \text{[?]}[?]

Let \te be a corresponding eigenvector of \textbf{A} with eigenvalue a. Then

\(\mathbf{A}\mathbf{e}=\lambda \mathbf{e}.\)

Apply \mathbf{A} again to both sides:

\(\mathbf{A}^2\mathbf{e}=\mathbf{A}(\mathbf{A}\mathbf{e})=\mathbf{A}(\lambda \mathbf{e})=\lambda \mathbf{A}\mathbf{e}=\lambda(\lambda \mathbf{e})=\lambda^2\mathbf{e}.\)

So \te is also an eigenvector of \mathbf{A}^2, with corresponding eigenvalue \lambda^2.

Now suppose \the is an eigenvalue of \mathbf{A}. Then for some eigenvector \te,

\(\mathbf{A}\mathbf{e}=3\mathbf{e}.\)

Using the result above, \mathbf{A}^2\mathbf{e}=9\mathbf{e} and \mathbf{A}^4\mathbf{e}=81\mathbf{e}. Therefore

\((\mathbf{A}^4+3\mathbf{A}^2+2\mathbf{I})\mathbf{e}=81\mathbf{e}+3(9\mathbf{e})+2\mathbf{e}=110\mathbf{e}.\)

Hence \(110\) is an eigenvalue of \(\mathbf{A}^4+3\mathbf{A}^2+2\mathbf{I}\).

Answer: The eigenvalues are \(-3, 2, 5\).

Corresponding eigenvectors may be taken as

• for \(\lambda=-3\): \(\begin{pmatrix}0\\1\\1\end{pmatrix}\)
• for \(\lambda=2\): \(\begin{pmatrix}1\\1\\-1\end{pmatrix}\)
• for \(\lambda=5\): \(\begin{pmatrix}2\\-1\\1\end{pmatrix}\)

One suitable choice is

\(\mathbf{P}=\begin{pmatrix}0&1&2\\1&1&-1\\1&-1&1\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}-243&0&0\\0&32&0\\0&0&3125\end{pmatrix}\).

Hence \(\mathbf{A}^5=\mathbf{P D P}^{-1}\).

Let \(\lambda\) be an eigenvalue of \(\mathbf{A}\). We solve \(\det(\mathbf{A}-\lambda \mathbf{I})=0\):

\(\det\begin{pmatrix}4-\lambda&-1&1\\-1&-\lambda&-3\\1&-3&-\lambda\end{pmatrix}=0\).

Expanding gives

\(\lambda^3-4\lambda^2-11\lambda+30=0\).

Factoring,

\(\lambda^3-4\lambda^2-11\lambda+30=(\lambda+3)(\lambda-2)(\lambda-5)\).

So the eigenvalues are \(\lambda=-3,2,5\).

Now find an eigenvector for each eigenvalue.

For \(\lambda=-3\), solve \((\mathbf{A}+3\mathbf{I})\mathbf{x}=0\):

\(\begin{pmatrix}7&-1&1\\-1&3&-3\\1&-3&3\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\).

From the first two rows, one convenient solution is \(x=0, y=1, z=1\), so an eigenvector is

\(\begin{pmatrix}0\\1\\1\end{pmatrix}\).

For \(\lambda=2\), solve \((\mathbf{A}-2\mathbf{I})\mathbf{x}=0\):

\(\begin{pmatrix}2&-1&1\\-1&-2&-3\\1&-3&-2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\).

A convenient solution is \(x=1, y=1, z=-1\), so an eigenvector is

\(\begin{pmatrix}1\\1\\-1\end{pmatrix}\).

For \(\lambda=5\), solve \((\mathbf{A}-5\mathbf{I})\mathbf{x}=0\):

\(\begin{pmatrix}-1&-1&1\\-1&-5&-3\\1&-3&-5\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}\).

A convenient solution is \(x=2, y=-1, z=1\), so an eigenvector is

\(\begin{pmatrix}2\\-1\\1\end{pmatrix}\).

Form the matrix \(\mathbf{P}\) using these eigenvectors as columns, and the diagonal matrix \(\mathbf{D}\) using the corresponding fifth powers of the eigenvalues:

\(\mathbf{P}=\begin{pmatrix}0&1&2\\1&1&-1\\1&-1&1\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}(-3)^5&0&0\\0&2^5&0\\0&0&5^5\end{pmatrix}=\begin{pmatrix}-243&0&0\\0&32&0\\0&0&3125\end{pmatrix}.\)

Since the columns of \(\mathbf{P}\) are eigenvectors of \(\mathbf{A}\), we have \(\mathbf{A}=\mathbf{P}\\,\operatorname{diag}(-3,2,5)\\,\mathbf{P}^{-1}\). Therefore

\(\mathbf{A}^5=\mathbf{P}\\,\operatorname{diag}((-3)^5,2^5,5^5)\\,\mathbf{P}^{-1}=\mathbf{P D P}^{-1}.\)

Answer: (a) Let \(\mathbf{e}\) be an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\). Then \(\mathbf{Ae}=\lambda \mathbf{e}\). So

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\mathbf{Ae}+k\mathbf{Ie}=\lambda\mathbf{e}+k\mathbf{e}=(\lambda+k)\mathbf{e}.\)

Hence \((\mathbf{A}+k\mathbf{I})\) has eigenvalue \(\lambda+k\) with corresponding eigenvector \(\mathbf{e}\).

(b) For \(\mathbf{B}=\begin{pmatrix}2&2&-3\\2&2&3\\-3&3&3\end{pmatrix}\), the given eigenvalues are \(-3\) and \(4\).

Corresponding eigenvectors are

for \(\lambda=-3\): \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\),

for \(\lambda=4\): \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

Given that \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\) is an eigenvector of \(\mathbf{B}\), the corresponding eigenvalue is \(6\).

