Assume a diagonal matrix $\mathbf X$ whose size $N\times N$ and its diagonal elements are $0.5 + 0.5i$, and the vector $\mathbf p$ of size $N\times 1$ whose elements have similar amplitude.

I have noticed that $$\operatorname{tr}\bigg\{\left(\mathbf{X^{-1}\cdot p\cdot p^H\cdot \left(X^H\right)^{-1}}\right)\bigg\}\tag{1}$$ is related into the amplitude of $\mathbf p$. where $\operatorname{tr}$ is the trace operator.

  • For example if $\mathbf p = 1$; $$\implies\operatorname{tr}\bigg\{\left(\mathbf{X^{-1}\cdot p\cdot p^H\cdot \left(X^H\right)^{-1}}\right)\bigg\} = 2N$$

  • On the other hand, if $\mathbf p=0.5$; $$\implies\operatorname{tr}\bigg\{\left(\mathbf{X^{-1}\cdot p\cdot p^H\cdot \left(X^H\right)^{-1}}\right)\bigg\} = \frac N2$$

  • And so on.

My question, is there a general formula expressing the relationship between the equation $(1)$ and the amplitude of $\mathbf p$ ?


1 Answer 1


The trace of a matrix just scales with the scaling of the matrix:


So if you scale the vector $\mathbf{p}$, the trace scales with the squared magnitude of the scaling constant. With $\mathbf{p}=c\mathbf{p}_0$ you have


  • $\begingroup$ Thanks for your reply, .. but do you mean that $tr{AP_0P_0^HA^H}$ will be constant ? and finally, is it $(A^H)^{-1}$ = $(A^{-1})^H$ ? $\endgroup$ Commented Nov 4, 2020 at 10:25
  • $\begingroup$ @New_student: $c$ is just a constant with which you scale the vector. So have a fixed vector $p_0$ and the final vector you use is $p=cp_0$. $A$ is just any matrix, it is irrelevant. In your case $A=X^{-1}$. And yes to your last question. $\endgroup$
    – Matt L.
    Commented Nov 4, 2020 at 10:29
  • $\begingroup$ Thanks again, In that case $tr(AP_0P_0^HA^H)$ if $A$ is diagonal with diagonal elements $0.5+0.5i$ and $P_0 = 1$ should be constant, right? $\endgroup$ Commented Nov 4, 2020 at 10:34
  • $\begingroup$ Could you please recheck the last equation? because when I made for example $tr(AP_0P_0^HA^H) = 8$ when $P_0 = 1$. Then I changed $P_0 = 0.5$ which means according to your last equality that $|0.5|^2tr(AP_0P_0^HA^H) = 2$ and that's right, but when made $P_0 = 0.25$, $|0.25|^2tr(AP_0P_0^HA^H) = 0.5$ however when doing it in MATLAB, it's $0.125$. I think something is missing in last equation. $\endgroup$ Commented Nov 4, 2020 at 10:43
  • $\begingroup$ @New_student: The equation looks correct. $\endgroup$
    – Matt L.
    Commented Nov 4, 2020 at 11:33

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