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Show that if $A$ is an $m \times n$ matrix and $B=U^*AV$, where $U$ and $V$ are unitary matrices of sizes $m \times m$ and $n \times n$ respectively, then $A^*A$ and $B^*B$ have the same trace.

Really what I want to know is if this part is acceptable, but I included the whole question above for context:

If I have $\text{tr}(V^* A^* A V)$, by rules of trace products, $\text{tr}(V^* A^* A V) = \text{tr}(V^* V A^* A )$.

and $V^* V =I$ ( by def. of unitary)

then $\text{tr}(V^* A^* A V) = \text{tr}(A^* A)$.

Am I missing something?

1 Answers 1

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$$B=U^*AV \to B^*B=(V^*A^*U)(U^*AV)=V^*A^*AV$$

So

$$tr (B^*B)=tr(V^*A^*AV)=tr(A^*AVV^*)=tr(A^*A)$$

Ps.: I'm using that $tr(XY)=tr(YX)$ and then $tr(V^*(A^*AV))=tr((A^*AV)V^*)$.

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    What you're saying is what I'm saying - yes? I don't see a difference, and I want to make sure I fully understand the concepts – 2017-01-26
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    @Snackbreak: you are right! I didn't check your question properly! – 2017-01-26
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    @Snackbreak: your solution is the same. It is right! – 2017-01-26
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    you are very welcome! – 2017-01-26