Is the largest diagonal diagonal entry of a symmetric, positive semidefinite matrix a lower bound on the largest eigenvalue?

Well, the question is as simple as that: Given a symmetric and positive semidefinite matrix. Is it true that the largest diagonal entry is always smaller or equal to the largest eigenvalue?

I was just getting a little frustrated while proving this for a specific kind of matrix (that happens to be symmetric and positve definite). So I'm wondering whether I can just skip the algebra and show that this is true in general. But a quick web search didn't reveal any promising references.

Can someone clear up for me whether this is true or not?

It's true. It's evident from the classic Courant-Fischer minimax principle, but you may also prove it with spectral decomposition. You only need to show that the largest eigenvalue of a positive semidefinite matrix $A$ is given by $\max_{\|u\|=1} u^\top Au$. Once this is proved, the assertion follows because every diagonal entry of $A$ is of the form $e_i^\top Ae_i$, where $\{e_1,\ldots,e_n\}$ is the standard basis of $\mathbb R^n$.
• Thank you! If I may ask one further question: I can see easily why $\max_{\|u\|=1} u^\top Au \leq \lambda_{\max}$ which is already sufficient to answer my question, but why do we have equality? – Amarus Nov 25 '16 at 11:07
• @Amarus If you take $u$ as a unit eigenvector corresponding to $\lambda_\max$, then ... – user1551 Nov 25 '16 at 11:19