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I encountered the definition of a normal matrix:

$$A^*A = AA^*$$

Whereas the definition of a hermitian matrix is:

$$A^{\dagger}=A$$

From this it follows that $A^{\dagger}A=AA^{\dagger}$, very similar to the definition of the normal matrix.

Does it work in the other way round? I.e., are there matrices for which it holds that $A^{\dagger}A=AA^{\dagger}$ but $A^{\dagger}=A$ does not?

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  • $\begingroup$ This even happens when the dimension is $1$, ie. scalars. $2\cdot 3 = 3\cdot 2$ but $2\neq 3$ $\endgroup$ – Callus Mar 9 at 13:40
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Yes, there are many such matrices. Hermitian matrices are just one type of normal matrix, but there are many other types of normal matrices. For example, you could consider skew-Hermitian matrices, which are not generally Hermitian. These matrices are ones that satisfy $A^* = -A$ (which implies $A^* A = AA^*$), and such $A$ is Hermitian if and only if $A = -A$, that is, iff $A = O$. For example take $A = \begin{bmatrix}0 & -1 \\ 1 & 0 \end{bmatrix}$, then $A$ is normal but $A^{*} \neq A$.

In fact, your question is answered in the Wikipedia page on normal matrices that you linked, in this section.

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