norm equivalence

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?

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: $\|A+B\|_p \le \|A\|_p + \|B\|_p$.
• 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. – Hans Mar 5 '13 at 4:18