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Let $A\in L(H)$ some Hermitian matrix, where $H$ is some finite dimensional Hilbertspace.

I want to show $$\left\|A\right\|_{tr} = \max_{U\in U(H)}|\text{tr}(UA)| \ \ \ (*)$$ where U is unitary, and $\left\|\cdot \right\|_{tr}$ is the trace-norm which is given by $\left\|A\right\|_{tr}= \text{tr}|A|$ where $|A|:=\sqrt{A^*A}$ where $|A|$ is the positive-semidefinite root of $A^*A$.

It appears that for hermitian A we have $\left\|A\right\|_{tr} = \sum_{i} |\lambda_i|$ (where $\lambda_i$ an eigenvalue of $A$).

To show (*) i can show '$\geq'$. Let $A = \sum_i \lambda_i \left|i\right\rangle \left\langle i \right|$ be its spectral decomposition, then $$|\text{tr}(UA)| = |\sum_i\lambda_i \text{tr}(U\left|i\right\rangle \left\langle i \right|)| =|\sum_i\lambda_i \left\langle i\right| U \left|i\right\rangle | \leq \sum_{i} |\lambda_i| $$

Is this right? By the same reasoning it seems to me impossible that '$\leq$' holds. A unitary $U$ maps an orthonormal basis $\{\left|i\right\rangle\}_{i\in I}$ into itself. Thus $\left\langle i\right| U \left|i\right\rangle$ seems to me be either 1 or 0. Where do i go wrong here? And how do I approach '$\leq$'.

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$\langle i|U|i\rangle$ are the diagonal entries of $U$ in your orthogonal basis. They could certainly be different from $0$ and $1$, but you are sure that they are less than $1$ in absolute value: $$ |\langle i|U|i\rangle|\leq\|U\|\,\|i\rangle\|^2=\|U\|=1. $$

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  • $\begingroup$ Thanks, i can solve it now. $\endgroup$ – DinkyDoe Apr 23 '14 at 15:48

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