Complex hermitian matrix: real eigenvalues and orthogonal eigenvectors?

Asked Aug 16 '14 at 22:39

Active Aug 17 '14 at 00:20

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Let $A\in\Bbb R^{n\times n} $be a symmetric matrix.

$1)$ Prove that its eigenvalues are real and that any two eigenvector associated with different eigenvalues are orthogonal.

$2)$ Is this true if $A\in \Bbb C ^{n\times n} $ and $A$ is hermitian?

I can solve $1)$

For $2)$ I don't see why the conclusions wouldn't hold: if it's hermitian its eigenvalues must be real and if it's hermitian the it's normal so the eigenvector for different eigenvalues will be orthogonal. However I feel this maybe a trick question. Is there anything I'm missing?

edited Aug 17 '14 at 00:20

Michael Hardy

asked Aug 16 '14 at 22:39

milo

1,047

This is known as the spectral theorem. No trick. – Klaas van Aarsen Aug 16 '14 at 22:43
I think you're absolutely correct. Doesn't look like a trick question to me. (Isn't the point of this question to prove what you stated?) – Tunococ Aug 16 '14 at 23:32
@milo : You don't need to write $nxn$; you can write $n\times n$. I edited accordingly. – Michael Hardy Aug 17 '14 at 00:21
Try to imitate this proof. – Mhenni Benghorbal Aug 17 '14 at 02:26

Complex hermitian matrix: real eigenvalues and orthogonal eigenvectors?

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