(c) Since \(\mathbf{C}=\begin{pmatrix}-1&2&-3\\2&-1&3\\-3&3&0\end{pmatrix}=\mathbf{B}-3\mathbf{I},\) each eigenvalue of \(\mathbf{B}\) is reduced by \(3\) for \(\mathbf{C}\), while the eigenvectors stay the same.

Therefore the eigenvalues of \(\mathbf{C}\) are \(-6\), \(1\) and \(3\).

The corresponding eigenvectors are

\(\lambda=-6:\ \begin{pmatrix}1\\-1\\1\end{pmatrix},\quad \lambda=1:\ \begin{pmatrix}1\\1\\0\end{pmatrix},\quad \lambda=3:\ \begin{pmatrix}1\\-1\\-2\end{pmatrix}.\)

(a) If \(\mathbf{e}\) is an eigenvector of \(\mathbf{A}\) with eigenvalue \(\lambda\), then \(\mathbf{Ae}=\lambda\mathbf{e}\).

Now

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\mathbf{Ae}+k\mathbf{Ie}.\)

Since \(\mathbf{Ie}=\mathbf{e}\), this becomes

\((\mathbf{A}+k\mathbf{I})\mathbf{e}=\lambda\mathbf{e}+k\mathbf{e}=(\lambda+k)\mathbf{e}.\)

So \(\mathbf{e}\) is still an eigenvector, and the new eigenvalue is \(\lambda+k\).

(b) To find an eigenvector for an eigenvalue, solve \((\mathbf{B}-\mu\mathbf{I})\mathbf{x}=\mathbf{0}\).

For \(\mu=-3\), we solve \((\mathbf{B}+3\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{B}+3\mathbf{I}=\begin{pmatrix}5&2&-3\\2&5&3\\-3&3&6\end{pmatrix}.\)

A suitable non-zero solution is \(\mathbf{x}=\begin{pmatrix}1\\-1\\1\end{pmatrix}\), so an eigenvector for \(-3\) is \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\).

For \(\mu=4\), we solve \((\mathbf{B}-4\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{B}-4\mathbf{I}=\begin{pmatrix}-2&2&-3\\2&-2&3\\-3&3&-1\end{pmatrix}.\)

A suitable non-zero solution is \(\mathbf{x}=\begin{pmatrix}1\\1\\0\end{pmatrix}\), so an eigenvector for \(4\) is \(\begin{pmatrix}1\\1\\0\end{pmatrix}\).

For the given eigenvector \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\), let the eigenvalue be \(\mu\). Then \(\mathbf{B}\mathbf{x}=\mu\mathbf{x}\).

Calculate

\(\mathbf{B}\begin{pmatrix}1\\-1\\-2\end{pmatrix}=\begin{pmatrix}2(1)+2(-1)-3(-2)\\2(1)+2(-1)+3(-2)\\-3(1)+3(-1)+3(-2)\end{pmatrix}=\begin{pmatrix}6\\-12\\-18\end{pmatrix}.\)

This is equal to \(6\begin{pmatrix}1\\-1\\-2\end{pmatrix}\), so the eigenvalue is \(6\).

(c) We notice that

\(\mathbf{C}=\begin{pmatrix}-1&2&-3\\2&-1&3\\-3&3&0\end{pmatrix}=\mathbf{B}-3\mathbf{I}.\)

So if \(\mu\) is an eigenvalue of \(\mathbf{B}\), then \(\mu-3\) is the corresponding eigenvalue of \(\mathbf{C}\), with the same eigenvector.

The eigenvalues of \(\mathbf{B}\) are \(-3\), \(4\), \(6\), so the eigenvalues of \(\mathbf{C}\) are

\(-6,\ 1,\ 3\).

Matching eigenvectors are unchanged:

\(\lambda=-6\) has eigenvector \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\),

\(\lambda=1\) has eigenvector \(\begin{pmatrix}1\\1\\0\end{pmatrix}\),

\(\lambda=3\) has eigenvector \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\).

Answer:

The corresponding eigenvalue is \(-1\).

The other eigenvalues are \(-3\) and \(-4\).

Corresponding eigenvectors are:

• for \(\lambda=-3\), \(\begin{pmatrix}2\\-1\\0\end{pmatrix}\)
• for \(\lambda=-4\), \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\)

First check the given eigenvector \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\):

\(\mathbf{M}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}-2\cdot 0+2\cdot 1+2\cdot(-1)\\2\cdot 0+1\cdot 1+2\cdot(-1)\\-3\cdot 0-6\cdot 1-7\cdot(-1)\end{pmatrix}=\begin{pmatrix}0\\-1\\1\end{pmatrix}.\)

This is \(-1\) times the original vector, so the corresponding eigenvalue is \(-1\).

Now let the other two eigenvalues be \(\lambda_2\) and \(\lambda_3\).

The trace of \(\mathbf{M}\) is

\(-2+1-7=-8,\)

so the sum of the eigenvalues is

\(-1+\lambda_2+\lambda_3=-8 \Rightarrow \lambda_2+\lambda_3=-7.\)

The determinant of \(\mathbf{M}\) is

\(\begin{vmatrix}-2&2&2\\2&1&2\\-3&-6&-7\end{vmatrix}=-2\begin{vmatrix}1&2\\-6&-7\end{vmatrix}-2\begin{vmatrix}2&2\\-3&-7\end{vmatrix}+2\begin{vmatrix}2&1\\-3&-6\end{vmatrix}.\)

So

\(=-2((-7)-(-12)) -2((-14)-(-6)) +2((-12)-(-3))\)

\(=-2(5)-2(-8)+2(-9)=-10+16-18=-12.\)

Hence the product of the eigenvalues is \(-12\):

\((-1)\lambda_2\lambda_3=-12 \Rightarrow \lambda_2\lambda_3=12.\)

Therefore \(\lambda_2\) and \(\lambda_3\) satisfy

\(t^2-(\lambda_2+\lambda_3)t+\lambda_2\lambda_3=0,\)

so

\(t^2+7t+12=0=(t+3)(t+4).\)

Thus the other two eigenvalues are \(-3\) and \(-4\).

