Does an n by n Hermitian matrix always has n independent eigenvectors? I am learning the MIT ocw 18.06 Linear Algebra, and I have learnt:
an arbitrary $n×n$ matrix A with n independent eigenvectors can be written as
$A=SΛS^{-1}$, and then for the Hermitian matrices, because the eigenvectors can be chosen orthonormal, it can be written as $A=QΛQ^H$ further.
I wonder does every $n×n$ Hermitian matrix has n independent eigenvectors and why?
Thank you!
P.S. 
MIT 18.06 Linear Algebra Lecture 25: Symmetric Matrices and Positive Definiteness.
You may wish to start from 4:20. From the course, I think the spectral theorem comes from diagonalizable matrix, it's just a special case, it's just the case eigenvectors are orthonormal. The eigenvectors of Hermitian matrices can be chosen orthnormal, but is every Hermitian matrix diagonalizable? If it is, why? 
 A: This is a theorem with a name: it is called the Spectral Theorem for Hermitian (or self-adjoint) matrices. As pointed out by Jose' Carlos Santos, it is a special case of the Spectral Theorem for normal matrices, which is just a little bit harder to prove.
Actually we can prove the spectral theorem for Hermitian matrices right here in a few lines.  
We are going to have to think about linear operators rather than matrices.  If $T$ is a linear operator on a finite dimensional complex inner product space $V$, its adjoint $T^*$ is  another linear operator determined by $\langle T v, w\rangle = \langle v, T^* w \rangle$ for all $v, w \in V$. (Note this is a basis-free description.)  $T$ is called Hermitian or self-adjoint if $T = T^*$.
Let $B$ be an $n$-by-$n$ complex matrix and $B^*$ the conjugate transpose matrix.  Let $T_B$ and $T_{B^*}$ be the corresponding linear operators.  Then $(T_B)^* = T_{B^*}$, so a a matrix is Hermitian if and only if the corresponding linear operator is Hermitian.
Let $A$ be a Hermitian linear operator on  a complex inner product space $V$ of dimension $n$.  We need to consider $A$--invariant subspaces of $V$, that is linear subspaces $W$ such that $A W \subseteq W$.  We should think about such a subspace as on an equal footing as our original space $V$.  In particular, any such subspace is itself an inner product space, $A_{|W} : W \to W$ is a linear operator on $W$, and  $A_{|W}$ is also Hermitian.
If $\dim W \ge 1$,  $A_{|W}$ has an least one eigenvector $w \in W$ -- because any linear operator at all acting on a (non-zero) finite dimensional complex vector space has at least one eigenvector. 
The basic phenomenon is this:    Let $W$ be any invariant subspace for $A$.    Then $W^\perp$ is also invariant under $A$. The reason is that if $w \in W$ and $x \in W^\perp$, then 
$$
\langle w, A x\rangle = \langle A^* w , x \rangle = \langle A w, x \rangle = 0,
$$
because $Aw \in W$ and $x \in W^\perp$.  Thus $A x \in W^\perp$.
Write $V  = V_1$. 
Take one eigenvector $v_1$ for $A$ in $V_1$.  Then $\mathbb C v_1$ is $A$--invariant.   Hence $V_2 = (\mathbb C v_1)^\perp$ is also $A$ invariant. Now just apply the same argument to $V_2$:  the restriction of $A$ to $V_2$ has an eigenvector $v_2$ and the perpendicular complement $V_3$  to $\mathbb C v_2$  in $V_2$ is $A$--invariant.   Continuing in this way, one gets a sequence of mutually orthogonal eigenvectors and a decreasing sequence of invariant subpsaces, $V =  V_1 \supset V_2 \supset V_3 \dots$ such that $V_k$ has dimension $n - k + 1$.  The process will only stop when we get to $V_n$ which has dimension 1.
A: Yes, it has. That is due to the spectral theorem: every normal $n\times n$ matrix is diagonalizable. And Hermitian matrices are normal.
