Exam-Style Problem

9231 P11 - Jun 2015 - Q10 - 12 marks

6313

The matrix \(\mathbf A\) is given by

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

The matrix \(\mathbf A\) has an eigenvector \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\). Find the corresponding eigenvalue.

The matrix \(\mathbf A\) also has eigenvalues \(4\) and \(6\). Find corresponding eigenvectors.

Hence find a matrix \(\mathbf P\) and a diagonal matrix \(\mathbf D\) such that \(\mathbf A=\mathbf P\mathbf D\mathbf P^{-1}\).

The matrix \(\mathbf B\) is such that \(\mathbf B=\mathbf Q\mathbf A\mathbf Q^{-1}\), where \(\mathbf Q=\begin{pmatrix}4&11&5\\1&4&2\\1&2&1\end{pmatrix}\).

By using the expression \(\mathbf P\mathbf D\mathbf P^{-1}\) for \(\mathbf A\), find the set of eigenvalues and a corresponding set of eigenvectors for \(\mathbf B\).

Solution

Answer: The eigenvalue corresponding to \(\begin{pmatrix}1\\-1\\1\end{pmatrix}\) is \(-3\).

For \(\lambda=4\), an eigenvector is \(\begin{pmatrix}1\\1\\0\end{pmatrix}\). For \(\lambda=6\), an eigenvector is \(\begin{pmatrix}1\\-1\\-2\end{pmatrix}\).

One suitable choice is \(\mathbf P=\begin{pmatrix}1&1&1\\-1&1&-1\\1&0&-2\end{pmatrix}\), \(\mathbf D=\begin{pmatrix}-3&0&0\\0&4&0\\0&0&6\end{pmatrix}\).

The eigenvalues of \(\mathbf B\) are \(-3\), \(4\), and \(6\), with corresponding eigenvectors \(\begin{pmatrix}2\\1\\0\end{pmatrix}\), \(\begin{pmatrix}15\\5\\3\end{pmatrix}\), and \(\begin{pmatrix}17\\7\\3\end{pmatrix}\), respectively.

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

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

So the corresponding eigenvalue is \(-3\).

For \(\lambda=4\), solve \((\mathbf A-4\mathbf I)\mathbf x=\mathbf0\). A corresponding eigenvector is

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

For \(\lambda=6\), solve \((\mathbf A-6\mathbf I)\mathbf x=\mathbf0\). A corresponding eigenvector is

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

Using these eigenvectors as the columns of \(\mathbf P\), in the same order as their eigenvalues, gives

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

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

Now \(\mathbf B=\mathbf Q\mathbf A\mathbf Q^{-1}\), so

\(\mathbf B=\mathbf Q\mathbf P\mathbf D\mathbf P^{-1}\mathbf Q^{-1}=(\mathbf Q\mathbf P)\mathbf D(\mathbf Q\mathbf P)^{-1}\).

Therefore \(\mathbf B\) has the same eigenvalues \(-3\), \(4\), and \(6\), and its corresponding eigenvectors are the columns of \(\mathbf Q\mathbf P\).

Calculate

\(\mathbf Q\mathbf P=\begin{pmatrix}-2&15&-17\\-1&5&-7\\0&3&-3\end{pmatrix}\).

So a corresponding set of eigenvectors for \(\mathbf B\) is

\(\begin{pmatrix}2\\1\\0\end{pmatrix}\), \(\begin{pmatrix}15\\5\\3\end{pmatrix}\), and \(\begin{pmatrix}17\\7\\3\end{pmatrix}\), for eigenvalues \(-3\), \(4\), and \(6\), respectively.