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I understand that Hermitian matrices has real eigenvalues.

Just to hit the point home, I have the following question.

Does every Hermitian matrix have 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.

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    $\begingroup$ Every complex matrix has at least one eigenvalue $\endgroup$
    – Ben Grossmann
    Dec 5, 2015 at 22:21
  • $\begingroup$ This is because every characteristic polynomial has at least one root $\endgroup$
    – Ben Grossmann
    Dec 5, 2015 at 22:22
  • $\begingroup$ great! That makes sense. Thank you! $\endgroup$
    – user247618
    Dec 5, 2015 at 22:23
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    $\begingroup$ Comment to the post (v1): Title question seems different from question in body. $\endgroup$
    – Qmechanic
    Dec 6, 2015 at 1:35
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    $\begingroup$ @BenGrossmann Yes, thanks to the robot that is randomly pushing old questions to the front. And indeed, that's the matrix. $\endgroup$
    – Marc van Leeuwen
    Oct 4, 2021 at 13:48

1 Answer 1

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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 $A$.

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