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Exercise 7. of Terence Tao's blog on random matrix, specifically on eigenvalues of Hermitian matrices http://terrytao.wordpress.com/2010/01/12/254a-notes-3a-eigenvalues-and-sums-of-hermitian-matrices/#more-3341 asks to

Show that the $p$-Schatten norms are indeed a norm on the space of Hermitian matrices for every $1\le p\le\infty$.

As I understand it, $p$-norm on the space of matrices of a matrix $A$ is $(\sum_{i,j}A_{ij}^p)^\frac{1}{p}$. To prove the proposition, I multiply a given Hermitian matrix $A$ with a test Hermitian matrix $B$ and take its trace, $$\text{tr}(BA) = \text{tr}(U^\dagger BU\Lambda) = \text{tr}(C\Lambda)$$, where $A=U\Lambda U^\dagger$ is the diagonalization of $A$ with $\Lambda$ the diagonal eigenvalue matrix, and $C$ is the diagonal part of $U^\dagger BU$. I attempt to make the connection between arbitrary test Hermitian matrix $B$ with $\sum_{i,j}|B_{ij}|^q=1$ and arbitrary test diagonal matrix $C$ with $\sum_i|C_{ii}|^q=1$. But I can not proceed.

Perhaps this is not the right route. Can anyone help?

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I'm not sure what you're trying to do, but in any case you have the definition of the Schatten norms wrong (if it was just $\left(\sum_{i,j} |A_{ij}|^p\right)^{1/p}$, that would be a norm because it's just treating the matrices as vectors in $\ell_p$). The true definition of the Schatten $p$-norm is

$$\|A\|_p = \left( \sum_{j} s_j(A)^p \right)^{1/p} $$

where $s_j(A)$ are the singular values of $A$. Equivalently, $$ \|A\|_p = \left(\text{trace}(|A|^p)\right)^{1/p} $$ where $|A| = |A^* A|^{1/2}$, and the $p$'th power is taken using functional calculus. In the case of a Hermitian matrix, you can just say

$$\|A\|_p = \left( \sum_j |e_j(A)|^p \right)^{1/p} $$ where $e_j(A)$ are the eigenvalues of $A$.

To show this is a norm, the only difficulty is to show the triangle inequality: $\|A+B\|_p \le \|A\|_p + \|B\|_p$.

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  • $\begingroup$ I misread the problem. I thought I was asked to show the equality of the $p$-Schatten norm with "$p$-Frobenius" norm of matrices. The triangular inequality can be deduced by the Lidskii inequality plus Hölder’s inequality, following exercise 6 of the above blog link. Thank you, Robert. $\endgroup$ – Hans Mar 5 '13 at 4:18

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