# Eigenvalues of Symmetric/Hermitian Matrices

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!

• Welcome to MSE!!! – user844292 Nov 15 '20 at 15:09
• Thank you very much – user849749 Nov 15 '20 at 15:17
• proof is clear and all good! – user844292 Nov 15 '20 at 15:18

## 1 Answer

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.