# Prove spectral norm compatible with vector norm

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).

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 $$||A x || _2 = \frac{ ||Ax ||_2 }{||x||_2} ||x || _2 \leq ||A || _2 ||x||_2$$

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.

$\|T\|=max\{\|T(x)\|\mid \|x\|=1\}$, 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.

• Please look at my comment to Jeb's answer - in your argument you are not using the spectral norm, but the operator norm... – peter a g Oct 12 '15 at 15:06
• math.stackexchange.com/questions/586663/… – Victor Rafael Oct 12 '15 at 15:10
• 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. – peter a g Oct 12 '15 at 15:43