How to prove symmetric matrix is orthogonally diagonalizable?

Question

I learned below theorem and there is a proof that orthogonally diagonalizable matrix is symmetric, but there is no proof that symmetric matrix are orthogonally diagonalizable.

Theorem 2. An $n\times n$ matrix $A$ is orthogonally diagonalizable if and only if $A$ is a symmetric matrix.

I searched proof in this website and found this proof, but I cannot understand why $\langle Ax,y \rangle = \langle x, A^Ty \rangle$.

How can I prove?

I think you are in search of https://en.wikipedia.org/wiki/Hermitian_adjoint — Sujit Bhattacharyya, Nov 29 '18 at 03:35

score 2 · Accepted Answer · answered Nov 29 '18 at 03:40

2

Writing the inner product $\langle v, w \rangle$ as $v^\top w$, we have $$\langle Ax, y\rangle = (Ax)^\top y = x^\top A^\top y = \langle x, A^\top y\rangle.$$

answered Nov 29 '18 at 03:40

angryavian

89,882

Then, inner product is thought as vector space in $\Bbb{R}^n$ not general vector space? – baeharam Nov 29 '18 at 03:56
You are multiplying the vectors by an $n \times n$ matrix $A$, so yes. – angryavian Nov 29 '18 at 04:05
Thank you, angryavian – baeharam Nov 29 '18 at 04:55

How to prove symmetric matrix is orthogonally diagonalizable?

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