91Ó°ÊÓ

$H=\left(\begin{array}{cc}2& 3-1\\ 3+i& -1\end{array}\right)$

A diagonalizable matrix is a geometric inhomogeneous dilation (or anisotropic scaling): it scales the space in the same way as a homogeneous dilation does, but by a different factor along each eigenvector axis, the factor determined by the corresponding eigenvalue. A faulty square matrix is one that cannot be diagonalized.

The matrix to be Hermitian then it's conjugate matrix should be equal to its transpose matrix.

$$\begin{gathered} H^{*}=\left(\begin{array}{cc}2& 3+1\\ 3-i& -1\end{array}\right) \\ {H}^T=\left(\begin{array}{cc}2& 3+1\\ 3-i& -1\end{array}\right) \\ {H}^{*}=H^T \end{gathered}$$

Thus, matrix H is Hermitian.

The eigen values and the eigen vectors are

$$\begin{gathered} det\left|H-\mathrm{λ}\mathrm{ι}\right|=0 \\ det\left|\begin{array}{cc}2-\mathrm{λ}& 3-i\\ 3+i& -1-\mathrm{λ}\end{array}\right|=0 \\ \left(2-\mathrm{λ}\right)\left(-1-\mathrm{λ}\right)-\left(3-i\right)\left(3+i\right)=0 \\ {\mathrm{λ}}^2-\mathrm{λ}-12=0 \\ \mathrm{λ}=-3,4 \end{gathered}$$

Thus, Eigen values are -3 and 4 .

For$\mathrm{λ}=-3$, an eigen vector satisfies the equation

$$\begin{gathered} \left(\begin{array}{cc}5& 3-i\\ 3+i& 2\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0or, \\ 5x+\left(3-i\right)y=0 \\ \left(3+i\right)x-5y=0 \end{gathered}$$

The above equations are satisfied x=2,y=-3-i. A choice for unit eigen vector islocalid="1658815461012" $\frac{1}{\sqrt{14}}\left(2,-3-i\right)$.

For$\mathrm{λ}=4$, the equations are found similarly,

$-2x+\left(3-i\right)y=0,\left(3+i\right)x-5y=0$

Which are satisfied by$y=2,x=3-i$. So a unit eigen vector is$\frac{1}{\sqrt{14}}\left(3-i,2\right)$

The eigen vectors are orthogonal such that the inner product is zero as shown.

$$\begin{gathered} =\left(2,-3-i\right)·\left(3-i,2\right) \\ =2\left(3-i\right)+2\left(-3+i\right) \\ =6-2i-6+2i \\ =0 \end{gathered}$$

The unit eigen vectors as a column of a matrix U which diagonalizes .

$U=\frac{1}{\sqrt{14}}\left(\begin{array}{cc}2& 3-i\\ -3-i& 2\end{array}\right),U^{-1}=\frac{1}{\sqrt{14}}\left(\begin{array}{cc}2& -3+i\\ 3+i& 2\end{array}\right)$$U=\frac{1}{\sqrt{14}}\left(\begin{array}{cc}2& 3-i\\ -3-i& 2\end{array}\right),U^{-1}=\frac{1}{\sqrt{14}}\left(\begin{array}{cc}2& -3+i\\ 3+i& 2\end{array}\right)$

Thus,$UH{U}^{-1}=\left(\begin{array}{cc}-3& 0\\ 0& 4\end{array}\right);H$ is diagonalizable.