Linear Algebra/Spectral Theorem

Given a Hermitian matrix $A$, $A$ is always diagonalizable. It is also the case that all eigenvalues of $A$ are real, and that all eigenvectors are mutually orthogonal. This is given by the "Spectral Theorem":

The columns of $U=\left(\begin{array}{cccc}{\overrightarrow{u}}_1& {\overrightarrow{u}}_2& \dots & {\overrightarrow{u}}_n\end{array}\right)$ are the eigenvectors of $U$, and the diagonal entries of $\mathrm{\Lambda }=\left(\begin{array}{cccc}{\lambda }_1& 0& \cdots & 0\\ 0& {\lambda }_2& \cdots & 0\\ ⋮& ⋮& \ddots & ⋮\\ 0& 0& \cdots & {\lambda }_n\end{array}\right)$ are the corresponding eigenvalues.

In essence $A$ can be decomposed into a "spectrum" of rank 1 projections: $A=\sum _{i=1}^n{\lambda }_i({\overrightarrow{u}}_i{\overrightarrow{u}}_i^H)$

The spectral theorem can in fact be proven without the need for the characteristic polynomial of $A$, or any of the derivative theorems.