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

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