matrix norms

**BBrown** · December 12th 2009, 04:30 PM

I am asked to show that

. I know that

and that

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!!!

**Opalg** · December 13th 2009, 12:45 PM

Originally Posted by BBrown

I am asked to show that

. I know that

and that

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.

**AidenMatthew** · December 15th 2009, 11:44 AM

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...

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