# The Unapologetic Mathematician

## 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