I understand that Hermitian matrices has real eigenvalues.
Just to hit the point home, I have the following question.
Does every Hermitian matrix has eigenvalues? Since the proof assumes that the eigenvalue exists, the proof does not imply that every Hermitian matrix must have some eigenvalues. It just says that if it has an eigenvalue, then the eigenvalue must be real.
Thanks!
Best Answer
Every complex $n \times n$ Hermitian matrix (or real symmetric matrix) has $n$ real eigenvalues. However, these eigenvalues might not be distinct. As a trivial example, $$ B = \left( \begin{array}{ccc} 5 & 0 & 0 \\ 0 & 5 & 0 \\ 0 & 0 & 1 \end{array} \right) $$ has three eigenvalues but two of them are equal.
For any Hermitian matrix $A$, there exists a complex matrix $U$ such that $U^tU = I$, (where $I$ the identity matrix) and $$ U^tAU = \Lambda $$ where $\Lambda$ is a real diagonal matrix which contains the eigenvalues of $A$. If $A$ is real symmetric then the matrix $U$ is real. The $k$th diagonal element of $\Lambda$ is the $k$th eigenvalue and it corresponds to the $k$th eigenvector given by the $k$th column of $U$.
Alternatively, $A$ can be written as $$ A = U \Lambda U^t $$ which is often known as the spectral decomposition of $A$.