Question

Determinant of an infinite dimensional matrix...!

Answer 1

The only one I know is that the matrix represented by: \[\Large A(x,y)=0^{|x-y|}\] is essentially the identity matrix, which I'm saying 0^0=1. So this has to have a determinant of 1. What can we do from here?

Answer 2

\[|A| = \prod_{i=1}^{\infty }1 = 1\]

Answer 3

?

Answer 4

Yes, that's exactly the problem I'm having right now. But this is sort of like in Hilbert space where we can define the dot product, so I'm imagining how do we normalize the determinant so that it isn't always 0, 1, or infinity.

Suppose we want the determinant of some scalar multiplied by this identity matrix, it should be s^n which will either go to 0 or infinity. So something has to normalize it. Or maybe it's just not going to happen.

Answer 5

let me tag @Zarkon @SithsAndGiggles

Answer 6

I suppose I want to look at other linear algebra structures that line up with this as being a dot product\[\int\limits_{-\infty}^\infty f(x)g(x) dx\]

Answer 7

In trying to find out alternate ways to the determinant I discovered this rule: \[\Large |\bar u \ \bar v^T|=0\] If the matrix can be written as two vectors multiplied together, then its determinant is zero.

Answer 8

...Which is no surprise considering this will always make a matrix of linearly dependent vectors. #_#

Answer 9

why don't u define a space instead of matrix :O
don't u think matrices is week way to deal with higher mathematics ?

Answer 10

@ikram002p You're right.

Answer 11

i'll work on this with u ltr , when exams done :)