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Hello this indeed a very short question from Algebra that I have no real idea on and figured it is simple but for some reason I cannot seem to find it. I am given $A$ and $B$ complex square matrices for which

$A^*A=B^*B-BB^*$ where * denotes the complex transpose. I am required to show $A=0$ but have no idea how to deduce this the only thing I figured out was that both sides of the equality I am given are Hermitian but cannot proceed. Any help appreciated. Thanks

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  • $\begingroup$ If $B$ is Hermitian then $B=B^*$. $\endgroup$ – zoli Jun 6 '15 at 11:34
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The matrix $A^*A$ is Hermitian non-negative, and it is enough to show that its only eigenvalue is $0$. By assumption, we have $Tr(A^*A)=0$. Recall that the trace is the sum of all eigenvalues. Now, if $\lambda>0$ is an eigenvalue, it follows that $A^*A$ has negative eigenvalues, which is impossible.

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Hint : $$Tr(A^*A)=Tr(B^*B-BB^*)$$

$$|a_{11}|^2+|a_{21}|^2+...+|a_{n1}|^2+|a_{12}|^2+...+|a_{n2}|^2+...+|a_{1n}|^2+...+|a_{nn}|^2$$$$=Tr(B^*B)-Tr(BB^*)=0$$

Finally you will get all the entries of the matrix is $0$ hence A is null matrix. I'm not so expert in Latex so I haven't typed whole thing. You can manage from this solution.

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