# Is there any matrix $A$ for which $A^{\dagger}A=AA^{\dagger}$ but $A\neq A^{\dagger}$?

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?

• This even happens when the dimension is $1$, ie. scalars. $2\cdot 3 = 3\cdot 2$ but $2\neq 3$ – Callus Mar 9 at 13:40

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 $$A^{*} \neq A$$.