Unitary matrix of a hermitian form matrix

  • Thread starter Locoism
  • Start date
  • #1
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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
 
Last edited:

Answers and Replies

  • #2
I like Serena
Homework Helper
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Hi Locoism! :smile:

Your v2 and v3 form a basis for the eigenspace of eigenvalue 2.
However, they are not orthogonal.
 
  • #3
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Ok but then do I construct an orthogonal basis from v2 and v3 (using gram-schmidt)?
 
  • #4
I like Serena
Homework Helper
6,577
176
Yep.
 

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