Span of an orthogonal basis of an hermitian matrix constructed using eigenvectors 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?
 A: 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. 
A: This is based on Schur triangularization theorem: every square matrix is unitarily similar to a triangular matrix. More explicitly, if $A$ is a square matrix (over the complex numbers), there exist $U$ unitary (that is, $UU^H=I$) and $T$ upper triangular such that
$$
A=UTU^H
$$
(Here $A^H$ denotes the Hermitian transpose.)
Suppose $A$ is normal, that is, $AA^H=A^HA$; then we have
$$
UTU^HUT^HU^H=UT^HU^HUTU^H
$$
so $UTT^HU^H=UT^HTU^H$ and, finally,
$$
TT^H=T^HT
$$
Actually, $T$ must be diagonal: write $T$ as a block matrix
$$
T=\begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix}
$$
where the upper left block is $1\times 1$ and the lower right block is $(n-1)\times(n-1)$; then
$$
TT^H=
\begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix}
\begin{bmatrix}\bar{t} & 0^H \\ v & T_1^H\end{bmatrix}=
\begin{bmatrix}
t\bar{t}+v^Hv & v^HT_1 \\
T_1v & T_1T_1^H
\end{bmatrix}
$$
whereas
$$
T^HT=
\begin{bmatrix}\bar{t} & 0^H \\ v & T_1^H\end{bmatrix}
\begin{bmatrix}t & v^H \\ 0 & T_1\end{bmatrix}=
\begin{bmatrix}
\bar{t}t & \bar{t}v^H \\
tv & vv^H+T_1^HT_1
\end{bmatrix}
$$
whence $v^Hv=0$, which implies $v=0$; therefore $T_1T^H=T_1^HT_1$ and induction ends the proof.
A Hermitian matrix is obviously normal.
A: You have, that the eigenvectors and generalized eigenvectors form a invariant subspace $U$, that is there exists a matrix $X$, s.t. $AU=UX$. Since there is definiatly a eigenvector in there, you have the first collumn of $X$ as a multiple of $e_1$. 
It is definiatly possible to form a orthogonal-basis of $U$, so that you can determine $X=U^HAU$. 
If $A=A^H$ you can conjugate-transpose that equation and get $X^H=(U^HAU)^H = U^HAU = X$. 
So $X$ has to be itself hermitian. 
Since the first collumn of $X$ has only one nonzero component in the diagonal, the same goes for the first collumn, meaning ever other vector of $U$ (forming $\underline{U}$) will be mapped into $\underline{U}$. 
From there on, you can use induction to prove by this, that $X$ is diagonal and every vector in $U$ is as eigenvector. 