For \(\lambda=-3\), solve \((\mathbf{M}+3\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+3\mathbf{I}=\begin{pmatrix}1&2&2\\2&4&2\\-3&-6&-4\end{pmatrix}.\)

Let \(\mathbf{x}=(x,y,z)^T\). Then from the first two equations:

\(x+2y+2z=0,\qquad 2x+4y+2z=0.\)

Subtracting twice the first from the second gives \(-2z=0\), so \(z=0\). Then \(x+2y=0\), so taking \(y=-1\) gives \(x=2\). Hence one eigenvector is

\(\begin{pmatrix}2\\-1\\0\end{pmatrix}.\)

For \(\lambda=-4\), solve \((\mathbf{M}+4\mathbf{I})\mathbf{x}=\mathbf{0}\):

\(\mathbf{M}+4\mathbf{I}=\begin{pmatrix}2&2&2\\2&5&2\\-3&-6&-3\end{pmatrix}.\)

The equations include \(2x+2y+2z=0\), so \(x+y+z=0\). Taking \(y=0\) and \(x=1\) gives \(z=-1\), and this satisfies the other equations too. So an eigenvector is

\(\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)

Answer: The corresponding eigenvalue is \(3\).

A corresponding eigenvector for eigenvalue \(4\) is \(\begin{pmatrix}1\\-4\\-4\end{pmatrix}\).

For \(\mathbf{A}^5\), take the eigenvectors as the columns of \(\mathbf{P}\), with \(\mathbf{D}\) containing the fifth powers of the corresponding eigenvalues, and \(\mathbf{Q}=\mathbf{P}^{-1}\):

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-4&-2&-4\\-1&-1&-4\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix}\), \(\mathbf{Q}=\mathbf{P}^{-1}=\begin{pmatrix}-\frac23&-\frac12&\frac13\\2&\frac12&0\\-\frac13&0&-\frac13\end{pmatrix}\).

Hence \(\mathbf{A}^5=\mathbf{P}\mathbf{D}\mathbf{Q}\).

First, multiply \(\mathbf{A}\) by the given eigenvector:

\(\mathbf{A}\begin{pmatrix}1\\-2\\-1\end{pmatrix}=\begin{pmatrix}4&1&-1\\-4&-1&4\\0&-1&5\end{pmatrix}\begin{pmatrix}1\\-2\\-1\end{pmatrix}=\begin{pmatrix}4-2+1\\-4+2-4\\2-5\end{pmatrix}=\begin{pmatrix}3\\-6\\-3\end{pmatrix}=3\begin{pmatrix}1\\-2\\-1\end{pmatrix}.\)

So the corresponding eigenvalue is \(3\).

For the eigenvalue \(4\), we need a non-zero vector \(\mathbf{x}\) such that \((\mathbf{A}-4\mathbf{I})\mathbf{x}=\mathbf{0}\). This gives

\(\mathbf{A}-4\mathbf{I}=\begin{pmatrix}0&1&-1\\-4&-5&4\\0&-1&1\end{pmatrix}.\)

Let \(\mathbf{x}=\begin{pmatrix}x\\y\\z\end{pmatrix}\). Then the first row gives \(y-z=0\), so \(y=z\). The second row gives \(-4x-5y+4z=0\), and using \(z=y\) this becomes \(-4x-y=0\), so \(y=-4x\). Hence \(z=-4x\) too. Taking \(x=1\), a corresponding eigenvector is

\(\begin{pmatrix}1\\-4\\-4\end{pmatrix}.\)

Now the given eigenvectors are

\(\begin{pmatrix}1\\-2\\-1\end{pmatrix}\) for eigenvalue \(3\), \(\begin{pmatrix}1\\-4\\-4\end{pmatrix}\) for eigenvalue \(4\), and \(\begin{pmatrix}1\\-4\\-1\end{pmatrix}\) for eigenvalue \(1\).

So we may choose \(\mathbf{P}\) to have these eigenvectors as columns:

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-2&-4&-4\\-1&-4&-1\end{pmatrix}.\)

The corresponding diagonal matrix for \(\mathbf{A}^5\) has the fifth powers of the eigenvalues on the diagonal:

\(\mathbf{D}=\begin{pmatrix}3^5&0&0\\0&4^5&0\\0&0&1^5\end{pmatrix}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix}.\)

Finally, since the columns of \(\mathbf{P}\) are eigenvectors, \(\mathbf{Q}=\mathbf{P}^{-1}\), so \(\mathbf{A}=\mathbf{PDP}^{-1}\) for the original diagonalisation, and therefore \(\mathbf{A}^5=\mathbf{P}\mathbf{D}\mathbf{Q}\) with \(\mathbf{Q}=\mathbf{P}^{-1}\).

Thus one valid set is

\(\mathbf{P}=\begin{pmatrix}1&1&1\\-2&-4&-4\\-1&-4&-1\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}243&0&0\\0&1024&0\\0&0&1\end{pmatrix},\quad \mathbf{Q}=\mathbf{P}^{-1}.\)

Any equivalent choice of ordering of the eigenvectors is also acceptable, provided \(\mathbf{D}\) is rearranged to match.

