Proving that $tr(A^*A)=tr(B^*B)$

Question

$A, B \in M_n(F)$ are unitarily equivalent. How do I use that to prove that $tr(A^*A)=tr(B^*B)$?

Additionally, how would I use that fact to prove that $\sum_{i,j}|A_{ij}|^2=\sum_{i,j}|B_{ij}|^2$?

Answer 1

If $A$ and $B$ are unitarily equivalent, then there exists a unitary matrix $U$ such that $A=UBU^{-1}=UBU^*$. But then $$ A^*=(UBU^*)^*=(U^*)^*B^*U^*=UB^*U^*. $$ So we have $$ A^*A=UB^*U^*UBU^*=UB^*BU^*. $$ Now what can you say about the trace of $A^*A=UB^*BU^*$ and the trace of $B^*B$?

For your second question, write $A^*A$ out in terms of the entries of the matrix $A=(A_{ij})$. For example, when $A$ is a $2\times 2$ matrix, this would be $$ \begin{pmatrix} \overline{A_{11}} & \overline{A_{21}} \\ \overline{A_{12}} & \overline{A_{22}} \end{pmatrix}\cdot\begin{pmatrix} A_{11} & A_{12} \\ A_{21} & A_{22} \end{pmatrix}= \begin{pmatrix} |A_{11}|^2+|A_{21}|^2 & \overline{A_{11}}A_{12}+\overline{A_{21}}A_{22} \\ \overline{A_{12}}A_{11}+\overline{A_{22}}A_{21} & |A_{12}|^2+|A_{22}|^2 \end{pmatrix}. $$ What do you get for general $n$?

Answer 2

Hint for the first part: if $B=U^*AU$ for unitary $U$, then $tr(A^*A)=tr(UU^*A^*UU^*A)=...$

Hint for the second part: the $ii$-th entry of the diagonal of $A^*A$ is equal to $\sum_j |A_{ji}|^2$, so calculating the trace yields...