Matrices III
Given two finite-dimensional vector spaces 
and 
, with bases 
and 
respectively, we know how to build a tensor product: use the basis 
.
But an important thing about the tensor product is that it’s a functor. That is, if we have linear transformations 
and 
, then we get a linear transformation 
. So what does this operation look like in terms of matrices?
First we have to remember exactly how we get the tensor product 
. Clearly we can consider the function 
. Then we can compose with the bilinear function 
to get a bilinear function from 
to 
. By the universal property, this must factor uniquely through a linear function 
. It is this map we call .
We have to pick bases 
of 
and 
of 
. This gives us a matrix coefficients 
for 
and 
for 
. To calculate the matrix for we have to evaluate it on the basis elements 
of 
. By definition we find:
=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'](http://s0.wp.com/latex.php?latex=%5Cleft%5BS%5Cotimes+T%5Cright%5D%28e_i%5Cotimes+f_j%29%3DS%28e_i%29%5Cotimes+T%28f_j%29%3D%5Cleft%28s_i%5Eke_k%27%5Cright%29%5Cotimes%5Cleft%28t_j%5Elf_l%27%5Cright%29%3Ds_i%5Ekt_j%5Ele_k%27%5Cotimes+f_l%27&bg=e6e6e6&fg=333333&s=0 "\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 
and the index pair 
is 
.
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) 
and 
, then we write the Kronecker product 
.
Like this:
Like Loading...
May 26, 2008 -
Posted by John Armstrong |
Algebra, Linear Algebra
[...] Like we saw with the tensor product of vector spaces, the dual space construction turns out to be a functor. In [...]
Pingback by Matrices IV « The Unapologetic Mathematician | May 28, 2008 |
[...] is one slightly touchy thing we need to be careful about: Kronecker products. When the upper index is a pair with and we have to pick an order on the set of such pairs. [...]
Pingback by Matrix notation « The Unapologetic Mathematician | May 30, 2008 |
[...] the monoidal product on objects by multiplication — — and on morphisms by using the Kronecker product. That is, if we have an matrix and an matrix , then we get the Kronecker [...]
Pingback by The Category of Matrices II « The Unapologetic Mathematician | June 3, 2008 |
[...] wait, there’s more! The functor is linear over , so it’s a functor enriched over . The Kronecker product of matrices corresponds to the monoidal product of linear transformations, so the functor is [...]
Pingback by The Category of Matrices III « The Unapologetic Mathematician | June 23, 2008 |
[...] can recognize this as a Kronecker product of two [...]
Pingback by More Commutant Algebras « The Unapologetic Mathematician | October 5, 2010 |
[...] want. And we know that when expressed in matrix form, the tensor product of linear maps becomes the Kronecker product of matrices. We write the character of as , that of as , and that of their tensor product as , [...]
Pingback by Outer Tensor Products « The Unapologetic Mathematician | November 4, 2010 |