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If I have an $N\times N$ symmetric matrix $Q$, what is the relationship between $\|Q\|$ and its maximum eigenvalue, where $\|\|$ is the $2$nd norm?

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The general answer to this is this $\|Q\| = \max \{|\lambda_{\max}(Q)|, |\lambda_{\min}(Q)|\}$. To see why this is true, consider $Q = I_{N}$ for the first case and $Q = -I_{N}$ for the second case.

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  • $\begingroup$ Those example don't really do anything to show why the result should hold. They are examples, though. $\endgroup$ Aug 13, 2017 at 21:34
  • $\begingroup$ It would be far clearer to write this as the maximum absolute value of the eigenvalues. $\endgroup$ Apr 8, 2022 at 0:41
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Let $Q=UDU^\top$ be the eigendecomposition of $Q$ with $U$ orthogonal and $D = \text{diag}(\lambda_1,\ldots,\lambda_n)$ where $\lambda_1 \ge \cdots \ge \lambda_n$.

$$\max_{v : \|v\|_2=1} \|Qv\|_2 = \max_{v : \|v\|_2=1} \sqrt{v^\top Q^\top Q v} = \max_{u : \|u\|_2=1} \sqrt{u^\top D^2 u} = \max_{u : \|u\|_2=1} \sqrt{\sum_{i=1}^n \lambda_i^2 u_i^2}$$

By Hölder's inequality and the constraint $\|u\|_2=1$, we have $$\max_{u : \|u_2\|_2=1} \sqrt{\sum_{i=1}^n \lambda_i^2 u_i^2} \le \sqrt{\max_i \lambda_i^2} = \max_i |\lambda_i|.$$ Equality is attained by taking $u=e_1$.

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