Matrix norms 1,2 equivalence

  • Thread starter garrus
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Homework Statement


Show that [itex]||A||_1 \le \sqrt{n} ||A||_2 , ||A||_2 \le \sqrt{n} ||A||_1[/itex] , where
[tex]||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|
[/tex]
with [itex]A,B\in \mathbb{R}^{n,n}, i,j\in[1...n] , \lambda_A[/itex]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 [itex]||x||_1 \le \sqrt{n} ||x||_2 , x\in \mathbb{R}^n[/itex] via the Cauchy Schwartz inequality.But since the max operator is not linear, isn't it a mistake to write
[tex]

||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}}


[/tex] ?
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| ?
 
Last edited:

Answers and Replies

  • #2
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Noone?

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

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

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