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Can someone please show me how to prove $||Ax||_2 \leq ||A||_2 ||x||_2$, where $||A||_2$ is the spectral norm and $ A \in \mathbb{R^{n \times n}} $ and $x \in \mathbb{R^n}$.

So far I tried to write the statement out in coordinates and then simplify, but now I'm stuck (I don't know what to do with the max eigenvalue).

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Isn't this obvious? By def'n of the spectral norm $$ ||A || _2 = \max_{||x||_2\neq 0} \frac{ ||Ax ||_2 }{ ||x||_2 } $$ Assume $||x||_2 \neq 0$, thus $$ ||A x || _2 = \frac{ ||Ax ||_2 }{||x||_2} ||x || _2 \leq ||A || _2 ||x||_2 $$

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The semi-positive definite operator $X = \bar A^t A$ is diagonalizable w.r.t. an o.n. basis $u_i$, with corresponding real (non-negative) eigen-values $\lambda_i$. Suppose $\lambda$ is the maximum eigen value of $X$. If $x = \sum \alpha_i u_i$, then $$ < Ax, Ax > = < x, X x> = \sum |\alpha_i|^2\lambda_i \le \lambda \sum |\alpha_i|^2 = \lambda <x,x>.$$

Note - The middle and last equalities hold because the $u_i$ are o.n.

Also - I hadn't noticed that the $A$ in the question was real. So we can take $X = A^t A$, and the argument is otherwise identical, as $X$ is diagonalizable over $\mathbb R$ w.r.t an o.n basis.

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$\|T\|=max\{\|T(x)\|\mid \|x\|=1\}$, for each x we know that $x/\|x\|$ is unitary, using the linearity of T we have:

$\|T(x/\|x\|)\|\le\|T\|$ =>$1/\|x\|\|T(x)\|\le\|T\|$ which results in the desirable property.

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  • $\begingroup$ Please look at my comment to Jeb's answer - in your argument you are not using the spectral norm, but the operator norm... $\endgroup$ – peter a g Oct 12 '15 at 15:06
  • $\begingroup$ math.stackexchange.com/questions/586663/… $\endgroup$ – Victor Rafael Oct 12 '15 at 15:10
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    $\begingroup$ You're right! And see my updated apologetic comment to Jeb.... Still there is an argument - which your link makes - that the two definitions (operator style - eigen-value style) are the same. $\endgroup$ – peter a g Oct 12 '15 at 15:43

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