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Given a $n \times n$ matrix $M$ such that $\|M\|_2 = c$ (where $\|\|_2$ denotes spectral norm, or operator norm), is it true that for all $i = 1...n$ it holds that $$\sqrt{(\sum_{k=0}^n M(i,k)^2)} \leq c$$ Or, does the norm (vector $l_2$ norm) of the rows of $M$ is bounded by its spectral norm? I believe so, but still not sure, will be glad for intuition explanation (or any hint).

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    $\begingroup$ $\|Me_i\| \le \|M\| \|e_i\|$ where $e_i = (0,\dotsc,0,1,0,\dotsc,0)^T$ $\endgroup$
    – user251257
    Mar 28 '16 at 19:30
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The same with the columns amounts to $\lVert Ae_i\lVert_2\leq \lVert A\lVert_2 \lVert e_i\lVert_2$. But this simply follows from the fact that the spectral norm is defined by

$$\lVert A\lVert_2=\sup_{u\neq0}\frac{\lVert Au\lVert_2}{\lVert u\lVert_2}$$

For the rows, simply use $A^T$ instead.

See

And also have a look at this MSE question for a proof that $\lVert A \lVert_2$ is indeed the spectral radius: Norm of a symmetric matrix equals spectral radius

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