1. ## matrix norms

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!!!

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

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

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 $\displaystyle y_i$ to be the complex number of absolute value 1 such that $\displaystyle \overline{x_i}y_i$ is real and positive.