# Rewriting a Trace of Matrix products

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 $trace{(AXBX^{H})}$.

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

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 $x_k$.

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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.

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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