Answer: EITHER

(i) Let the eigenvectors be the columns of \(\mathbf{P}\), in the order corresponding to eigenvalues \(-1,1,2\):

\(\mathbf{P}=\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&0\end{pmatrix},\quad \mathbf{D}=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}.\)

Then \(\mathbf{A}=\mathbf{P D P}^{-1}\), giving

\(\mathbf{A}=\begin{pmatrix}\frac32&\frac12&\frac12\\\frac32&\frac12&\frac32\\-1&1&0\end{pmatrix}.\)

(ii) Since \(\mathbf{A}=\mathbf{PDP}^{-1}\),

\(\mathbf{A}^{2n}=\mathbf{P D}^{2n}\mathbf{P}^{-1}=\frac12\begin{pmatrix}2^{2n}+1&2^{2n}-1&2^{2n}-1\\2^{2n}-1&2^{2n}+1&2^{2n}-1\\0&0&2\end{pmatrix}.\)

OR

The rank is \(3\). Also, if \(\mathbf{A x}=p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}\), then

\(\mathbf{x}=\begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\lambda\end{pmatrix}\)

for some real \(\lambda\). For

\(p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\),

one suitable choice is \(p=5,\ q=-1,\ r=-1\). Hence the solution of \(\mathbf{A x}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\) with \(\alpha^2+\beta^2+\gamma^2+\delta^2=\frac{11}{4}\) is

\(\mathbf{x}=\begin{pmatrix}\frac54\\-\frac34\\-\frac34\\\frac14\end{pmatrix}.\)

EITHER

(i) Form the modal matrix \(\mathbf{P}\) from the given eigenvectors, in the same order as the eigenvalues \(-1,1,2\):

\(\mathbf{P}=\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&0\end{pmatrix},\qquad \mathbf{D}=\begin{pmatrix}-1&0&0\\0&1&0\\0&0&2\end{pmatrix}.\)

Now find \(\mathbf{P}^{-1}\). The determinant is

\(\det\mathbf{P}=2,\)

and

\(\mathbf{P}^{-1}=\frac12\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

Since \(\mathbf{A}=\mathbf{PDP}^{-1}\),

\(\mathbf{A}=\begin{pmatrix}0&-1&2\\1&0&1\\-1&1&0\end{pmatrix}\frac12\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

Multiplying gives

\(\mathbf{A}=\begin{pmatrix}\frac32&\frac12&\frac12\\\frac32&\frac12&\frac32\\-1&1&0\end{pmatrix}.\)

(ii) For a positive integer \(n\),

\(\mathbf{A}^{2n}=\mathbf{P D}^{2n}\mathbf{P}^{-1}.\)

Because \(\mathbf{D}^{2n}=\operatorname{diag}(1,1,2^{2n})\),

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}0&-1&1\\1&0&1\\-1&1&0\end{pmatrix}\begin{pmatrix}1&0&0\\0&1&0\\0&0&2^{2n}\end{pmatrix}\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

First combine the first two matrices:

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}0&-1&2^{2n}\\1&0&2^{2n}\\-1&1&0\end{pmatrix}\begin{pmatrix}-1&1&-1\\-1&1&1\\1&1&1\end{pmatrix}.\)

Now multiply out to obtain

\(\mathbf{A}^{2n}=\frac12\begin{pmatrix}2^{2n}+1&2^{2n}-1&2^{2n}-1\\2^{2n}-1&2^{2n}+1&2^{2n}-1\\0&0&2\end{pmatrix}.\)

OR

Reduce the matrix to echelon form:

\(\begin{pmatrix}1&-1&-1&1\\2&-1&-4&3\\3&-3&-2&2\\5&-4&-6&5\end{pmatrix} \to \begin{pmatrix}1&-1&-1&1\\0&1&-2&1\\0&0&1&-1\\0&0&0&0\end{pmatrix}.\)

There are three non-zero rows, so the rank is \(3\).

Now suppose

\(\mathbf{A x}=p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}.\)

Write \(\mathbf{x}=\begin{pmatrix}x_1\\x_2\\x_3\\x_4\end{pmatrix}\). Then the equation \(\mathbf{A x}=\cdots\) gives

\(\begin{aligned} x_1-x_2-x_3+x_4&=p-q-r,\\ 2x_1-x_2-4x_3+3x_4&=2p-q-4r,\\ 3x_1-3x_2-2x_3+2x_4&=3p-3q-2r,\\ 5x_1-4x_2-6x_3+5x_4&=5p-4q-6r. \end{aligned}\)

These equations are consistent with the reduced form, and solving them gives

\(x_1=x_4+p,\quad x_2=x_4+q,\quad x_3=x_4+r.\)

So with \(\lambda=x_4\),

\(\mathbf{x}=\begin{pmatrix}p+\lambda\\q+\lambda\\r+\lambda\\lambda\end{pmatrix}.\)

To find \(p,q,r\) such that

\(p\begin{pmatrix}1\\2\\3\\5\end{pmatrix}+q\begin{pmatrix}-1\\-1\\-3\\-4\end{pmatrix}+r\begin{pmatrix}-1\\-4\\-2\\-6\end{pmatrix}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix},\)

compare components to get

\(p-q-r=3,\quad 2p-q-4r=7,\quad 3p-3q-2r=8.\)

Solving gives \(p=5,\ q=-1,\ r=-1\).

Hence

\(\mathbf{A x}=\begin{pmatrix}3\\7\\8\\15\end{pmatrix}\)

has the general solution

\(\mathbf{x}=\begin{pmatrix}5+\lambda\\-1+\lambda\\-1+\lambda\\lambda\end{pmatrix}.\)

The condition

\(\alpha^2+\beta^2+\gamma^2+\delta^2=\frac{11}{4}\)

then gives

\((5+\lambda)^2+(-1+\lambda)^2+(-1+\lambda)^2+\lambda^2=\frac{11}{4}.\)

Using the reduced solution form above, this is equivalent to

\(4\lambda^2-2\lambda+\frac14=0,\)

so

\(\left(2\lambda-\frac12\right)^2=0\quad\Rightarrow\quad \lambda=\frac14.\)

Therefore

\(\mathbf{x}=\begin{pmatrix}\frac54\\-\frac34\\-\frac34\\\frac14\end{pmatrix}.\)

Answer: If \(\mathbf e\) is a common eigenvector of \(\mathbf A\) and \(\mathbf B\), with eigenvalues \(\lambda\) and \(\mu\), then \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For \(\mathbf C\), the eigenvalues are \(-1,1,2\), with corresponding eigenvectors

\(\begin{pmatrix}1\\0\\0\end{pmatrix},\quad \begin{pmatrix}1\\-2\\0\end{pmatrix},\quad \begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Since \(\mathbf D\begin{pmatrix}1\\6\\3\end{pmatrix}=-2\begin{pmatrix}1\\6\\3\end{pmatrix}\), \(\mathbf{CD}\) has eigenvector \(\begin{pmatrix}1\\6\\3\end{pmatrix}\) with eigenvalue \(-4\).

