Question: Verify that each of the following matrices is Hermitian. Find its eigenvalues and eigenvectors, write a unitary matrix U which diagonalizes H by a similarity transformation, and show that${\mathit{U}}^{\mathbf{-}\mathbf{1}}\mathbf{+}\mathit{H}\mathit{U}$ is the diagonal matrix of eigenvalues.$\left(\begin{array}{cc}-2& 3+4i\\ 3-4i& -2\end{array}\right)$

$\left(\begin{array}{cc}-2& 3+4\mathrm{i}\\ 3-4\mathrm{i}& -2\end{array}\right)$

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.

The square $2\times 2$matrix,

$H=\left(\begin{array}{cc}-2& 3+4\mathrm{i}\\ 3-4\mathrm{i}& -2\end{array}\right)\left(*\right)$

The matrix provided in equation (*) is Hermitian$H^{†}=H$

Where,

$H^{†}$, complex conjugate of transpose of matrix H

By take complex conjugate of each element and then transpose the resultant matrix.

The relation for$H^{†}$

$H^{†}={\left(H*\right)}^{†}$

$$\begin{gathered} U^{-1}HU=\frac{1}{2}\left(\begin{array}{cc}-\left(\frac{3}{5}-\frac{4}{5}i\right)& 1\\ \frac{3}{5}-\frac{4}{5}i& 1\end{array}\right)\left(\begin{array}{cc}-2& 3+4i\\ 3-4i& -2\end{array}\right)\left(-\begin{array}{cc}\left(\frac{3}{5}+\frac{4}{5}i\right)& \frac{3}{5}+\frac{4}{5}\\ 1& 1\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}\frac{21}{5}-\frac{28}{5}i& -7\\ \frac{9}{5}-\frac{12}{5}i& 3\end{array}\right)\left(\begin{array}{cc}-2& 3+4i\\ 3-4i& -2\end{array}\right) \\ =\frac{1}{2}\left(\begin{array}{cc}-14& 0\\ 0& 6\end{array}\right) \\ =\left(\begin{array}{cc}-7& 0\\ 0& 3\end{array}\right) \end{gathered}$$

Therefore, the diagonal elements are the eigen values of H.

Hence, H is diagonalized by a unitary similarity Transformation.

$U^{-1}+HU$is a diagonal matrix of eigen values,$\lambda =-7,3$