Let $T \colon V \to W$ be a linear transformation, and if $\dim(V) < \dim(W)$, T cannot be onto.

Question

Let $V$, $W$, be two finite dimensional vector space. Prove that if $T \colon V \to W$ is a linear transformation from $V$ to $W$ and $\dim(V) < \dim(W)$, then $T$ is not onto.

$$\dim(V) = \operatorname{rank}(T) + \operatorname{nullity}(T) = \dim(\operatorname{Range}(T)) + \dim(\operatorname{ker}(T))$$

Since $\dim(V) < \dim(W)$ , can I safely assume that the inequality $(\operatorname{rank}(T) < \dim(W))$ is true for whatever $\operatorname{nullity}(T)$ may be? And if so, then $\operatorname{Range}(T)$ is guaranteed to be a proper subset of $W$, correct?

Thank you, I'm very new to this stuff.

Answer 1

You cannot assume $\text{rank}\; T<\dim W $ and you must prove this. This follows from dimension theorem as you mentioned, since $$\text{rank}\; T=\dim \text{range} \;T=\dim V-\dim \text{null}\;T\leq\dim V<\dim W$$

So, $\text{range}\;T \neq W$, and hence result follows!