1. ## Diagonalize unitary matrix

Problem statement:
Diagonalize this unitary matrix $\textbf{V}$ to reach $\textbf{V}=\textbf{U}\Lambda\textbf{U}^*$.

$
V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}
1 & 1-i \\
1+i & -1 \\
\end{array}\right]
$

Pathetic attempt:
The trace of $\textbf{V}$ is zero and since it is unitary it has eigenvalues with absolute value one. This mean that $\lambda = 1,-1$

What is confusing me is the fact that I am supposed to make a matrix with orthonormal column vectors out of a unitary matrix (which has orthonormal column vectors)...

Any hints are greatly appreciated, thanks!

2. Originally Posted by Mollier
Problem statement:
Diagonalize this unitary matrix $\textbf{V}$ to reach $\textbf{V}=\textbf{U}\Lambda\textbf{U}^*$.

$
V=\frac{1}{\sqrt{3}}\left[\begin{array}[pos]{cc}
1 & 1-i \\
1+i & -1 \\
\end{array}\right]
$

Pathetic attempt:
The trace of $\textbf{V}$ is zero and since it is unitary it has eigenvalues with absolute value one. This mean that $\lambda = 1,-1$

What is confusing me is the fact that I am supposed to make a matrix with orthonormal column vectors out of a unitary matrix (which has orthonormal column vectors)...

Any hints are greatly appreciated, thanks!
Find the eigenvectors corresponding to eigenvalues 1 and -1. Because those are distinct eigenvalues, the eigenvectors will be orthogonal. Make their lengths 1 and the matrix having those unit length eigenvectors as columns will diagonalize the matrix.

3. To find the eigenvectors for $\lambda=1$, do you solve this system by elimination?

$
\left[\begin{array}[pos]{cc}
\frac{1}{\sqrt{3}}-1 & \frac{1-i}{\sqrt{3}} \\
\frac{1+i}{\sqrt{3}} & -\frac{1}{\sqrt{3}}-1 \\
\end{array}\right] \textbf{x} = \textbf{0}
$

I write from row 2:
$x = \frac{1+\sqrt{3}}{1+i}y$

Then from row 1:
$
-\frac{2}{\sqrt{3}(1+i)}y + \frac{2}{\sqrt{3}(1+i)}y = 0
$

Basically I can choose y to be whatever I want, but I guess I would get the "prettiest" solution if I choose $y = 1+i$.

Is this the right way to go about it?
Thanks.