If $A \in \mathbb{C}^{n \times n}$ is hermitian, then all it's eigenvalues are real and eigenvectors of different eigenspaces are orthogonal. Left to prove is the fact that there exists an orthonormal Basis $\{ \ v_1 \ v_2 \ \dots \ v_n \ \}$ of eigenvectors of $A$.
In every eigenspace, I can construct an orthonormal basis using the Gram-Schmidt algorithm. Since the eigenvectors of different eigenspaces are orthogonal, all left to prove is that this basis spans $\mathbb{C}^{n \times n}$ – the geometric multiplicity of every eigenvalue must be equal to it's algebraic multiplicity.
Let's try induction over $n$:
$n = 1: \xi_1 v_1 = 0 \implies \xi_1 = 0 \quad \text{since, per Definition} \quad v_1 \neq 0$
This is where I am stuck – I don't know how to prove that adding an eigenvector from an orthogonal basis from one of the eigenspaces still implies that $\Sigma_{i = 1}^{n+1} \xi_i v_i = o\implies \forall i: \xi_i = 0$. It is easy enough when for $1 \leq i \leq n: \lambda_{n+1} \neq \lambda_i$. Can somebody point me in the right direction for the other case?
Best Answer
Here is a standard argument.
Let $\lambda$ be an eigenvalue of $A$, and $v$ be an associated eigenvector. Set $E=span\left\{ v\right\} $ and $E^{\perp}$ be the orthogonal complement, i.e., \begin{alignat*}{1} E^{\perp} & =\left\{ u\in\mathbb{C}^{n}\mid\left\langle u,v\right\rangle =0\right\} . \end{alignat*} $E$ and $E^{\perp}$ are both invariant under $A$. To see that $E^{\perp}$ is $A$-invariant, let $u\in E^{\perp}$ then \begin{align*} \left\langle Au,v\right\rangle & =\left\langle u,A^{*}v\right\rangle =\left\langle u,Av\right\rangle =\left\langle u,\lambda v\right\rangle =0. \end{align*}
Now the restriction of $A$ to $E^{\perp}$ is also Hermitian, so by repeating the above argument, there exist $\lambda_{1},\cdots,\lambda_{n}$ and $v_{1},\cdots,v_{n}$ such that $Av_{j}=\lambda_{j}v_{j}$.
The $\lambda_{j}$'s are not necessarily distinct. By construction, $v_{j}$'s are mutually orthogonal.