Can Matrix Norms be Used to Bound the Eigenvalues of a Matrix?

[garrus]

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}| \\ <br /> ||A||_2 = (p(A^TA))^\frac{1}{2} \\<br /> p(B) = \max|\lambda_B|<br />
with A,B\in \mathbb{R}^{n,n}, i,j\in[1...n] , \lambda_Athe 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
<br /> <br /> ||A||_1 = \max_{1\le j\le n} \sum_{i=1}^{n}|a_{ij}| =<br /> \max_{1\le j\le n}\sum_{i=1}^{n}|a_{ij}| * 1 \le<br /> \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| ?

 

[garrus]

Noone?

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

<br /> \|Ax\|\ge \frac{\|x\|}{10} \iff \|A\| \|x\| \ge\|Ax\|\ge \frac{\|x\|}{10} \Rightarrow<br /> \|A\| \ge \frac{1}{10} (*)<br />
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}\| &gt; 10
<br /> \|A^{-1}\| &gt; 10 \iff ||A|| \|A^{-1}\| &gt; 10 \|A\| \\ but\\<br /> 1 = \|A A^{-1}\| \le \|A\| \|A^{-1}\| \\<br /> (*) \Rightarrow 10 \|A\| \ge 1<br />
, which is a contradiction, so \|A^{-1}\| &gt; 10 is false.Thus \|A^{-1}\| \le 10