$\operatorname{span}\{x_1,\ldots,x_n\} = \mathbb{C}^n$? If $A$ is a $n \times n$ Hermitian matrix, then is it right that,
\begin{equation}
\operatorname{span}\{x_1,\ldots,x_n\} = \mathbb{C}^n \text{?}
\end{equation}
where $x_i \ne 0$ is the eigenvector associated with the eigenvalue $\lambda_i$ of $A$. That is, $A x_i = \lambda_i x_i, i=1,\ldots,n$.
 A: This is correct (as long as the span is taken over $\mathbb C$ and $\{x_1,...,x_n \}$ is linearly independent). Hermitian matrices are always diagonalizable, which is equivalent to the statement that the eigenvectors form a basis for the vector space. For a proof of this look at the Spectral theorem, which not only tells us that there is a basis consisting of eigenvectors of $A$ but also that there is an orthogonal basis consisting of eigenvectors of $A$.
A: A matrix is diagonalisable if and only if the generalised eigenspaces all equal the (non-generalised) eigenspaces, as the generalised eigenspaces contain the eigenspaces, but direct sum to the entire space. This is therefore equivalent to showing $\mathrm{ker} (M - \lambda I) =\mathrm{ker} (M - \lambda I)^2$ for all eigenvalues $\lambda$. Note that $\mathrm{ker} (M - \lambda I) \subseteq \mathrm{ker} (M - \lambda I)^2$ is always true.
Suppose $\lambda$ is an eigenvalue for $A$, the Hermitian matrix. Then $\lambda \in \mathbb{R}$ (a nice exercise, if you're not familiar). Suppose $v \in \mathrm{ker} (M - \lambda I)^2$. Then,
$$0 = v^*(M - \lambda I)^2v = (v^*(M^* - \overline{\lambda} I^*)) (M - \lambda I)v = ((M - \lambda I)v)^* (M - \lambda I)v = \|(M - \lambda I)v\|^2$$
Hence $v \in \mathrm{ker} (M - \lambda I)$, so we get $\mathrm{ker} (M - \lambda I) = \mathrm{ker} (M - \lambda I)^2$, and so $M$ is diagonalisable.
