Diagonalizing a Hermitian Matrix: A

[pivoxa15]

Find the diagonal form of the Hermitian matrix

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

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,

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

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

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

which clearly isn't correct.

 

[matt grime]

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.

 

[greisen]

hey
rigth method but wrong eigenvalues

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

this will help

 

[Hurkyl]

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?)

 

[pivoxa15]

hey
rigth method but wrong eigenvalues

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

this will help

That could be my mistake.