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During the last lecture one of my professors claimed that the hermitian matrix is the ONLY complex matrix which was diagonizable.

This seems strange to mee (not to say a very very strong claim to make), I know that Hermitian matrices are the only ones diagonizable by unitary transformations, but that they're the only diagonizable ones seems strange.

Could someone elaborate on this ? Or provide some kind of proof for this ?

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    $\begingroup$ as far as I know, you may look for spectral decomposition on normal matrices first in which they can be unitarily diagonalized with complex eigenvalues along their diagonals. For spectral decomposition on Hermitian matrix, the corresponding diagonal elements are real eigenvalues of the original one. For spectral decomposition on unitary matrices, magnitude of elements along diagonal of the diagonal matrix must be one. $\endgroup$ – nam Mar 29 '14 at 16:00
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    $\begingroup$ Isn't $A=\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right]$ diagonal but not Hermitian, since $A^\ast=\left[ \begin{matrix} 1-i & 0 \\ 0 & 1-i \end{matrix}\right] \neq A$? $\endgroup$ – Hayden Mar 29 '14 at 16:02
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    $\begingroup$ Diagonalizable over what field? $\endgroup$ – Git Gud Mar 29 '14 at 16:03
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    $\begingroup$ All matrices with distinct eigenvalues are diagonalizable, no matter what type. However, if the eigenvalues are degenerate, the multiplicity of eigenvectors may not be enough, if the matrix is not normal. It's also possible that he talked about unitary diagonalizability. $\endgroup$ – orion Mar 29 '14 at 16:10
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    $\begingroup$ @Nick No. Hayden's example is unitary diagonalizable ($A=IAI$). Any normal matrix is unitarily diagonalizable, though. $\endgroup$ – Git Gud Mar 29 '14 at 21:15
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As requested, I'll provide my example and a little bit of exposition:

The claim your professor makes, namely

All diagonalizable (in $\mathbb{C}$) matrices are Hermitian.

is false. Let us quickly recall what it means for a square complex matrix $A$ to be Hermitian: it means that $A=A^\ast$, where $A^\ast$ is the conjugate transpose of $A$, A simple counter-example then is the following: $\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right]$ which is clearly diagonalizable (it is diagonal already) but is not Hermitian because $A\neq A^\ast$.

Perhaps your professor meant to make a weaker claim:

All matrices that are unitary diagonalizable are Hermitian.

Again, this is false, as the above example points out: $$\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right] =\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right] \left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right] \left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix} \right]$$ The identity matrix $\left[ \begin{matrix} 1 & 0 \\ 0 & 1 \end{matrix}\right]$ is indeed a unitary matrix, so that $\left[ \begin{matrix} 1+i & 0 \\ 0 & 1+i \end{matrix}\right]$ has a unitary diagonalization.

In general, the Spectral Theorem tells us that all normal complex matrices are unitary diagonalizable, and there are normal matrices that are not Hermitian.

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