Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Consider 3 $N \times N$ complex matrices $A$,$B$ and $X$. $A$ and $B$ are hermitian matrices. Let$X=[x_1,x_2..x_N]$ where $x_i$'s the $N\times 1$ column vectors of $X$. I am interested in the term $trace(AXBX^{H})$. Is there anyway, I can write it in terms of columns of $X$. To point out a example for another case, $trace(AX)=\sum_{i=1}^{N}a_i^{H}x_i$ where $a_i$ are the columns of $A$ (hermitian matrix). Similarly $trace(BX^{H})=trace(X^{H}B)=\sum_{i}^{N}x_i^Hb_i$ where $b_i$ are columns of $B$. Can anyone come up with a similar presentation for $trace{(AXBX^{H})}$.

share|cite|improve this question
up vote 2 down vote accepted

$$ \operatorname{tr}AXBX^H=\sum_{ijkl}A_{ij}X_{jk}B_{kl}X_{il}^*=\sum_{ijkl}A_{ij}(x_k)_jB_{kl}(x^H_l)_i=\sum_{lk}B_{kl}(x_l^HAx_k)\;. $$

Thus this is a linear combination of values of the quadratic form defined by $A$, with $B_{kl}$ specifying the coefficient of the value for $x_l$ and $x_k$.

share|cite|improve this answer
I didn't understand your arguments, but the last step did provide me insight to prove what I needed. Thanks a lot!! – dineshdileep Nov 3 '12 at 15:38

From the cyclic property of trace $$\operatorname{trace}(AXBX^H)=\operatorname{trace}(X^HAXB)$$ If $ZAZ^H = D$ diagonal, then we have $\operatorname{trace}(X^HZ^HDZXB)$

and with $Y=D^{1 \over 2}ZX$ $$ = \operatorname{trace}(Y^HYB)$$

This is a form of two hermitian matrices and may be more to your liking. Of course if use the diagonalization of $B$ instead of $A$ then the cyclic property of trace need not be used.

share|cite|improve this answer
Unfortunately MSE doesn't have a option to accept 2 answers, otherwise I would have definitely accepted yours. Yes, your formulation looks neat, but not enough for the problem I am working on. Thanks a lot!! – dineshdileep Nov 3 '12 at 15:39
I like many of your questions precisely because I am not quite sure of your problems that you work on, but your questions often contain many of the same concepts I find in my work. I usually learn much from them! I hope you keep it up! – adam W Nov 3 '12 at 16:19

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.