Axler 3rd Theorem 3.60 Prove that L(V,W) is isomorphic to F^{m,n}

Question

My problem is the injectivity part of the proof. He assumes $M(T)=0$ and concludes that $T=0$ (the zero map). He then concludes that $M$ is injective because $\ker T=\{0\}$. I'm missing something very basic here. If $T$ is the zero map and $M(T)=0$, isn't it the case then that $\ker T = V$?

Answer 1

If $M(T)=0$, then $T=0$ and hence $\operatorname{Null}(T)=V$. So $T\in L(V,W)$ is not injective, as you said yourself. But we are trying to show that $M$ is injective. So what we actually showed here is that $\operatorname{Null}(M)=0\subseteq L(V,W)$, and this implies (according to 3.16 in your reference) that $M$ is injective.

The point is that you are applying the result 3.16 but to the linear map $M\in L(L(V,W),F^{m,n})$. The $T$ in 3.16 is $M$ in 3.60 and the $T$ in 3.60 is just an element of the vector space on the domain of $M$, which happens to be a linear map denoted by $T$. The confusion is understandable!

Answer 2

Thanks Pedro that makes sense now.

By the definition of Null space the Null Space of $M$ is all $T\in L(V,W)$ in the domain that M maps to $F^{m,n}= 0$ (the $m\times n$ null matrix) in the codomain.

And we have shown that the (only) $T$ that does this is $T=0$ (the zero map).

Thus we can conclude that $\operatorname{null}(M)=0,$ which is why $M$ is injective.

Answer 3

Note that for a linear map $M$, $\ker M=0\iff M$ is injective... (Proof?)

In this case, $M$ happens to be a map whose domain consists of linear transformations.

Now one way to show a linear transformation $T\in L(V,W)$ is zero, is to show $\ker T=V$...

Of course, the null space is just another name for the kernel...