Be $n \in \mathbb N$ and $A \in Mat(n \times n, \mathbb C)$ with real entries and be $A^t = A$.

Question

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Be $n \in \mathbb N$ and $A \in Mat(n \times n, \mathbb C)$ with real entries and be $A^t = A$.

(i) Are all eigenvalues of $A$ real numbers?

(ii) Be $(v_{1},...,v_n),(w_{1},...,w_n)\in \mathbb R^n$ eigenvectors of $A$ regarding the eigenvalue $\lambda$ and $\mu$. Suppose that $\lambda \neq \mu $. Show that $\sum_{i=1}^n v_iw_i = 0$.

Thanks in advance!

Answer 1

We have $A^* = (\bar A)^t = A^t = A$ and so $A$ is self-adjoint.

(i) If $Av=\lambda v$ with $v \ne 0$, then $$ \lambda \langle v,v \rangle = \langle \lambda v,v \rangle = \langle A v,v \rangle = \langle v, A^* v \rangle = \langle v, A v \rangle = \langle v, \lambda v\rangle = \bar\lambda \langle v,v \rangle $$ implies $\lambda = \bar\lambda$ and so $\lambda$ is real.

(ii) $$ \lambda \langle v,w \rangle = \langle \lambda v,w \rangle = \langle A v,w \rangle = \langle v, A^* w \rangle = \langle v, A w \rangle = \langle v, \mu w\rangle = \bar\mu \langle v,w \rangle = \mu \langle v,w \rangle $$ implies $\langle v,w \rangle=0$ since $\lambda \ne \mu$.