Letxbe am- vector,Abe an m x n matrix, show that

(1) infinity normx<= 2nd normx<=square root (m)*infinity normx

(2) [1/square root(n)] * infinitynormA <=2nd norm A

<=

square root (m)*infinity norm A

Printable View

- Sep 18th 2008, 12:57 AMtinngVector Norm and Matrix Norm
Let

*x*be a*m*- vector,*A*be an m x n matrix, show that

(1) infinity norm*x*<= 2nd norm*x*<=*square root (m)******infinity norm*x*

(2) [1/square root(n)] * infinity*norm*A <=2nd norm A

<=

*square root (m)******infinity norm A - Sep 18th 2008, 10:49 AMMoo
Hello,

Get back to the definitions of the norms :

$\displaystyle \left\|x \right\|_p=\left(\sum_{i=1}^m |x_i|^p \right)^{1/p}$

$\displaystyle \left\|x \right\|_\infty=\max_{1 \le i \le m} |x_i|$

You'll have to use two inequalities :

- Triangle inequality : $\displaystyle |x+y| \le |x|+|y|$

- Cauchy-Schwarz inequality : $\displaystyle \left(\sum_{i=1}^m x_iy_i\right)^2 \le \left(\sum_{i=1}^m x_i^2 \right)\left(\sum_{i=1}^m y_i^2 \right)$