Exam-Style Problem

9231 P12 - Jun 2014 - Q9

6506

The matrix \(\mathbf{M}\), where
\(\mathbf{M}=\left(\begin{array}{rrr} -2 & 2 & 2 \\ 2 & 1 & 2 \\ -3 & -6 & -7 \end{array}\right),\)
has an eigenvector \(\left(\begin{array}{r}0 \\ 1 \\ -1\end{array}\right)\). Find the corresponding eigenvalue.

It is given that if the eigenvalues of a general \(3 \times 3\) matrix \(\mathbf{A}\), where
\(\mathbf{A}=\left(\begin{array}{lll} a & b & c \\ d & e & f \\ g & h & i \end{array}\right),\)
are \(\lambda_{1}, \lambda_{2}\) and \(\lambda_{3}\) then
\(\lambda_{1}+\lambda_{2}+\lambda_{3}=a+e+i\)
and
the determinant of \(\mathbf{A}\) has the value \(\lambda_{1} \lambda_{2} \lambda_{3}\).
Use these results to find the other two eigenvalues of the matrix \(\mathbf{M}\), and find corresponding eigenvectors.

Solution

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

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

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

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

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

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

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

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

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