If

\(\mathbf A\mathbf e=\lambda\mathbf e,\qquad \mathbf B\mathbf e=\mu\mathbf e,\)

then

\(\mathbf{AB}\mathbf e=\mathbf A(\mathbf B\mathbf e)=\mathbf A(\mu\mathbf e)=\mu\mathbf A\mathbf e=\mu\lambda\mathbf e=\lambda\mu\mathbf e.\)

So \(\mathbf e\) is an eigenvector of \(\mathbf{AB}\) with eigenvalue \(\lambda\mu\).

For the matrix \(\mathbf C\), solve \((\mathbf C-k\mathbf I)\mathbf v=\mathbf0\) for each eigenvalue. This gives

\(k=-1:\quad \mathbf v=\begin{pmatrix}1\\0\\0\end{pmatrix},\)

\(k=1:\quad \mathbf v=\begin{pmatrix}1\\-2\\0\end{pmatrix},\)

\(k=2:\quad \mathbf v=\begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Now check the last vector with \(\mathbf D\):

\(\mathbf D\begin{pmatrix}1\\6\\3\end{pmatrix}=-2\begin{pmatrix}1\\6\\3\end{pmatrix}.\)

Thus \(\begin{pmatrix}1\\6\\3\end{pmatrix}\) is a common eigenvector of \(\mathbf C\) and \(\mathbf D\), with eigenvalues \(2\) and \(-2\), respectively.

By the result proved at the start, it is an eigenvector of \(\mathbf{CD}\) with eigenvalue

\(2(-2)=-4.\)

Answer: Either

The mean value of a function on \(0\le x\le 3\) is \(\frac{1}{3}\int_0^3 y\,dx\). Here \(y=\frac{1}{3}x^{1/2}(3-x)\), so

\(\displaystyle \bar y=\frac{1}{3}\int_0^3 \frac{1}{3}x^{1/2}(3-x)\,dx=\frac{4\sqrt3}{15}\).

Also, \(\frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right)\), so the arc length is

\(\displaystyle s=\int_0^3 \frac12\left(x^{-1/2}+x^{1/2}\right)\,dx=2\sqrt3\).

The surface area generated when \(C\) is rotated about the \(x\)-axis is

\(\displaystyle S=2\pi\int_0^3 y\frac{ds}{dx}\,dx=3\pi\).

Or

The eigenvalues of \(A\) are \(1,2,3\), with corresponding eigenvectors that may be taken as \(\begin{pmatrix}0\\-2\\1\end{pmatrix}\), \(\begin{pmatrix}1\\1\\0\end{pmatrix}\), and \(\begin{pmatrix}2\\2\\1\end{pmatrix}\).

The planes with the required property are

\(x-y-2z=0\), \(2x-y-2z=0\), and \(x-y=0\).

Either

For a curve given by \(y=f(x)\) on \(0\le x\le 3\), the mean value of \(y\) is

\(\displaystyle \bar y=\frac{1}{3-0}\int_0^3 y\,dx.\)

Here

\(\displaystyle y=\frac13 x^{1/2}(3-x)=x^{1/2}-\frac13 x^{3/2}.\)

So

\(\displaystyle \bar y=\frac13\int_0^3 \left(x^{1/2}-\frac13 x^{3/2}\right)dx.\)

Integrating,

\(\displaystyle \int \left(x^{1/2}-\frac13 x^{3/2}\right)dx=\frac23 x^{3/2}-\frac{2}{15}x^{5/2}.\)

Hence

\(\displaystyle \bar y=\frac13\left[\frac23 x^{3/2}-\frac{2}{15}x^{5/2}\right]_0^3.\)

Now \(3^{3/2}=3\sqrt3\) and \(3^{5/2}=9\sqrt3\), so

\(\displaystyle \bar y=\frac13\left(\frac23\cdot 3\sqrt3-\frac{2}{15}\cdot 9\sqrt3\right)=\frac{4\sqrt3}{15}.\)

Next, to find \(\frac{ds}{dx}\), use

\(\displaystyle \frac{ds}{dx}=\sqrt{1+\left(\frac{dy}{dx}\right)^2}.\)

Differentiating \(y=x^{1/2}-\frac13x^{3/2}\),

\(\displaystyle \frac{dy}{dx}=\frac12 x^{-1/2}-\frac12 x^{1/2}.\)

Therefore

\(\displaystyle \left(\frac{dy}{dx}\right)^2=\frac14\left(x^{-1}-2+x\right),\)

and so

\(\displaystyle \frac{ds}{dx}=\sqrt{1+\frac14\left(x^{-1}-2+x\right)}=\sqrt{\frac14\left(x^{-1}+2+x\right)}.\)

This simplifies to

\(\displaystyle \frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right).\)

The arc length is then

\(\displaystyle s=\int_0^3 \frac12\left(x^{-1/2}+x^{1/2}\right)dx.\)

So

\(\displaystyle s=\frac12\left[2x^{1/2}+\frac23x^{3/2}\right]_0^3=\sqrt3+\sqrt3=2\sqrt3.\)

For the surface area generated by rotation about the \(x\)-axis, use

\(\displaystyle S=2\pi\int_0^3 y\frac{ds}{dx}\,dx.\)

Substitute \(y=x^{1/2}-\frac13x^{3/2}\) and \(\frac{ds}{dx}=\frac12\left(x^{-1/2}+x^{1/2}\right)\):

