Exam-Style Problem

9231 P13 - Jun 2016 - Q11E - 14 marks

6338

EITHER

It is given that \(1\) and \(4\) are eigenvalues of the matrix \(\mathbf A\), where

\(\mathbf A=\begin{pmatrix}1&-3&-3\\-8&6&-3\\8&-2&7\end{pmatrix}.\)

Find eigenvectors corresponding to each of these eigenvalues.

Given further that \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) is an eigenvector of \(\mathbf A\), find the corresponding eigenvalue.

Write down matrices \(\mathbf P\) and \(\mathbf D\) such that \(\mathbf P^{-1}\mathbf A\mathbf P=\mathbf D\), where \(\mathbf D\) is a diagonal matrix, and find \(\mathbf P^{-1}\).

Write down a matrix \(\mathbf C\) such that \(\mathbf C^2=\mathbf D\), and deduce a matrix \(\mathbf B\) such that \(\mathbf B^2=\mathbf A\).

Solution

Answer: Eigenvectors may be taken as \(\begin{pmatrix}1\\1\\-1\end{pmatrix}\) for \(\lambda=1\), and \(\begin{pmatrix}1\\1\\-2\end{pmatrix}\) for \(\lambda=4\).

The vector \(\begin{pmatrix}0\\1\\-1\end{pmatrix}\) has eigenvalue \(9\).

One suitable choice is

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

Take \(\mathbf C=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix}\). Then one suitable square root is

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

For \(\lambda=1\), solve \((\mathbf A-\mathbf I)\mathbf v=0\). This gives an eigenvector proportional to \(\begin{pmatrix}1\\1\\-1\end{pmatrix}\).

For \(\lambda=4\), solve \((\mathbf A-4\mathbf I)\mathbf v=0\). This gives an eigenvector proportional to \(\begin{pmatrix}1\\1\\-2\end{pmatrix}\).

Now multiply \(\mathbf A\begin{pmatrix}0\\1\\-1\end{pmatrix}\):

\(\begin{pmatrix}1&-3&-3\\-8&6&-3\\8&-2&7\end{pmatrix}\begin{pmatrix}0\\1\\-1\end{pmatrix}=\begin{pmatrix}0\\9\\-9\end{pmatrix}=9\begin{pmatrix}0\\1\\-1\end{pmatrix}.\)

So the corresponding eigenvalue is \(9\).

Use the three eigenvectors as the columns of \(\mathbf P\), in the same order as the diagonal entries of \(\mathbf D\):

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

Direct calculation gives

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

Since the diagonal entries of \(\mathbf D\) are \(1\), \(4\), and \(9\), choose

\(\mathbf C=\begin{pmatrix}1&0&0\\0&2&0\\0&0&3\end{pmatrix},\qquad \mathbf C^2=\mathbf D.\)

Then

\(\mathbf B=\mathbf P\mathbf C\mathbf P^{-1}\)

satisfies \(\mathbf B^2=\mathbf A\), because

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

Multiplying the matrices gives

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