# Matrix norms 1,2 equivalence

## Homework Statement

Show that $||A||_1 \le \sqrt{n} ||A||_2 , ||A||_2 \le \sqrt{n} ||A||_1$ , where
$$||A||_1 = \max_{1\le j\le n}\sum_{i=1}^n |a_{ij}| \\ ||A||_2 = (p(A^TA))^\frac{1}{2} \\ p(B) = \max|\lambda_B|$$
with $A,B\in \mathbb{R}^{n,n}, i,j\in[1...n] , \lambda_A$the eigenvalues of matrix A

## Homework Equations

wiki page on matrix norms

## The Attempt at a Solution

I figured i could go the same way in proving that $||x||_1 \le \sqrt{n} ||x||_2 , x\in \mathbb{R}^n$ via the Cauchy Schwartz inequality.But since the max operator is not linear, isn't it a mistake to write
$$||A||_1 = \max_{1\le j\le n} \sum_{i=1}^{n}|a_{ij}| = \max_{1\le j\le n}\sum_{i=1}^{n}|a_{ij}| * 1 \le \max_{1\le j\le n} (\sum_{i=1}^{n}|a_{ij}|^2 * \sum_{i=1}^{n}1^2)^{\frac{1}{2}}$$ ?
Any hints?

edit:
Slightly off topic question: In triangle inequality, it holds that
|a - b| >= |a| - |b| . can we also bound it from below, by |a-b| = |a +(-b)| <=|a|+|b| ?

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Noone?

One another norm problem, i'm given.If you could verify / correct:
$$A\in\mathbb{R}^{n,n} , x\in \mathbb{R}^n. \|Ax\|\ge \frac{\|x\|}{10}$$
and ask to show that $\|A^{-1}\| \le 10$
where $||\cdot||$ a norm and the corresponding matrix norm derived by it.

$$\|Ax\|\ge \frac{\|x\|}{10} \iff \|A\| \|x\| \ge\|Ax\|\ge \frac{\|x\|}{10} \Rightarrow \|A\| \ge \frac{1}{10} (*)$$
Only way i managed was to assert a value and lead to contradiction,but i suspect there must be a more elegant way.Let $\|A^{-1}\| > 10$
$$\|A^{-1}\| > 10 \iff ||A|| \|A^{-1}\| > 10 \|A\| \\ but\\ 1 = \|A A^{-1}\| \le \|A\| \|A^{-1}\| \\ (*) \Rightarrow 10 \|A\| \ge 1$$
, which is a contradiction, so $\|A^{-1}\| > 10$ is false.Thus $\|A^{-1}\| \le 10$

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