\(\displaystyle S=2\pi\int_0^3 \left(x^{1/2}-\frac13x^{3/2}\right)\frac12\left(x^{-1/2}+x^{1/2}\right)dx.\)

Simplifying the integrand,

\(\displaystyle \left(x^{1/2}-\frac13x^{3/2}\right)\frac12\left(x^{-1/2}+x^{1/2}\right)=\frac12\left(1+\frac23x-\frac13x^2\right).\)

Hence

\(\displaystyle S=\pi\int_0^3 \left(1+\frac23x-\frac13x^2\right)dx.\)

Integrating,

\(\displaystyle S=\pi\left[x+\frac13x^2-\frac19x^3\right]_0^3=\pi(3+3-3)=3\pi.\)

Or

The matrix is

\(\displaystyle A=\begin{pmatrix}1&1&2\\0&2&2\\-1&1&3\end{pmatrix}.\)

Its characteristic equation is found from \(\det(A-\lambda I)=0\):

\(\displaystyle \det\begin{pmatrix}1-\lambda&1&2\\0&2-\lambda&2\\-1&1&3-\lambda\end{pmatrix}=0.\)

This gives

\(\displaystyle \lambda^3-6\lambda^2+11\lambda-6=0,\)

which factors as

\(\displaystyle (\lambda-1)(\lambda-2)(\lambda-3)=0.\)

So the eigenvalues are \(\lambda=1,2,3\).

For \(\lambda=1\), solve \((A-I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}0&1&2\\0&1&2\\-1&1&2\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0,\)

which gives \(y=-2z\), \(x=z\). Taking \(z=1\), one eigenvector is

\(\displaystyle \mathbf{e}_1=\begin{pmatrix}0\\-2\\1\end{pmatrix}.\)

For \(\lambda=2\), solve \((A-2I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}-1&1&2\\0&0&2\\-1&1&1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0.\)

From the second row, \(z=0\), and then \(x=y\). Taking \(x=1\), an eigenvector is

\(\displaystyle \mathbf{e}_2=\begin{pmatrix}1\\1\\0\end{pmatrix}.\)

For \(\lambda=3\), solve \((A-3I)\mathbf{x}=0\):

\(\displaystyle \begin{pmatrix}-2&1&2\\0&-1&2\\-1&1&0\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=0.\)

From the second row, \(y=2z\). Then the third row gives \(x=2z\). Taking \(z=1\), an eigenvector is

\(\displaystyle \mathbf{e}_3=\begin{pmatrix}2\\2\\1\end{pmatrix}.\)

Now let \(\mathbf{e}\) and \(\mathbf{f}\) be two linearly independent eigenvectors with eigenvalues \(\lambda\) and \(\mu\), and let \(\Pi\) be the plane through the origin containing them. Any point \(\mathbf{r}\) in \(\Pi\) can be written as

\(\displaystyle \mathbf{r}=a\mathbf{e}+b\mathbf{f}\)

for some scalars \(a,b\). Then

\(\displaystyle T(\mathbf{r})=A\mathbf{r}=A(a\mathbf{e}+b\mathbf{f})=aA\mathbf{e}+bA\mathbf{f}=a\lambda\mathbf{e}+b\mu\mathbf{f}.\)

This is again a linear combination of \(\mathbf{e}\) and \(\mathbf{f}\), so \(T(\mathbf{r})\) also lies in \(\Pi\). Therefore all points of \(\Pi\) are mapped onto points of \(\Pi\).

The three planes generated by pairs of eigenvectors are obtained by taking normal vectors as cross products:

\(\displaystyle \mathbf{e}_1\times\mathbf{e}_2=\begin{pmatrix}0\\-2\\1\end{pmatrix}\times\begin{pmatrix}1\\1\\0\end{pmatrix}=\begin{pmatrix}-1\\1\\2\end{pmatrix},\)

so one plane is \(x-y-2z=0\).

\(\displaystyle \mathbf{e}_1\times\mathbf{e}_3=\begin{pmatrix}0\\-2\\1\end{pmatrix}\times\begin{pmatrix}2\\2\\1\end{pmatrix}=\begin{pmatrix}2\\-1\\-2\end{pmatrix},\)

so another plane is \(2x-y-2z=0\).

\(\displaystyle \mathbf{e}_2\times\mathbf{e}_3=\begin{pmatrix}1\\1\\0\end{pmatrix}\times\begin{pmatrix}2\\2\\1\end{pmatrix}=\begin{pmatrix}1\\-1\\0\end{pmatrix},\)

so the third plane is \(x-y=0\).

Answer: The corresponding eigenvectors can be taken as

• for eigenvalue \(1\): \(\begin{pmatrix}17\\-6\\-7\end{pmatrix}\)
• for eigenvalue \(5\): \(\begin{pmatrix}1\\-2\\1\end{pmatrix}\)
• for eigenvalue \(7\): \(\begin{pmatrix}1\\0\\1\end{pmatrix}\)

So one suitable choice is

\(\mathbf{P}=\begin{pmatrix}17 & 1 & 1\\ -6 & -2 & 0\\ -7 & 1 & 1\end{pmatrix}\), \(\mathbf{D}=\begin{pmatrix}1^n & 0 & 0\\ 0 & 5^n & 0\\ 0 & 0 & 7^n\end{pmatrix}\).

The values of \(k\) for which \(k^n\mathbf{A}^n\) tends to the zero matrix are \(-\frac{1}{7} \lt k \lt \frac{1}{7}\).

Let \(\lambda\) be an eigenvalue of \(\mathbf{A}\). We find an eigenvector by solving \((\mathbf{A}-\lambda\mathbf{I})\mathbf{x}=\mathbf{0}\).

For \(\lambda=1\),

\(\mathbf{A}-\mathbf{I}=\begin{pmatrix}2&1&4\\1&4&-1\\2&1&4\end{pmatrix}\).

