# Let $A \in \mathbb{C}^{n \times n}$ be hermitian. Prove all eigenvalues of $A$ are real…

Let $A \in \mathbb{C}^{n \times n}$ be hermitian. Prove all eigenvalues of $A$ are real and that eigenvectors corresponding to different eigenvalues are orthogonal.

(1.) The eigenvalues of A are real.

proof

Let $\lambda$ be an arbitrary eigenvalue of $A$, then there exists an $x \in \mathbb{C}^n$ such that $Ax = \lambda x$. Now

$$x^*Ax = x^*(Ax) = x^*\lambda x = \lambda||x||^2 \,\,\,\,\,\,\,\,\text { and } \\ x^*Ax = x^*A^*x = (Ax)^*x = (\lambda x)^*x= \lambda ^*||x||^2$$

So

$$\lambda ||x||^2 = \lambda^*||x||^2 \\ \lambda = \lambda^*$$

Which is only possible if $Im(\lambda) = 0$.

(2.) The eigenvectors corresponding to different eigenvalues are orthogonal.

proof

Let $\lambda_x, \lambda_y$ be distinct eigenvalues of $A$ corresponding to vectors $x,y$. Then

$$(Ax)^*(Ay) = {\lambda}_{x}^* {\lambda}_{y} x^*y = {\lambda}_{x} {\lambda}_{y} x^*y$$

But

$$(Ax)^*(Ay) = x^*A^*Ay = x^*A^*A^*y = (AAx)^*y = (\lambda_x^2 x)^*y = \lambda_x^2 x^*y$$

which implies

$$\lambda_x^2 x^*y = {\lambda}_{x} {\lambda}_{y} x^*y$$

And since (by assumption) $\lambda_x \neq \lambda_y$ the equality only holds if $x^*y = 0$.

Better ways to do it?

• In your last deduction, there is also the case $\lambda_x=0$, which a priori could allow for the possibility of $x^*y\neq0$. Of course, you can rescue your argument by exchanging the roles of $\lambda_x$ and $\lambda_y$, and of $x$ and $y$. – user561348 May 20 '18 at 0:50
• Usually people prove it $\lambda_xx^*y=\lambda_x^*x^*y=(Ax)^*y=x^*(Ay)=\lambda_yx^*y$. This gives you $(\lambda_x-\lambda_y)x^*y=0$. – user561348 May 20 '18 at 0:53
• @arugula Yep, that way is much nicer. – Zduff May 20 '18 at 0:59

$$\lambda_xx^*y = \lambda^*_xx^*y = (Ax)^*y = x^*A^*y = x^*(Ay) = \lambda_yx^*y$$
$$(\lambda_x - \lambda_y) x^*y = 0$$
And by assumption $\lambda_x \neq \lambda_y$ so it must be that $x^*y = 0$.