# Hermitian and Unitary matrices

1. Mar 27, 2010

### snakebite

1. The problem statement, all variables and given/known data

Hello,

the problem is asking me to find a unitary matrix U such that (U bar)^T(H)(U) is diagonal. And we have H = [{7,2,0},{2,4,-2},{0,-2,5}]

3. The attempt at a solution
I don't know where to start. I tried getting the eigenvalues of the matrix A but that lead to very long and complicated equations with 3rd roots which i doubt is the correct way.
Other than that I have no idea what to start off with

Thanks you

2. Mar 27, 2010

### vela

Staff Emeritus
What is the matrix A? Do you mean H? If so, are you sure you have the right H? That one leads to messy results, so maybe you are doing the problem correctly.

3. Mar 28, 2010

### snakebite

Yea I'm sorry there is no A i meant H. And yes i'm sure i have to correct matrix H on. So the correct way would be to just find the eigenvectors and U would just be a matrix with the eigenvectors as the columns?

4. Mar 28, 2010

### vela

Staff Emeritus
Yes, with the eigenvector normalized.

5. Mar 29, 2010

### snakebite

I did what you said on another example where H = [{-2,1+i,1-i},{1-i,-1,-2i},{1+i,2i,-1}]
I got the eigenvalues which are 2, -3, -3 (I'm sure about this) and the normalized eigenvectors are
e1=[{(1-i)/$$\sqrt{10}$$,-i$$\sqrt{2/5}$$,$$\sqrt{2/5}$$}]
e2 = [{(-1-i)/$$\sqrt{3}$$,1/$$\sqrt{3}$$, 0}]
e3 = [{(-1+i)/$$\sqrt{3}$$,0,1/$$\sqrt{3}$$,}]

Once again I am positive about these eigenvalues,
So let U = [e1,e2,e3] (column of U are the normalized eigenvectors)
When I do the operation
(U bar)^T.U I get [{1,0,0},{0,1,-2i/3},{0,2i/3,1}] which is not quite the identity matrix (which i should be getting)

Is there something i should do to the eigenvectors to get U to be unitary?

Thanks again

6. Mar 29, 2010

### vela

Staff Emeritus
Since e2 and e3 have the same eigenvalue, any linear combination of e2 and e3 is also an eigenvector of H, and the usual process for finding eigenvectors won't guarantee orthogonality. You need to make them orthogonal by hand, using the Gram-Schmidt process. (In contrast, if the eigenvalues were all distinct, you get orthogonality automatically.)