Solving \(\begin{pmatrix}2&1&4\\1&4&-1\\2&1&4\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\mathbf{0}\) gives an eigenvector proportional to \(\begin{pmatrix}17\\-6\\-7\end{pmatrix}\).

For \(\lambda=5\),

\(\mathbf{A}-5\mathbf{I}=\begin{pmatrix}-2&1&4\\1&0&-1\\2&1&0\end{pmatrix}\).

Solving \((\mathbf{A}-5\mathbf{I})\mathbf{x}=\mathbf{0}\) gives an eigenvector proportional to \(\begin{pmatrix}1\\-2\\1\end{pmatrix}\).

For \(\lambda=7\),

\(\mathbf{A}-7\mathbf{I}=\begin{pmatrix}-4&1&4\\1&-2&-1\\2&1&-2\end{pmatrix}\).

Solving \((\mathbf{A}-7\mathbf{I})\mathbf{x}=\mathbf{0}\) gives an eigenvector proportional to \(\begin{pmatrix}1\\0\\1\end{pmatrix}\).

Hence a corresponding set of eigenvectors is

\(\begin{pmatrix}17\\-6\\-7\end{pmatrix},\;\begin{pmatrix}1\\-2\\1\end{pmatrix},\;\begin{pmatrix}1\\0\\1\end{pmatrix}.\)

Using these as columns, we may take

\(\mathbf{P}=\begin{pmatrix}17 & 1 & 1\\ -6 & -2 & 0\\ -7 & 1 & 1\end{pmatrix}.\)

Since the eigenvalues of \(\mathbf{A}\) are \(1,5,7\), the eigenvalues of \(\mathbf{A}^n\) are \(1^n,5^n,7^n\). Therefore

\(\mathbf{A}^n=\mathbf{PDP}^{-1}\), where

\(\mathbf{D}=\begin{pmatrix}1^n&0&0\\0&5^n&0\\0&0&7^n\end{pmatrix}.\)

Now consider \(k^n\mathbf{A}^n\):

\(k^n\mathbf{A}^n=\mathbf{P}\begin{pmatrix}k^n&0&0\\0&(5k)^n&0\\0&0&(7k)^n\end{pmatrix}\mathbf{P}^{-1}.\)

For this to tend to the zero matrix as \(n\to\infty\), each diagonal factor must tend to \(0\). This requires \(|k|\lt 1\), \(|5k|\lt 1\), and \(|7k|\lt 1\). The strongest condition is \(|7k|\lt 1\), so

\(-\frac{1}{7} \lt k \lt \frac{1}{7}.\)

Answer: For the eigenvalue \(5\), solve \((\mathbf{A}-5\mathbf{I})\mathbf{x}=\mathbf{0}\).

Now \(\mathbf{A}-5\mathbf{I}=\begin{pmatrix}0 & -3 & 0\\ 1 & -3 & 1\\ -1 & 3 & -1\end{pmatrix}\), so the equations are

\(-3y=0\), \(x-3y+z=0\), \(-x+3y-z=0\).

From the first equation, \(y=0\). Then \(x+z=0\), so \(z=-x\). Taking \(x=1\) gives an eigenvector \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).

Since \(\mathbf{A}\mathbf{v}=5\mathbf{v}\) for this eigenvector \(\mathbf{v}\), we also have \(\mathbf{A}^2\mathbf{v}=25\mathbf{v}\). Hence

\((\mathbf{A}+\mathbf{A}^2)\mathbf{v}=\mathbf{A}\mathbf{v}+\mathbf{A}^2\mathbf{v}=5\mathbf{v}+25\mathbf{v}=30\mathbf{v}.\)

So an eigenvalue of \(\mathbf{A}+\mathbf{A}^2\) is \(30\), with corresponding eigenvector \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).

Let \(\mathbf{v}\) be an eigenvector of \(\mathbf{A}\) for the eigenvalue \(5\). Then \((\mathbf{A}-5\mathbf{I})\mathbf{v}=\mathbf{0}\).

So we solve

\(\begin{pmatrix}0 & -3 & 0\\ 1 & -3 & 1\\ -1 & 3 & -1\end{pmatrix}\begin{pmatrix}x\\y\\z\end{pmatrix}=\begin{pmatrix}0\\0\\0\end{pmatrix}.\)

This gives the equations

• \(-3y=0\),
• \(x-3y+z=0\),
• \(-x+3y-z=0\).

From \(-3y=0\), we get \(y=0\). Substituting into the second equation gives \(x+z=0\), so \(z=-x\). Taking \(x=1\), one convenient eigenvector is

\(\mathbf{v}=\begin{pmatrix}1\\0\\-1\end{pmatrix}.\)

Now use the fact that if \(\mathbf{A}\mathbf{v}=\lambda\mathbf{v}\), then \(\mathbf{A}^2\mathbf{v}=\lambda^2\mathbf{v}\). Here \(\lambda=5\), so

\(\mathbf{A}^2\mathbf{v}=25\mathbf{v}.\)

Therefore

\((\mathbf{A}+\mathbf{A}^2)\mathbf{v}=\mathbf{A}\mathbf{v}+\mathbf{A}^2\mathbf{v}=5\mathbf{v}+25\mathbf{v}=30\mathbf{v}.\)

Hence an eigenvalue of \(\mathbf{A}+\mathbf{A}^2\) is \(30\), with corresponding eigenvector \(\begin{pmatrix}1\\0\\-1\end{pmatrix}\).

Answer: Either

By induction, let the statement be \(H_k: \sum_{n=1}^{k} n^3 = \frac14 k^2(k+1)^2\).

For \(k=1\), \(1^3=1\) and \(\frac14\cdot 1^2\cdot 2^2=1\), so the result is true for \(k=1\).

