# Diagonalize unitary matrix

• Dec 8th 2009, 09:21 PM
Mollier
Diagonalize unitary matrix
Problem statement:
Diagonalize this unitary matrix $\displaystyle \textbf{V}$ to reach $\displaystyle \textbf{V}=\textbf{U}\Lambda\textbf{U}^*$.

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

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

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!
• Dec 9th 2009, 07:21 AM
HallsofIvy
Quote:

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

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

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

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.
• Dec 9th 2009, 08:08 PM
Mollier
To find the eigenvectors for $\displaystyle \lambda=1$, do you solve this system by elimination?

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

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

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

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

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