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## Homework Statement

Given the matrix H=

\begin{array}{cc}

4 & 2+2i & 1-i \\

2-2i & 6 & -2i \\

1+i & 2i & 3 \\

\end{array}

Find a unitary matrix U such that U*HU is diagonal

(U* is the conjugate transpose of U, and U* = U

^{-1})

## The Attempt at a Solution

I find the eigenvalues

λ1 = 9

λ2 = 2

λ3 = 2

and the corresponding eigenvectors

v1 = [ 1-i , -2i , 1 ]

v2 = [ -1+i , 0 , 2 ]

v3 = [ -1-i , 1 , 0 ]

Normalizing them and using them as column vectors for U I get

\begin{array}{cc}

(1-i)/\sqrt{7} & (-1+i)/\sqrt{6} & (-1-i)/\sqrt{3} \\

-2i/\sqrt{7} & 0 & 1/\sqrt{3} \\

1/\sqrt{7} & 2/\sqrt{6} & 0 \\

\end{array}

but this doesn't work. I try U*U and I try U*HU but neither come out right.

What am I missing?

*edit* I've just noticed that U isn't in hermitian form because of v1. Is there a way of transforming v1 so I get a real number for u1,1? Like multiplying it my the conjugate of 1-i? I'm sure it has something to do with the repeated eigenvalue, but I don't know what to do

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