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Let $||A||_1=tr((A^* A)^{1/2})$ In my linear algebra book, we have the following relations

For arbitrary unitary matrices U and V let $||UAV^*||_1=||A||_1$, $||A||_1=\sigma_1+...+\sigma_k$ and $||A||_1=||A^*||_1$

Could you explain me why these relations hold?

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Note how the spacing around the addition symbols gets messed up if you use ... instead of one of the predefined commands, \ldots or \dotso in this case. –  joriki Nov 6 '12 at 2:01
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For the first, note that $(UAV^*)^*(UAV) = V A^* U^* U A V^* = VA*A V^*$. Now $(A^* A)^{1/2}$ is by definition the positive semidefinite square root of $A^* A$. If this is $P$, then $VPV^*$ is also positive semidefinite and $(VPV^*)^2 = VP^2V^* = VA^*AV^*$, so $VPV^* = (VA^*AV^*)^{1/2}$. Finally, $\text{tr}(VPV^*) = \text{tr}(PV^*V) = \text{tr}(P)$ since $\text{tr}(CD) = \text{tr}(DC)$ for any matrices $C,D$ such that both $CD$ and $DC$ exist.

$\|A\|_1 = \text{tr}(P)$ is the sum of the eigenvalues of $P$, but the nonzero eigenvalues of $P$ are the singular values of $A$.

$\|A\|_1 = \|A^*\|_1$ because $AA^*$ and $A^*A$ have the same nonzero eigenvalues.

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