A well known theorem of Linear Algebra is the following one:
Suppose $V$ is finite-dimensional and $T\in\mathcal{L}(V)$ (i.e. a linear map $T:V\to V$). Then the following are equivalent:
- $T$ is invertible;
- $T$ is injective;
- $T$ is surjective.
I saw a solution for a particular exercise, where I had to show that a function $\Gamma:\mathcal{L}(V,W)\to\mathcal{L}(V',W')$ is indeed an isomorphism. For proving that, the author said $\dim \mathcal{L}(V,W)=\dim \mathcal{L}(V',W')$ which is true and easy to show, but right after that, used the theorem I stated before but $\Gamma$ isn't an operator. I doubt about it but at the same time since we have an isomorphism, it is the same as thinking it like a linear operator.
For instance, thinking about $2\times 2$ matrices is the same as $\mathbb{R}^4$, so you could use the theorem above. Is my reasoning correct, or you can't use the theorem?
Thanks in advance.
Note: $T\in\mathcal{L}(V,W)$ means $T:V\to W$.
Clearly, $(1)$ implies $(2)$. We also have that $(2)\Rightarrow(3)$ due to the Rank-Nullity theorem and the fact that a linear transformation is injective iff its kernel is trivial. The same theorem ensures us that $(3)\Rightarrow(1)$ once a linear transformation is surjective iff its image equals the counter-domain.
For reference, such theorem can be stated as follows:
(Rank-Nullity) Theorem
Given a linear transformation $T:V\to W$ from the finite-dimensional vector space $V$ to the finite dimensional vector space $W$, the following claim holds: \begin{align*} \dim V = \dim\ker(T) + \dim T(V) \end{align*}
Hopefully this helps!