# $L^p$ norm of diagonal is $\leq$ Schatten $L^p$ norm

Let $A = (a_{ij})$ be an $n\times n$ symmetric matrix. Its Schatten norm is defined by $\|A\|_{S^p}^p = \sum_{j=1}^n |\lambda_j|^p$, where $\lambda_j$ are the eigenvalutes of $A$. I am trying to prove that $\sum_{j=1}^n |a_{jj}|^p \leq \|A\|_{S^p}^p$.

I know we can find an orthogonal matrix $Q$ with $Q^T A Q = diag(\lambda_j)$. So, we need to show that the $L^p$ norm of the diagonal of $Q^T A Q$ is at least the $L^p$ norm of the diagonal of $A$. However, I cannot find a way to do this. Any suggestions?

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I fixed the question. –  user15464 Dec 29 '12 at 18:35
If $A$ is positive then this is true for any non-decreasing function in place of $\phi(t)=t^p$. It's a consequence of the Schur-Horn theorem. In general, I think you should interpret $|\lambda_j|$ as singular values of $A$, and try the min-max principle. –  user53153 Dec 29 '12 at 19:59