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 '$\leq$'.


$\langle i|U|i\rangle$ 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: $$ |\langle i|U|i\rangle|\leq\|U\|\,\|i\rangle\|^2=\|U\|=1. $$

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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.