# matrix norms

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• Dec 12th 2009, 04:30 PM
BBrown
matrix norms
I am asked to show that http://s3.amazonaws.com/answer-board...3950009460.gif. I know that http://s3.amazonaws.com/answer-board...5825005037.gif and that http://s3.amazonaws.com/answer-board...1450006019.gif but from there, I get stuck because if I'm looking at y = 1, wouldn't the norm always be the same... ? I'm not sure what I'm missing, any help would be greatly appreciated!!!
• Dec 13th 2009, 12:45 PM
Opalg
Quote:

Originally Posted by BBrown
I am asked to show that http://s3.amazonaws.com/answer-board...3950009460.gif. I know that http://s3.amazonaws.com/answer-board...5825005037.gif and that http://s3.amazonaws.com/answer-board...1450006019.gif but from there, I get stuck because if I'm looking at y = 1, wouldn't the norm always be the same... ? I'm not sure what I'm missing, any help would be greatly appreciated!!!

Are the scalars here supposed to be real or complex? If they are real then the argument goes like this.

If $\|y\|_\infty=1$ then $|y_i|\leqslant1$ for all i. Therefore $|x^Hy| = |x_1y_1+\ldots+x_ny_n|\leqslant |x_1||y_1|+\ldots+|x_n||y_n|\leqslant |x_1|+\ldots+|x_n| = \|x\|_1$. Thus $\textstyle\max_{\|y\|_\infty=1}|x^Hy|\leqslant \|x\|_1$.

For the reverse inequality, define y by $y_i = \text{sign}(x_i)$ for each i (in other words +1 if $x_i$ is positive, –1 if it is negative, and 0 if it is 0). Then $\|y\|_\infty=1$ and $|x^Hy| = |x_1|+\ldots+|x_n| = \|x\|_1$.

If the scalars are complex then the argument is similar, with complex conjugates thrown in where necessary. For the second part, you have to define $y_i$ to be the complex number of absolute value 1 such that $\overline{x_i}y_i$ is real and positive.
• Dec 15th 2009, 11:44 AM
AidenMatthew
reply
I don't understand how you're getting your second inequality (should it be an inequality?)

I don't see the reasoning behind your conclusion to the second part...