Repeat Problem 25 for Problem 19.

For a matrix A, all possible values of $\mathit{\lambda }$which satisfies the equation$\left|A-\lambda I\right|\mathbf{=}\mathbf{0}$ , where I is the identity matrix, are considered as the eigenvalues of a matrix.

The given matrix is $M=\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)$. The characteristic equation is $\left|M-\lambda I\right|\mathbf{=}\mathbf{0}$.

Solve for $\lambda$as follows:

$\begin{array}{rcl}\left|\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\right|& =& 0\\ \left|\begin{array}{cc}5-\lambda & 1\\ 4& 2-\lambda \end{array}\right|& =& 0\\ \left(5-\lambda \right)\left(2-\lambda \right)-\left(4\right)\times \left(1\right)& =& 0\\ 10-7\lambda +{\lambda }^2-4& =& 0\end{array}$

Solve further,

$\begin{array}{rcl}{\lambda }^2-7\lambda +6& =& 0\\ \left(\lambda -6\right)\left(\lambda -1\right)& =& 0\\ \lambda & =& 1\\ \lambda & =& 6\end{array}$

Therefore, the eigenvalues are 1 and 6 .

Now, find the eigenvectors for each eigenvalue by solving $\begin{array}{rcl}\left|M-\lambda I\right|X& =& 0\end{array}$for X.

$\begin{array}{rcl}\left|M-I\right|X& =& 0\\ \left|\begin{array}{cc}5-1& 1\\ 4& 2-1\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left(\begin{array}{cc}4& 1\\ 4& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

The above equation implies that $\begin{array}{rcl}4x+y& =& 0\end{array}$which is equivalent to$\begin{array}{rcl}y& =& -4x\end{array}$.

$\begin{array}{rcl}X& =& \left(\begin{array}{c}x\\ -4x\end{array}\right)\\ X& =& x\left(\begin{array}{c}1\\ -4\end{array}\right)\end{array}$

Therefore, the eigenvector corresponding to the eigenvalue 1 is $\begin{array}{rcl}& & \left(\begin{array}{c}1\\ -4\end{array}\right)\end{array}$.

For $\begin{array}{rcl}\lambda & =& 6\end{array}$,

$\begin{array}{rcl}\left|M-6I\right|X& =& 0\\ \left|\begin{array}{cc}5-6& 1\\ 4& 2-6\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left|\begin{array}{cc}-1& 1\\ 4& -1\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

The above equation implies that $\begin{array}{rcl}-x+y& =& 0\end{array}$which is equivalent to $\begin{array}{rcl}y& =& x\end{array}$.

$\begin{array}{rcl}X& =& \left(\begin{array}{c}x\\ y\end{array}\right)\\ X& =& x\left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}$

Therefore, the eigenvector corresponding to the eigenvalue 6 is $\begin{array}{rcl}& & \left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}$.

Now, we have to multiply the eigenvector $\begin{array}{rcl}& & \left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}$by $\begin{array}{rcl}& & \frac{1}{\sqrt{2}}\end{array}$to make it of unit length and the eigenvector $\begin{array}{rcl}& & \left(\begin{array}{c}1\\ -4\end{array}\right)\end{array}$by $\begin{array}{rcl}& & \frac{1}{\sqrt{17}}\end{array}$to make it of unit length.

So, the eigenvectors of unit length of $\begin{array}{rcl}& & M\end{array}$are $\begin{array}{rcl}& & \frac{1}{\sqrt{17}}\left(\begin{array}{c}1\\ -4\end{array}\right),\frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)\end{array}$.

Thus, the matrix C which diagonalizes the matrix M is $\begin{array}{rcl}& & \left(\begin{array}{cc}\frac{1}{\sqrt{17}}& \frac{1}{\sqrt{2}}\\ -\frac{4}{\sqrt{17}}& \frac{1}{\sqrt{2}}\end{array}\right)\end{array}$.

