Question

For each of the matrices A in Exercises 7 through 11, find an orthogonal matrix S and a diagonal matrix Dsuch that S^-1AS = D. Do not use technology.

11.$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]$

Short answer

The diagonal matrix is $D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right]$and the orthogonal matrix is $S=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$.

Step-by-step solution

Define symmetric matrix

In linear algebra, a symmetric matrix is a square matrix that does not change when its transpose is calculated. A symmetric matrix is defined as one whose transpose is identical to the matrix itself.

A square matrix of size n x n is symmetric if B^T= B.

Find the eigenvalues of the given matrix

Given,

$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]$$A=\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]$

According to theorem 7.2.1,

$\det \left(A-\lambda I_n\right)=0$

The formula for finding eigenspace is

localid="1668584723060" $E_{\lambda }=\ker \left(A-\lambda I_n\right)=\left\{\overrightarrow{v}\text{in}{\mathrm{ℝ}}^n:A\overline{v}=\lambda \overline{v}\right\}$

Now we need to find an orthonormal eigenbasis for the given matrix A.

$$\begin{gathered} det(A-\lambda I_n)=0\to \left|\begin{array}{ccc}1-\lambda & 0& 1\\ 0& 1-\lambda & 0\\ 1& 0& 1-\lambda \end{array}\right|=0 \\ \left(1-\lambda \right)\left|\begin{array}{cc}1-\lambda & 0\\ 0& 1-\lambda \end{array}\right|+\left|\begin{array}{cc}0& 1-\lambda \\ 1& 0\end{array}\right|=0 \\ (1-\lambda ){(-\lambda +1)}^2+(\lambda -1)=0 \\ {(-\lambda +1)}^3+\lambda -1=0 \\ 1-3\lambda +3{\lambda }^2-{\lambda }^3+\lambda -1=0 \\ -{\lambda }^3+3{\lambda }^2-2\lambda =0 \\ -\lambda (\lambda -1)(\lambda -2)=0 \end{gathered}$$

The eigenvalues of the given matrix A are 0,1,2.

Find the eigenvectors

Eigenvector for $\lambda =0$

$$\begin{gathered} \left|A-0I\right|\overrightarrow{X}=0\to \left[\begin{array}{ccc}1-0& 0& 1\\ 0& 1-0& 0\\ 1& 0& 1-0\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ \left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ \left[\begin{array}{c}{x}_1+x_2\\ x_2\\ x_1+x_3\end{array}\right]=0 \\ \overrightarrow{v_1}=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right] \end{gathered}$$

Eigenvector for$\lambda =1$

width="302">A-IX→=0→1-10101-10101-1x1x2x3=0001000100x1x2x3=0x30x1=0v2→=010

Eigenvector for $\lambda =2$

$$\begin{gathered} \left[A-2I\right]\overrightarrow{x}=0\to \left[\begin{array}{ccc}1-2& 0& 1\\ 0& 1-2& 0\\ 1& 0& 1-2\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ \left[\begin{array}{ccc}-1& 0& 1\\ 0& -1& 0\\ 1& 0& -1\end{array}\right]\left[\begin{array}{c}{x}_1\\ x_2\\ x_3\end{array}\right]=0 \\ \left[\begin{array}{c}{x}_3-x_1\\ -x_2\\ x_1-x_3\end{array}\right]=0 \\ \overrightarrow{v_3}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right] \end{gathered}$$

Hence the eigenvectors are $\overrightarrow{v_1}=\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right],\overrightarrow{v_2}=\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right],\overrightarrow{v_3}=\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]$.

Find the orthogonal matrix

The orthonormal eigenbasis can be calculated as follows:

$$\begin{gathered} \overrightarrow{\overrightarrow{u_1}}=\frac{1}{\overrightarrow{v_1}}\overrightarrow{v_1}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ -1\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ 0\\ -\frac{1}{\sqrt{2}}\end{array}\right] \\ \overrightarrow{u_2}=\frac{1}{\overrightarrow{v_2}}\overrightarrow{v_2}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}0\\ 1\\ 0\end{array}\right] \\ \overrightarrow{u_3}=\frac{1}{\overrightarrow{v_3}}\overrightarrow{v_3}=\frac{1}{\sqrt{2}}\left[\begin{array}{c}1\\ 0\\ 1\end{array}\right]=\left[\begin{array}{c}\frac{1}{\sqrt{2}}\\ 0\\ \frac{1}{\sqrt{2}}\end{array}\right] \\ S=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right] \end{gathered}$$

Hence the orthogonal matrix is $S=\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]$.

Find the diagonal matrix

Now we have to find the diagonal matrix as below:

$D=S^{-1}AS$

role="math" localid="1668587676039" $$\begin{gathered} {\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]}^{-1}\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right] \\ =\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& -\frac{1}{\sqrt{2}}\\ 0& 1& 0\\ \frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right]\left[\begin{array}{ccc}1& 0& 1\\ 0& 1& 0\\ 1& 0& 1\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right] \\ =\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ \sqrt{2}& 0& \sqrt{2}\end{array}\right]\left[\begin{array}{ccc}\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\\ 0& 1& 0\\ -\frac{1}{\sqrt{2}}& 0& \frac{1}{\sqrt{2}}\end{array}\right] \\ =\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right] \end{gathered}$$

Hence the diagonal matrix is $D=\left[\begin{array}{ccc}0& 0& 0\\ 0& 1& 0\\ 0& 0& 2\end{array}\right]$.