I don't know how to show that $A\in M_2$ is unitarily similar to $\begin{bmatrix}\lambda_1&x\\0&\lambda_2\end{bmatrix}$, where $\lambda_1$ and $\lambda_2$ is its eigenvalues, $x^2=\text{tr}AA^*-|\lambda_1|^2-|\lambda_2|^2$.
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Hint:
Note that if $A=U MU^\ast$ (where $M$ is the matrix you wrote) then
$$AA^\ast=(UMU^\ast)(UMU^\ast)^\ast=UMU^\ast U M^\ast U^\ast=UMM^\ast U^\ast$$
So,
$$\text{tr}(AA^\ast)=\text{tr}(UMM^\ast U^\ast)=\text{tr}(MM^\ast)$$
Now, what is $\text{tr}(MM^\ast)$?
Alex Youcis
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More hints here are helpful, I do know that if $A$ is unitarily similar to $M$, then $$\text{tr}AA^=\text{tr}MM^.$$But this doesn't guarantee that $A$ is unitarily similar to $M$ – user95640 Oct 03 '13 at 22:08
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@user95640 No more hints are helpful. Compute $\text{tr}(MM^\ast)$ and then stare at it.. – Alex Youcis Oct 03 '13 at 22:08
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$\text{tr}MM^=|\lambda_1|^2+|\lambda_2|^2+x^2$, since $x^2=\text{tr}AA^-|\lambda_1|^2-|\lambda_2|^2$, we have $\text{tr}MM^=\text{tr}AA^$, right? And then.... – user95640 Oct 03 '13 at 22:16
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I'm wondering whether I should figure out what $U$ is. – user95640 Oct 03 '13 at 22:18
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Oh, I see, Specht's theorem right? $\text{tr}W(A,A^)=\text{tr}W(B,B^)$ – user95640 Oct 03 '13 at 22:22
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@user95640 I must be misunderstanding your question. I thought the problem was to show that if $A$ is unitarily similar to $\begin{pmatrix}\lambda_1 & x\ 0 & \lambda_2\end{pmatrix}$ then the Frobenius norm of $A$ is $|x|^2+|\lambda_1|^2+|\lambda_2|^2$. Was it not? – Alex Youcis Oct 04 '13 at 08:02