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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 $A$.

Sunni
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    I would say $A$ belongs to the commutant of ${B,B^{\ast}}$. I don't know if there is a more specialized terminology. – t.b. Sep 06 '11 at 00:27
  • I would like to know if someone used this condition in his/her work. – Sunni Sep 06 '11 at 01:07
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    Well, it's used a lot in the theory of operator algebra, e.g. http://en.wikipedia.org/wiki/Von_Neumann_bicommutant_theorem which is a cornerstone of the theory. – t.b. Sep 06 '11 at 01:14

2 Answers2

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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 $H=\mathbb{C}^{n}$.

Edvin Goey
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  • http://en.wikipedia.org/wiki/Fuglede's_theorem – t.b. Sep 06 '11 at 01:33
  • Yep, I was thinking about this theorem it can be found in "Putnam's generalization" section. – Edvin Goey Sep 06 '11 at 01:41
  • What do you mean by "normal operators from $\mathcal{B}(H,K)$"? – Jonas Meyer Sep 06 '11 at 02:54
  • 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 $NA=AM$. – Edvin Goey Sep 06 '11 at 14:23
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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.

Jonas Meyer
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