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I don't understand to one part of proof of this theorem:

All eigenvalues of each hermitian matrix A are real numbers and, moreover, there exists unitary matrix R such, that

[tex]

R^{-1}AR

[/tex]

is diagonal

Proof:By induction with respect to n (order of matrix A)

For n = 1 it's obvious.

Suppose that the theorem holds for 1, 2, ..., n-1

We know that [itex]\exists[/itex] eigenvalue [itex]\lambda[/itex] and appropriate eigenvector [itex]x \in \mathbb{C}[/itex].

Using Steinitz's theorem, we can extend [itex]x[/itex] to orthonormal base of [itex]\mathbb{C}^{n}[/itex].

Suppose that [itex]||x|| = 1[/itex] and construct matrix [itex]P_n[/itex] from vectors of this base ([itex]P_n[/itex] will have these vectors in its columns).

[itex]P_n[/itex] is unitary [itex]\Leftarrow P_{n}^{H}P_n = I[/itex], because standard inner product of two different vectors in the orthonormal base is zero and inner product of two identical vectors is 1.

This holds:

[tex]

\left(P_{n}^{H}A_{n}P_{n}\right)^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H} = P_{n}^{H}A_{n}P_{n}

[/tex]

Last line is what I don't understand, probably it's trivial but I can't see that

[tex]

\left(P_{n}^{H}A_{n}P_{n}\right)^{H} = \left(P_{n}^{H}\right)^{H}A_{n}^{H}P_{n}^{H} = P_{n}^{H}A_{n}^{H}\left(P_{n}^{H}\right)^{H}

[/tex]

(the second equality)

Thank you for the explanation.

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# Something about hermitian matrixes

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