# Matrices and norms

1. Dec 17, 2007

### mathboy

I'm trying to understand one step in the following proof to the following problem:

http://img402.imageshack.us/img402/264/82127528zq9.jpg [Broken]

Last edited by a moderator: May 3, 2017
2. Dec 17, 2007

### Kreizhn

I'm not sure if this will help you explicitly, but there's a little identity that might contribute to your solution

We know that

$$\displaystyle\left( \sum_i a_i \right) ^2 \geq 0$$

by non-negativity of a square, with equality holding iff the summand is identically 0. Furthermore, we can expand this to

$$\displaystyle\left( \sum_i a_i \right) ^2 = \sum_i a_i^2 + 2 \sum_{i<j} a_i a_j$$

Thus

$$\sum_i a_i^2 + 2 \sum_{i<j} a_i a_j \geq 0$$

Also, this can probably be generalized but off the top of my head I'm not too sure how, but

$$(x-y)^2 \geq 0$$

$$\Rightarrow x^2+y^2-2xy \geq 0$$

$$\Rightarrow 2xy \leq x^2+y^2$$

3. Dec 17, 2007

### mathboy

Thanks, but I already knew all that. So far I have

[si [sj(A_ij)y_j]^2]^(1/2) <= [si [sjN y_j]^2]^(1/2) = nN [[sj(y_j)]^2]^(1/2)

but leaves me wondering what to do with [sj(y_j)]^2

4. Dec 17, 2007

### Kreizhn

Then the only thing left that I can think of using is the identity above that I gave, namely

$$\displaystyle\left( \sum_i a_i \right) ^2 = \sum_i a_i^2 + 2 \sum_{i<j} a_i a_j$$

Then as long as you can make a non-negativity argument about

$$\sum_{i<j} a_i a_j$$

You'll be good to go

5. Dec 17, 2007

### mathboy

Still can't get it. I'm just trying to understand one step in the following proof to the following problem:

http://img402.imageshack.us/img402/264/82127528zq9.jpg [Broken]

Last edited by a moderator: May 3, 2017
6. Dec 17, 2007

### Kreizhn

In all honesty, I'm wondering if there just isn't a typo. It seems like the exponents should all be nested one parentheses earlier, though I might be missing something

7. Dec 17, 2007

### mathboy

It's an online solution by a professor (whom I don't know personally) to Spivak's "Calculus on Manifolds". If it is a typo, what is the proper way to finish it off? I've checked that there is no mistake before the inequality sign.

I'm thinking that his n should be mn
I think I've corrected the professor's solution, and I think his M is supposed to be N[mn]^(1/2)

Last edited: Dec 17, 2007
8. Dec 17, 2007

### morphism

I think this is correct. He probably made the same mistake I did and treated both summations as if they run up to n, and not that one runs up to n and the other to m.