Diagonalizing a Hermitian Matrix: A

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Find the diagonal form of the Hermitian matrix

A=\left(<br /> \begin{array}{cc}<br /> 2 &amp; 3i\\<br /> -3i &amp; 2<br /> \end{array}<br /> \right)<br />

The spectral theorem could be used with PAP*=D where D is diagonal matrix and P is a unitary matrix.

I put the columns of P as the eigenvectors (with unit length) of A,

P=\frac{1}{\sqrt{2}}\left(<br /> \begin{array}{cc}<br /> i &amp; -i\\<br /> 1 &amp; 1<br /> \end{array}<br /> \right)<br />

I have checked that P is unitary with P^{-1}=P^{*} and the diagonal entries of D should be 5 and -1. But I got

D=\left(<br /> \begin{array}{cc}<br /> 2 &amp; -3\\<br /> -3 &amp; 2<br /> \end{array}<br /> \right)<br />

which clearly isn't correct.
 
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Yes, you're right it isn't correct. I'm, nots sure what you want anyone here to do. You have the method correct, so just make sure you'renot making any dumb mistakes in multiplying out matrices.
 
hey
rigth method but wrong eigenvalues

P = 1/sqrt(2) [i -1;i 1]

this will help
 
The spectral theorem could be used with PAP*=D where D is diagonal matrix and P is a unitary matrix.
I can never remember for sure, but isn't it supposed to be A = PDP*? (and thus P*AP = D?)
 
greisen said:
hey
rigth method but wrong eigenvalues

P = 1/sqrt(2) [i -1;i 1]

this will help

That could be my mistake.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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