Two linear transformations defined on same vector spaces with finite dimension. Can we say that they will have same rank?
I am given a problem that $f,g:V\to W$ be two linear transformation and $f$ and $g$ are respectively injective and surjective. To prove $\dim(V)=\dim(W)$.
I am thinking it like that since f is injective $\operatorname{rank}(f)=\dim(V)$ and since $g$ is surjective $\operatorname{rank}(g)=\dim(W)$.
Then to equate both of them.....but I do not understand whether it is correct or wrong.
Since $f$ is injective, $\dim(V) \leq \dim(W)$. On the other hand, since $g$ is surjective, $\dim(V) \geq \dim(W)$. Hence $\dim(V) = \dim(W)$.
The above inequalities follow from the "Rank-Nullity Theorem":
$$ \dim(\mathrm{im}(f)) = \dim(V) - \dim(\mathrm{ker}(f)) $$
To obtain the inequalities mentioned above, note that $\dim(\mathrm{ker}(f)) = 0$ since $f$ is injective, and $\dim(\mathrm{im}(g)) = \dim(W)$ since $g$ is surjective.
They do have the same rank, because, as you say, $\dim(V) = \mathrm{rank}(f)$ and $\dim(W) = \mathrm{rank}(g)$. But you can prove that $\dim(V) = \dim(W)$ without this fact (by my reasoning above).