Hilbert-Schmidt Norm: Calculation & Solution

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If the norm of blah is zero, then blah is zero. Is blah zero in this case?
 
morphism said:
If the norm of blah is zero, then blah is zero. Is blah zero in this case?

You're right something is wrong.

But is the integral set up with the correct boundaries?
 
Ok presuming the boundaries are ok I end up with:

||A||_{HS} = \frac{2 (b-a)^n}{((n-1)!)^2 (2n-1)(2n)}

Is this correct?
 
Did you remember to take the square root?
 
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You might want to include some characteristic function like \chi_{\{s\leq t\}} in your kernel function.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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