The Unapologetic Mathematician

Mathematics for the interested outsider

Matrices III

Given two finite-dimensional vector spaces U and V, with bases \left\{e_i\right\} and \left\{f_j\right\} respectively, we know how to build a tensor product: use the basis \left\{e_i\otimes f_j\right\}.

But an important thing about the tensor product is that it’s a functor. That is, if we have linear transformations S:U\rightarrow U' and T:V\rightarrow V', then we get a linear transformation S\otimes T:U\otimes V\rightarrow U'\otimes V'. So what does this operation look like in terms of matrices?

First we have to remember exactly how we get the tensor product S\otimes T. Clearly we can consider the function S\times T:U\times V\rightarrow U'\times V'. Then we can compose with the bilinear function U'\times V'\rightarrow U'\otimes V' to get a bilinear function from U\times V to U'\otimes V'. By the universal property, this must factor uniquely through a linear function U\otimes V\rightarrow U'\otimes V'. It is this map we call S\otimes T.

We have to pick bases \left\{e_k'\right\} of U' and \left\{f_l'\right\} of V'. This gives us a matrix coefficients s_i^k for S and t_j^l for T. To calculate the matrix for S\otimes T we have to evaluate it on the basis elements e_i\otimes f_j of U\otimes V. By definition we find:

\left[S\otimes T\right](e_i\otimes f_j)=S(e_i)\otimes T(f_j)=\left(s_i^ke_k'\right)\otimes\left(t_j^lf_l'\right)=s_i^kt_j^le_k'\otimes f_l'

that is, the matrix coefficient between the index pair (i,j) and the index pair (k,l) is s_i^kt_j^l.

It’s not often taught anymore, but there is a name for this operation: the Kronecker product. If we write the matrices (as opposed to just their coefficients) \left(s_i^k\right) and \left(t_j^l\right), then we write the Kronecker product \left(s_i^k\right)\boxtimes\left(t_j^l\right)=\left(s_i^kt_j^l\right).

May 26, 2008 Posted by | Algebra, Linear Algebra | 6 Comments