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Math Help - matrix norms

  1. #1
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    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!!!
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  2. #2
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    Quote Originally Posted by BBrown View Post
    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.
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  3. #3
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    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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