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For two square complex matrices $A, B$ of the same order, what is the name of the condition

$AB=BA$, $AB^*=B^*A$?

Is this condition popular? Here $A^*$ means the transpose conjugate of $A$.

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    Well, it's used a lot i$n$ the theory of operator algebra, e.g. http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem which is a cornerstone of the theory.2011-09-06

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If you assume that $B$ is a normal matrix i.e $B^{*}B=BB^{*}$ and $A$ and $B$ commute, then you will automatically get that $B^{*}A=AB^{*}$. It is called Fuglede-Putnam Theorem and it is even true for normal operators from $\mathcal{B}(H)$ - all bounded linear operators on $H$, where $H$ is a Hilbert spaces. In your case $H=\mathbb{C}^{n}$.

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    I had a mistake cos the theorem in full generality involves possibly different two normal operators $N$ and $M$ and one can be on $H$ and second on $K$ since $A$ is the bounded operator from $K$ to $H$ such that $NA=AM$.2011-09-06
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$A$ and $B$ "doubly commute," some say. You will find at least some evidence of this by searching for "doubly commuting matrices". Leaving out "matrices" will find many doubly commuting operators on Hilbert space, meaning operators that commute with each other and with each other's adjoints.