Question

Find a unitary matrix U which diagonalizes A in (11.31) and verify that ${\mathit{U}}^{\mathbf{-}\mathbf{1}}\mathit{A}\mathit{U}$ is diagonal with the eigenvalues down the main diagonal.

Short answer

Unitary matrix

U=$\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& i\sqrt{2}& -i\sqrt{2}\\ \sqrt{2}& -1& -1\end{array}\right)$

Step-by-step solution

Given Information

$A=\frac{1}{2}\left(\begin{array}{ccc}1& \sqrt{2}& 1\\ -\sqrt{2}& 0& \sqrt{2}\\ 1& -\sqrt{2}& 1\end{array}\right)$

Unitary Matrix

In linear algebra, a complex square matrix $\mathit{U}$ is unitary if its conjugate transpose ${\mathit{U}}^{\mathbf{*}}$ is also its inverse, that is, if

${\mathit{U}}^{\mathbf{*}}\mathit{U}\mathbf{=}\mathit{U}{\mathit{U}}^{\mathbf{*}}\mathbf{=}\mathit{U}{\mathit{U}}^{\mathbf{-}\mathbf{1}}\mathbf{=}\mathit{I}\mathbf{,}$

Where, $\mathit{l}$ is the identity matrix.

Eigenvalue equation

$Ar=\lambda r,$

Thus,

$\left|\begin{array}{ccc}\frac{1}{2}-\lambda & \frac{1}{\sqrt{2}}& \frac{1}{2}\\ -\frac{1}{\sqrt{2}}& -\lambda & \frac{1}{\sqrt{2}}\\ \frac{1}{2}& -\frac{1}{\sqrt{2}}& \frac{1}{2}-x\end{array}\right|=0$

$$\begin{gathered} -\lambda {\left(\frac{1}{2}-\lambda \right)}^2+\frac{1}{4}+\frac{1}{4}+\frac{\lambda }{4}+\frac{1}{2}\left(\frac{1}{2}-\lambda \right)+\frac{1}{2}\left(\frac{1}{2}-\lambda \right) \\ =-\lambda {\left(\frac{1}{2}-\lambda \right)}^2+\frac{1}{2}+\frac{\lambda }{4}+\left(\frac{1}{2}-\lambda \right) \\ =-\lambda {\left(1-2\lambda \right)}^2+2+\lambda +2\left(1-2\lambda \right) \end{gathered}$$

$$\begin{gathered} =4{\lambda }^2-4{\lambda }^3-4\lambda +4 \\ ={\lambda }^3-{\lambda }^2+\lambda -1 \\ ={\lambda }^2(\lambda -1)+\lambda -1 \\ =(\lambda -1)({\lambda }^2+1)=0 \end{gathered}$$

The eigenvalues

${\lambda }_1=1,$

And

${\lambda }_2=i,$

And

${\lambda }_3=-i,$

The eigen values

$v_{-i}=\frac{1}{2}\left(\begin{array}{c}1\\ -i\sqrt{2}\\ -1\end{array}\right)$

Unitary matrix U which diagonalizes A ,

$U=\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& i\sqrt{2}& -i\sqrt{2}\\ \sqrt{2}& -1& -1\end{array}\right)$

The inverse of the equation

$U^{-1}=U^{{}^{+}}=\frac{1}{2}\left(\begin{array}{ccc}\sqrt{2}& 0& \sqrt{2}\\ 0& -i\sqrt{2}& -1\\ \sqrt{2}& -i\sqrt{2}& -1\end{array}\right)$

$$\begin{gathered} U^{-1}AD=D \\ =\frac{1}{8}\left(\begin{array}{ccc}\sqrt{2}& 0& \sqrt{2}\\ 1& -i\sqrt{2}& -1\\ 1& i\sqrt{2}& -1\end{array}\right)\left(\begin{array}{ccc}1& \sqrt{2}& 1\\ -\sqrt{2}& 0& \sqrt{2}\\ 1& -\sqrt{2}& -1\end{array}\right)\left(\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& i\sqrt{2}& -i\sqrt{2}\\ \sqrt{2}& -1& -1\end{array}\right) \\ =\frac{1}{8}\left(\begin{array}{ccc}2\sqrt{2}& 0& 2\sqrt{2}\\ 2i& 2\sqrt{2}& -2i\\ -2i& 2\sqrt{2}& 2i\end{array}\right)\left(\begin{array}{ccc}\sqrt{2}& 1& 1\\ 0& i\sqrt{2}& -i\sqrt{2}\\ \sqrt{2}& -1& -1\end{array}\right) \\ =\left(\begin{array}{ccc}1& 0& 0\\ 0& i& 0\\ 0& 0& -1\end{array}\right) \end{gathered}$$