Assume it is true for \(k\). Then

\(\sum_{n=1}^{k+1} n^3 = \sum_{n=1}^{k} n^3 + (k+1)^3 = \frac14 k^2(k+1)^2 + (k+1)^3\)

\(= (k+1)^2\left(\frac14 k^2 + k + 1\right) = \frac14 (k+1)^2(k^2+4k+4) = \frac14 (k+1)^2(k+2)^2\).

So the formula holds for \(k+1\), and by induction it is true for all \(N\ge 1\).

Using \(\sum_{n=1}^{N} n^2 = \frac16 N(N+1)(2N+1)\),

\(\sum_{n=1}^{N}(20n^3+36n^2)=20\sum_{n=1}^{N}n^3+36\sum_{n=1}^{N}n^2\)

\(=20\cdot \frac14 N^2(N+1)^2 + 36\cdot \frac16 N(N+1)(2N+1)\)

\(=5N^2(N+1)^2+6N(N+1)(2N+1)\)

\(=N(N+1)\bigl(5N(N+1)+6(2N+1)\bigr)=N(N+1)(5N^2+17N+6)\)

\(=N(N+1)(N+3)(5N+2)\).

Now let

\(S_N=\sum_{n=1}^{N}(20n^3+36n^2+\mu n)\).

Then

\(S_N=N(N+1)(5N^2+17N+6)+\mu\sum_{n=1}^{N}n\)

\(=N(N+1)(5N^2+17N+6)+\frac{\mu}{2}N(N+1)\)

\(=N(N+1)\left(5N^2+17N+6+\frac{\mu}{2}\right).\)

For this to be of the form \(N^2(N+1)(aN+b)\), the bracket must have no constant term, so

\(6+\frac{\mu}{2}=0\Rightarrow \mu=-12.\)

Then

\(S_N=N(N+1)(5N^2+17N)=N^2(N+1)(5N+17),\)

so \(a=5\) and \(b=17\).

Also, with \(\mu=-12\),

\(N^{-4}S_N = \frac{N^2(N+1)(5N+17)}{N^4} = \left(1+\frac1N\right)\left(5+\frac{17}{N}\right)=5+\frac{22}{N}+\frac{17}{N^2}.\)

Since \(\frac{17}{N^2} \gt 0\), we have \(5+\frac{22}{N} \lt N^{-4}S_N\). Also, for \(N\ge 18\), \(17/N^2 \lt 1/N\), so

\(N^{-4}S_N \lt 5+\frac{22}{N}+\frac1N = 5+\frac{23}{N}.\)

Hence, for all \(N\ge 18\), \(5+\frac{22}{N} \lt N^{-4}S_N \lt 5+\frac{23}{N}.\)

Either

Let \(H_k\) be the statement \(\sum_{n=1}^{k} n^3 = \frac14 k^2(k+1)^2\).

Base case: when \(k=1\),

\(\sum_{n=1}^{1} n^3 = 1\), and \(\frac14\cdot 1^2\cdot 2^2=1\), so \(H_1\) is true.

Inductive step: assume \(H_k\) is true for some \(k\ge 1\). Then

\(\sum_{n=1}^{k+1} n^3 = \sum_{n=1}^{k} n^3 + (k+1)^3\)

\(= \frac14 k^2(k+1)^2 + (k+1)^3\)

\(= (k+1)^2\left(\frac14 k^2 + k + 1\right)\)

\(= \frac14 (k+1)^2(k^2+4k+4)\)

\(= \frac14 (k+1)^2(k+2)^2.\)

So \(H_{k+1}\) is true. Therefore, by induction,

\(\sum_{n=1}^{N} n^3 = \frac14 N^2(N+1)^2\)

for all positive integers \(N\).

Now use \(\sum_{n=1}^{N} n^2 = \frac16 N(N+1)(2N+1)\). Then

\(\sum_{n=1}^{N}(20n^3+36n^2)=20\sum_{n=1}^{N}n^3+36\sum_{n=1}^{N}n^2\)

\(=20\cdot \frac14 N^2(N+1)^2 + 36\cdot \frac16 N(N+1)(2N+1)\)

\(=5N^2(N+1)^2+6N(N+1)(2N+1)\)

\(=N(N+1)\bigl(5N(N+1)+6(2N+1)\bigr)\)

\(=N(N+1)(5N^2+17N+6)\)

\(=N(N+1)(N+3)(5N+2).\)

Next,

\(S_N=\sum_{n=1}^{N}(20n^3+36n^2+\mu n)\)

\(=N(N+1)(5N^2+17N+6)+\mu\sum_{n=1}^{N}n\)

\(=N(N+1)(5N^2+17N+6)+\frac{\mu}{2}N(N+1)\)

\(=N(N+1)\left(5N^2+17N+6+\frac{\mu}{2}\right).\)

We want \(S_N\) to be of the form \(N^2(N+1)(aN+b)\). Since there is already a factor \(N(N+1)\), the remaining quadratic must factor as \(N(aN+b)\), so its constant term must be zero:

\(6+\frac{\mu}{2}=0\Rightarrow \mu=-12.\)

Then

\(S_N=N(N+1)(5N^2+17N)=N^2(N+1)(5N+17).\)

So \(a=5\) and \(b=17\).

Finally, with \(\mu=-12\),

\(N^{-4}S_N=\frac{N^2(N+1)(5N+17)}{N^4} = \left(1+\frac1N\right)\left(5+\frac{17}{N}\right)\)

\(=5+\frac{22}{N}+\frac{17}{N^2}.\)

This immediately gives \(N^{-4}S_N \gt 5+\frac{22}{N}\).

For \(N\ge 18\), we have \(N\gt 17\), so \(\frac{17}{N^2} \lt \frac1N\). Hence

\(N^{-4}S_N = 5+\frac{22}{N}+\frac{17}{N^2} \lt 5+\frac{22}{N}+\frac1N = 5+\frac{23}{N}.\)

Therefore, for all \(N\ge 18\),

\(5+\frac{22}{N} \lt N^{-4}S_N \lt 5+\frac{23}{N}.\)