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I am new to stack exchange so please excuse me as I write in JAX I have a question on linear algebra that goes like this : If $A$ is an $n\times n$ real symmetric matrix or complex Hermitian matrix, then its eigenvalues are real.

I want to prove this theorem but I hope someone can this proof and correct me if I am wrong : We have that By the Schur’s decomposition, $A=QRQ^{*}$ however $A=A^{*}=(QRQ^{*})^{*}=QR^{*}Q^{*}$ where we have that $R$ an upper triangular matrix with eigenvalues of $A$ on its diagonal entries, so $R^{*}$ is a lower triangular matrix thus $A-A^{*}=Q(R-R^{*})Q^{*}=0\implies R-R^{*}=0\implies r_{i,i}=\overline{r_{i,i}}$. Therefore, $R$ is a diagonal matrix and its diagonal entries are real. I would like to thank whoever checks my proof. Best regards!

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    $\begingroup$ Welcome to MSE!!! $\endgroup$ – user844292 Nov 15 '20 at 15:09
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    $\begingroup$ Thank you very much $\endgroup$ – user849749 Nov 15 '20 at 15:17
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    $\begingroup$ proof is clear and all good! $\endgroup$ – user844292 Nov 15 '20 at 15:18
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Yes, yes, the proof given by our OP Alexandre von Maylen looks fine. However, it leans heavily on the Schur decomposition which is itself non-trivial. A simple proof may be had by proceeding directly from first principles and definitions, to wit:

We are given

$Ax = \mu x \tag 1$

and

$A^\dagger = A; \tag 2$

we may assume that

$\langle x, x \rangle = 1, \tag 3$

with $\langle \cdot, \cdot \rangle$ the standard Hermitian inner product on $\Bbb C^n$; thus we have

$\mu = \mu \langle x, x \rangle = \langle x, \mu x \rangle$ $= \langle x, Ax \rangle = \langle A^\dagger x, x \rangle = \langle Ax, x\rangle$ $= \overline{\langle x, Ax \rangle} = \overline{\langle x, \mu x \rangle} = \overline{\mu \langle x, x\rangle}= \bar \mu. \tag 4$

Note that

$\langle x, Ax \rangle = \langle A^\dagger x, x \rangle \tag 5$

is essentially the definition of $A^\dagger$, which coincides with $A^T$ since $A$ is specified to be a real $n \times n$ matrix.

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