Now, check if Mis symmetric or not. Here,$\begin{array}{rcl}M& =& \left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)\end{array}$and$\begin{array}{rcl}{M}^T& =& \left(\begin{array}{cc}5& 4\\ 1& 2\end{array}\right)\end{array}$ . So, the matrix Mis not symmetric because$\begin{array}{rcl}M& \ne & M^T\end{array}$.

Check if Cis orthogonal or not. Here,$\begin{array}{rcl}C& =& \left(\begin{array}{cc}\frac{1}{\sqrt{17}}& \frac{1}{\sqrt{2}}\\ -\frac{4}{\sqrt{17}}& \frac{1}{\sqrt{2}}\end{array}\right)\end{array}$and $\begin{array}{rcl}{C}^T& =& \left(\begin{array}{cc}\frac{1}{\sqrt{17}}& -\frac{4}{\sqrt{17}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\end{array}$.

Now, find $\begin{array}{rcl}& & C{C}^T\end{array}$:

$\begin{array}{rcl}C{C}^T& =& \left(\begin{array}{cc}\frac{1}{\sqrt{17}}& \frac{1}{\sqrt{2}}\\ -\frac{4}{\sqrt{17}}& \frac{1}{\sqrt{2}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{17}}& -\frac{4}{\sqrt{17}}\\ \frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\\ & =& \left(\begin{array}{cc}\frac{19}{34}& \frac{9}{34}\\ \frac{9}{34}& \frac{49}{34}\end{array}\right)\end{array}$

Thus,$\begin{array}{rcl}& & C\end{array}$is not orthogonal because $\begin{array}{rcl}C{C}^T& \ne & I\end{array}$.

Now, the determinant of $\begin{array}{rcl}& & C\end{array}$is $\begin{array}{rcl}& & \frac{5}{\sqrt{34}}\end{array}$which implies $\begin{array}{rcl}& & C\end{array}$is an invertible matrix. Find the inverse of role="math" localid="1664362646892" $\begin{array}{rcl}& & C\end{array}$:

$\begin{array}{rcl}{C}^{-1}& =& \frac{1}{\left|C\right|}\text{adj}\hspace{0.17em}C\\ & =& \frac{1}{{\displaystyle \frac{5}{\sqrt{34}}}}\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{4}{\sqrt{17}}& \frac{1}{\sqrt{17}}\end{array}\right)\\ & =& \frac{\sqrt{34}}{5}\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{4}{\sqrt{17}}& \frac{1}{\sqrt{17}}\end{array}\right)\end{array}$

Now, Find$\begin{array}{rcl}& & C^{-1}MC\end{array}$

$\begin{array}{rcl}{C}^{-1}MC& =& \frac{\sqrt{34}}{5}\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{4}{\sqrt{17}}& \frac{1}{\sqrt{17}}\end{array}\right)\left(\begin{array}{cc}5& 1\\ 4& 2\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{17}}& \frac{1}{\sqrt{2}}\\ -\frac{4}{\sqrt{17}}& \frac{1}{\sqrt{2}}\end{array}\right)\\ & =& \frac{\sqrt{34}}{5}\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& -\frac{1}{\sqrt{2}}\\ \frac{4}{\sqrt{17}}& \frac{1}{\sqrt{17}}\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{17}}& \frac{6}{\sqrt{2}}\\ -\frac{4}{\sqrt{17}}& \frac{6}{\sqrt{2}}\end{array}\right)\\ & =& \frac{\sqrt{34}}{5}\left(\begin{array}{cc}\frac{5}{\sqrt{34}}& 0\\ 0& \frac{30}{\sqrt{34}}\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 0\\ 0& 6\end{array}\right)\\ & =& D\end{array}$

Here, the diagonal entries of$\begin{array}{rcl}& & D\end{array}$ are the eigenvalues of$\begin{array}{rcl}& & M\end{array}$ . Thus, the matrix$\begin{array}{rcl}& & C\end{array}$ diagonalizes$\begin{array}{rcl}& & M\end{array}$ .