# Thread: Normalizing a 4x4 matrix with unknown functions as elements

1. ## Normalizing a 4x4 matrix with unknown functions as elements

Hi All!

This is my first time putting up my own thread on MF. I can usually find what I'm looking for, but this time: no go.

As the title says, I'm trying to find the normalization constant of this 4x4 matrix (g and f are functions):
(1-g^2) 0 0 0
0 (1+f^2) (-g^2-f^2) 0
0 (-g^2-f^2) (1+f^2) 0
0 0 0 (1-g^2)

It's a matrix that's in a research paper which gives the normalization constant (they use the word 'factor'. I'm not sure if that means something diff.) as: N=4-2g^2+2f^2.

1]I've been looking up online and found that N can be found with: N=\sqrt{\sum{X^2}} where X represents the elements of the matrix.

2]I also found somewhere which said that I need to find the determinant.

I'm not sure who's right, but I'm not getting what's in the paper.
For method [1] I'm getting as far as: N^2 = 4(1+f^4+f^2g^2+f^2) and got stuck trying to find the square root (it's been a while since I've done multinomial theorem). So I backtracked to see if their N^2 matches my N^2. But their N^2=16+4g^4+4f^4+16g^2-8g^2f^2+16f^2.
and method [2] is giving me something so long, with so many variables of (g^2), (f^2), (g^4), (f^4),(g^2f^4) (and it keeps going for about 3tysomething variables) that I've given up.

So I'm wrong all over the place.

Can someone help me out?

booklist05 confused

2. I don't know what is meant by the normalisation constant of a matrix. But it looks from your suggested definitions 1] and 2] as though it is meant to be some function of the eigenvalues of a (positive?) matrix. The definition 1] gives the root mean square sum of the eigenvalues. The suggestion in 2] is to find the determinant (product of the eigenvalues).

It looks to me as though the formula for N given here ( $N=4-2g^2+2f^2$) is simply that N is the sum of the eigenvalues. This is equal to the trace of the matrix, which is the sum of its diagonal elements. That is easily seen to be $4-2g^2+2f^2$. Or is that too simple-minded a suggestion?

3. Omg! can it really be that simple? Later on in the paper, they take each of the elements and divide them each by N. This is why I think they call it "normalization constant".

I'm still in shock. so simple? Well since it works, that's what I'm going with. No one else I've asked has been able to give me an answer.

Thanks!