pivoxa15
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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)<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 & -i\\<br /> 1 & 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 & -3\\<br /> -3 & 2<br /> \end{array}<br /> \right)<br />
which clearly isn't correct.
A=\left(<br /> \begin{array}{cc}<br /> 2 & 3i\\<br /> -3i & 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 & -i\\<br /> 1 & 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 & -3\\<br /> -3 & 2<br /> \end{array}<br /> \right)<br />
which clearly isn't correct.
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