# Why is tensor product of linear maps defined as $(S\otimes T)(v\otimes w)=S(v)\otimes T(w)$?

In my understanding, the definition of tensor product of linear maps cannot be directly derived from the definition of tensor product of vector spaces (or modules), since it's not clear what is the domain, range, and the map of the result product. Originally I thought this is just a definition. But then I learned that if we represent $S$ and $T$ by matrices, then the matrix describing the tensor product $S \otimes T$ is the Kronecker product of the two matrices.

However, Kronecker product of matrices can be developed totally without the concept of tensor product of linear maps. For example, we can find basis of the matrix spaces (say $V$ and $W$). Then we can get natural basis for $V \otimes W$. And we can easily get the definition of Kronecker product from these basis.

So my question is: is this coincidence just a coincidence or there is some deep reason behind the definition so that it has to be defined like this. I can't even see why the domain of $S \otimes T$ should be $V\otimes W$, given $S$ and $T$ are linear maps over $V$ and $W$.

(I know that the definition of tensor product of linear maps is very natural, and I can't imagine other definitions of it. I just thought there must be a formal reason that the definition has to be that. )

## 2 Answers

At some level, the tensor product of maps is, like most things in math, a convenient choice of definition. However, it arises naturally, in a precise sense.

The standard definition of a tensor product of two spaces, $V \otimes W$, actually provides more than a vector space constructed from $V$ and $W$. It is a universal construction, meaning that it satisfies a particular property, and is the best choice of a vector space that does so. There is a bilinear map $i: V \times W \rightarrow V \otimes W$ taking $(v,w) \mapsto v \otimes w$. Now for any bilinear map $f: V \times W \rightarrow U$, there exists a unique linear map $\tilde{f}: V\otimes W \rightarrow U$ such that $\tilde{f}\circ i = f$.

Now if $S: V \rightarrow V'$, $T: W \rightarrow W'$, we want to define a new map $X \otimes T: V \otimes W \rightarrow V' \otimes W'$. We choose this domain and range because we want the tensor product of maps to be compatible with the tensor product of spaces (in a precise sense: we want the association of a tensor product of spaces and maps to be a "bifunctor" - see here).

Now to actually define the map, we appeal to the universal property. Let $(S \times T)(v,w) = S(v) \otimes T(w)$. This map is bilinear from $V \times W$ to $V' \otimes W'$, so it induces the map $(S \otimes T)(v,w): V \otimes W \rightarrow V' \otimes W'$.

These concepts become a bit more clear if you're familiar with the language of category theory - natural constructions, universal properties, functors, etc.

• If the $f$ and $g$ are two linear maps, then $f \otimes g$ is bilinear, right? – Fareed AF Jul 21 at 15:08

I'm not entirely sure what your question really means but my understanding of the situation is as follows: the general definition of tensor products and linear maps on them are intended to be basis-free, i.e. independent of particular bases as far as possible. But since one would like to define linear maps on tensor products in some economical way other than by defining them on a basis and then extending them by linearity, defining them on the elementary tensors $v\otimes w$ seems to be the simplest workable alternative. In the case of matrices, bases are pretty much in one's face (or perhaps even unavoidable) whether one wants them or not.

• You are entitled to your opinion but I question whether an opinion which adds little or nothing to the discussion needs to be aired here. And yes, the answer above has not escaped my notice. – InTransit Jul 18 '14 at 14:42