What does it mean when $\dim(V)=\operatorname{rank}(T)$

Question

I have a question relating to a linear transformation and have ended up with the result that $\dim(V)=\operatorname{rank}(T)$.

I got to this because I'm told that $V$ and $W$ are finite dimensional vector spaces, and the linear map $T:V \rightarrow W$ is one-to-one.

Following from this I can determine that if the map is one-to-one then $\operatorname{nullity}(T)=0$ so the Rank-Nullity Theorem leaves me with $\dim(V)=\operatorname{rank}(T)$.

I was wondering if there is a way to determine what $\dim(V)$ and $\operatorname{rank}(T)$ are if I have also been told that $\dim(W)=5$.

Answer 1

Since - as you stated correctly - $dim(V)$ and $rank(T)$ are the same, it suffices to determine one of them. But unfortunately the only thing you can say about $V$ is, that the dimension is bounded by $5$, since $T$ is one-to-one. This also follows directly by the Rank Nullity Theorem.

Answer 2

let $V,W$ be finite dimensional vector spaces.

If there is an isomorphism (one to one function) $T: V \to W$ then $dim(V)=dim(W)$.

by that definition, if $dim(W)=5$, then $dim(V)=5$, and so $rank(T)=5$

Please note that a similar definition for rank of a linear transformation is rank = dimension of the image. Since in our case $T$ is one-to-one, we can infer $Image(T)=W$. and so $rank(T)=dim(W)$.

Edit:

I just realized your question said that $T$ is one-to-one function, not correspondence. what i said is true if its one-to-one correspondence (IE injective and surjective, aka bijective). in our case, what we can say is that $dim(V)$ is at most 5