# Kronecker product of infinite dimensional matrices

1. Aug 23, 2012

### yphink

Hi there,
I was recently working with Kronecker product of matrices, and a question came up that I'm not sure how to answer. Is the matrix that represents a Kronecker product of two infinite dimensional matrices well defined? If yes, are some of the properties of the Kronecker product listed in http://en.wikipedia.org/wiki/Kronecker_product satisfied for the infinite dimensional matrices:
1. $A \otimes (B + C) = A \otimes B + A \otimes C$
2. $k (A \otimes B) = (kA \otimes B)$
3. $(A \otimes B)(C \otimes D) = AC \otimes BD$

On one hand, it seems tricky to define such a product matrix because it is supposed to look like a block matrix, but how can it exist if each block is going to be of an infinite size?
On the other hand, looking on the literature about the tensor product of linear operators, it seems possible to define a product linear operator, even if the vector space that these operators work on are infinite dimensional. In this case, if I'm not mistaken, infinite dimensional matrices are instances of these linear operations.

Any clarifications are much appreciated!

Yuri

2. Aug 23, 2012

### chiro

Hey yphink and welcome to the forums.

Before worrying about Kronecker products, you should take a look at operators on Hilbert-Spaces and see what the issues are for multiplication. As a starting point you will want to look at what is known as Banach-Space theory which is really the starting point behind the theory of analyzing these infinite-dimensional operators.

Because of the nature of infinity being involved there are complications in the same way that you get issues when dealing with infinite series and trying to figure out whether they converge or not.

3. Aug 24, 2012

### yphink

Thank you chiro for your reply. What mainly bothers me though is that, as I said, the tensor product of two linear maps exist even if these are maps in infinite dimensional spaces. Now, the Kronecker product is a representation of a tensor product wrt the standard basis (quoting wikipedia). So shouldn't it be the case that such a Kronecker product (matrix) exists? Am I missing something here?

Thank you

4. Aug 24, 2012

### chiro

I see what you are saying, but the key thing about an operator making "sense" are the properties of convergence and this kind of thing is what is looked at in the fields mentioned above.

Also in that wiki, it says that (A X B)(C X D) = AC X BD (X is tensor product) or something similar and AC is a multiplication, then it means that you need convergence criteria for the operator AC (and BD) to make sense.

In terms of this making sense, you would probably have to assess if the operators A,B,C,D make "sense" and whether that implies that the tensor product produces an operator that also "makes sense".

I'm guessing you will need to resort to norms and spectra to do this and this is exactly what these operator algebras as they are called look at.

5. Aug 24, 2012

### yphink

OK, I guess I should give more details.
Suppose we are given an infinite dimensional vector space V with a basis satisfying the following property: all the elements in V can be represented by a finite linear combinations of the elements of the basis (I think this condition resolves convergence problems that you have raised). Now I should be able to talk about linear maps in this space, and their matrix representation, e.g.
$A: V → V, B: V → V$,
where $AB$ or $BA$ are well defined too. This is what I understand from reading Lectures in Abstract Algebra, Vol. 2 by Nathan Jacobson (http://reslib.com/book/Lectures_in_Abstract_Algebra__Vol__2#1), chapter 9 on infinite dimensional vector spaces.
From here, I want to talk about the matrix that represents the kronecker product $A \otimes B$, but I'm not sure if it is well defined.
The interesting thing is, that later in chapter 9 of this book (page 256 middle paragraph) the author says that we can talk about Kronecker product of arbitrary vector spaces. However, at that point I could not follow the details due to insufficient background, so I don't know if it indeed answers my question.

Thank you again for your help.