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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?

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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 $\mathbb R^n$.

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  • $\begingroup$ 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? $\endgroup$ Commented Nov 25, 2016 at 11:07
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    $\begingroup$ @Amarus If you take $u$ as a unit eigenvector corresponding to $\lambda_\max$, then ... $\endgroup$ Commented Nov 25, 2016 at 11:19

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