Question

Find the $\mathit{C}$matrix which diagonalizes the matrix $\mathit{M}$of problem 18. Observe that $\mathit{M}$is not symmetric, and$\mathit{C}$ is not orthogonal (see section 11). However,$\mathit{C}$ does have an inverse; find ${\mathit{C}}^{\mathbf{-}\mathbf{1}}$and show that${\mathit{C}}^{\mathbf{-}\mathbf{1}}\mathit{M}\mathit{C}\mathbf{=}\mathit{D}$ .

Short answer

The matrix$C$ is $\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& 1\end{array}\right)$and $C^{-1}$is$\left(\begin{array}{cc}\sqrt{2}& 0\\ -1& 1\end{array}\right)$ .

Step-by-step solution

Definition of eigenvalue

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

Find a matrix which diagonalize the given matrix

The given matrix is$M=\left(\begin{array}{cc}1& 0\\ 3& -2\end{array}\right)$. The characteristic equation is localid="1664357190245" $\left|M-\lambda I\right|=0$.

Solve for $\lambda$,

$\begin{array}{rcl}\left|\begin{array}{c}\left(\begin{array}{cc}1& 0\\ 3& -2\end{array}\right)-\lambda \left(\begin{array}{cc}1& 0\\ 0& 1\end{array}\right)\end{array}\right|& =& 0\\ \left|\begin{array}{cc}1-\lambda & 0\\ 3& -2-\lambda \end{array}\right|& =& 0\\ \left(1-\lambda \right)\left(-2-\lambda \right)-\left(3\right)\times 0& =& 0\\ \left(1-\lambda \right)\left(-2-\lambda \right)& =& 0\\ & & \end{array}$

The above equation implies that localid="1664354980467" $\begin{array}{rcl}\lambda & =& 1,-2\end{array}$.Therefore, the eigenvalues are 1 and -2.

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

For $\begin{array}{rcl}\lambda & =& 1\end{array}$,

$\begin{array}{rcl}\left|M-I\right|X& =& 0\\ \left|\begin{array}{cc}1-1& 0\\ 3& -2-1\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left|\begin{array}{cc}0& 0\\ 3& -3\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

The above equation implies that localid="1664357181610" $\begin{array}{rcl}3x-3y& =& 0\end{array}$ which is equivalent to localid="1664357168539" $\begin{array}{rcl}y& =& x\end{array}$. Now, we have $\begin{array}{rcl}X& =& \left(\begin{array}{c}x\\ y\end{array}\right)\end{array}$which becomes:

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

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

For $\begin{array}{rcl}\lambda & =& -2\end{array}$,

$\begin{array}{rcl}\left|M-\left(-2\right)I\right|X& =& 0\\ \left|\begin{array}{cc}1+2& 0\\ 3& -2+2\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\\ \left|\begin{array}{cc}3& 0\\ 3& 0\end{array}\right|\left(\begin{array}{c}x\\ y\end{array}\right)& =& \left(\begin{array}{c}0\\ 0\end{array}\right)\end{array}$

The above equation implies that $\begin{array}{rcl}3x& =& 0\end{array}$which is equivalent to $\begin{array}{rcl}x& =& 0\end{array}$.

$\begin{array}{rcl}X& =& \left(\begin{array}{c}0\\ y\end{array}\right)\\ X& =& y\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}$

Therefore, the eigenvector corresponding to the eigenvalue -2 is $\begin{array}{rcl}& & \left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}$.

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 $\begin{array}{rcl}& & \left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}$is of unit length already.

So, the eigenvectors of unit length of $\begin{array}{rcl}& & M\end{array}$are $\begin{array}{rcl}& & \frac{1}{\sqrt{2}}\left(\begin{array}{c}1\\ 1\end{array}\right)\left(\begin{array}{c}0\\ 1\end{array}\right)\end{array}$.

Thus, the matrix $\begin{array}{rcl}& & C\end{array}$which diagonalizes the matrix $\begin{array}{rcl}& & M\end{array}$is $\begin{array}{rcl}& & \left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& 1\end{array}\right)\end{array}$.

Now, check if M is symmetric or not. Here, $\begin{array}{rcl}M& =& \left(\begin{array}{cc}1& 0\\ 3& -2\end{array}\right)\end{array}$and $\begin{array}{rcl}{M}^T& =& \left(\begin{array}{cc}1& 3\\ 0& -2\end{array}\right)\end{array}$ . So, the matrix Mis not symmetric because $\begin{array}{rcl}M& \ne & M^T\end{array}$.

Check if C is orthogonal or not. Here,$\begin{array}{rcl}C& =& \left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& 1\end{array}\right)\end{array}$and $\begin{array}{rcl}{C}^T& =& \left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ 0& 1\end{array}\right)\end{array}$.

Now, find CC^T:

$\begin{array}{rcl}C{C}^T& =& \left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\\ 0& 1\end{array}\right)\\ & =& \left(\begin{array}{cc}\frac{1}{2}& \frac{1}{2}\\ \frac{1}{2}& \frac{3}{2}\end{array}\right)\end{array}$

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

Now, the determinant of C is $\begin{array}{rcl}& & \frac{1}{\sqrt{2}}\end{array}$which implies C is an invertible matrix. Find the inverse of C:

localid="1664357237303" $\begin{array}{rcl}{C}^{-1}& =& \frac{1}{\left|C\right|}\text{adj}C\\ & =& \frac{1}{{\displaystyle \frac{1}{\sqrt{2}}}}\left(\begin{array}{cc}1& 0\\ -\frac{1}{\sqrt{2}}& \frac{1}{\sqrt{2}}\end{array}\right)\\ & =& \left(\begin{array}{cc}\sqrt{2}& 0\\ -1& 1\end{array}\right)\end{array}$

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

$\begin{array}{rcl}{C}^{-1}MC& =& \left(\begin{array}{cc}\sqrt{2}& 0\\ -1& 1\end{array}\right)\left(\begin{array}{cc}1& 0\\ 3& -2\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& 1\end{array}\right)\\ & =& \left(\begin{array}{cc}\sqrt{2}& 0\\ -1& 1\end{array}\right)\left(\begin{array}{cc}\frac{1}{\sqrt{2}}& 0\\ \frac{1}{\sqrt{2}}& -2\end{array}\right)\\ & =& \left(\begin{array}{cc}1& 0\\ 0& -2\end{array}\right)\\ & =& D\end{array}$

Here, the diagonal entries of D are the eigenvalues of M. Thus, the matrix C diagonalizes M .