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${\mathit{U}}^{\mathbf{-}\mathbf{1}}\mathit{H}\mathit{U}$ is the diagonal matrix of eigenvalues.

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

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

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 matrix H to be Hermitian then it's conjugate matrix $H^{*}$should be equal to its transpose matrix$H^T$.

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

Thus, matrix H is Hermitian.

The eigen values and the eigen vectors

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

Thus, Eigen values are and .

For$\lambda =1$ , an eigen vector satisfies the equation

$$\begin{gathered} \left(\begin{array}{cc}1& i\\ -i& 1\end{array}\right)\left(\begin{array}{c}x\\ y\end{array}\right)=0 \\ x+iy=0 \\ -ix+y=0 \end{gathered}$$

The above equations are satisfied $x=-i,y=1$. A choice for unit eigen vector is role="math" localid="1658818325461" $\frac{1}{\sqrt{2}}(-i,1).$.

For$\lambda =3$ , the equations are found similarly,

$-x+iy=0,-x-iy=0$

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

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

$$\begin{gathered} =\left(-i,1\right).\left(i,1\right) \\ =\left(i.i\right)+\left(1.1\right) \\ =-1+1 \\ =0 \end{gathered}$$

Now, by writing the unit eigen vectors as a column of a matrix U which diagonalizes H.

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

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