Let $V, W$ be vector spaces and $A\in\operatorname{End}(V), B\in\operatorname{End}(W)$ endomorphisms. We can define a linear map $A\tilde\otimes B \in \operatorname{End}(V\otimes W)$ by

$$(A\tilde\otimes B)(v\otimes w) := Av \otimes Bw,$$

and another linear map $$\Gamma: \operatorname{End}(V) \otimes \operatorname{End}(W) \rightarrow \operatorname{End}(V\otimes W)$$


$$\Gamma(A\otimes B) := A\tilde\otimes B.$$

(Note: Both linear maps are defined by their values on the elementary tensors $x\otimes y$; it was part of a previous exercise to show that both maps are uniquely defined that way)

Problem: Show that if either $V$ or $W$ has a finite dimension, $\Gamma$ is an isomorphism.

My attempt:

Dimensional considerations seem to be of no use since one of $V, W$ may still be infinite-dimensional, so I tried the "straightforward" way of showing that $\Gamma$ is both injective and surjective:

Injectivity is rather easy and does not require $V$ or $W$ to be finite-dimensional (granted I did not make any mistakes; anyway, I'll skip that part).

Surjectivity, on the other hand, I don't know how to handle. It seemed trivial at first as all I need to show is that the set

$$M:=\left\{ A \tilde\otimes B: A\in\operatorname{End}(V), B\in\operatorname{End}(W) \right\}$$

spans the vector space $\operatorname{End}(V\otimes W)$, but frankly, I just have no idea how to decompose an element of this vector space into a linear combination of elements of $M$.

I'd be delighted about any hint to the right direction. Thanks!


Hint: Suppose that $V$ is finite dimensional and that $A_1,\dots,A_n$ is a basis of End$(V)$. Note that for any $B$ in End$(W)$, we have $$ \left(\sum a_i A_i\right) \otimes B= \sum a_i (A_i \otimes B) $$ Now, we need to show that the image of $\Gamma$ spans the codomain. In particular, it suffices to show that a spanning set of the codomain is in the image.

  • $\begingroup$ Thanks a lot for your answer. Unfortunately, I have come that far already. The relation you gave suggests that $\left\{ A_i \tilde\otimes B : 1 \leq i \leq n, B \in \right\}$ spans the image $\operatorname{Im}(\Gamma)$ - is that what I was supposed to conclude? Please bear with me, but I still don't see how this makes $\Gamma$ an isomorphism.. $\endgroup$ – Andy Brandi Sep 8 '15 at 13:07
  • $\begingroup$ Yes, we are supposed to conclude exactly that. I'll expand a bit. $\endgroup$ – Omnomnomnom Sep 8 '15 at 13:17
  • $\begingroup$ I'll clarify more later if I have time. I can't do much for now. $\endgroup$ – Omnomnomnom Sep 8 '15 at 13:23

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