# Proving dimension formula in linear algebra

Let $V$ and $W$ be finite dimensional vector spaces and let $T:V \to W$ be a linear transformation.

(a) Prove that if $\dim(V) < \dim(W)$ then $T$ cannot be onto.

(b) Prove that if $\dim(V) > \dim(W)$ then $T$ cannot be one-to-one.

What I tried:

(a) Proving by contradiction. Suppose that $T$ is onto. Then, since we are also given that $T$ is linear, then $T$ has to be one-to-one. Thus $T$ is both one-to-one and onto which means $\dim(V) = \dim(W)$ hence contradiction the fact that $\dim(V) < \dim(W)$.

(b) Again proving by contradiction, suppose that $T$ is one-to-one. Then we know that $\dim N(T) = 0$.

And since $\dim R(T) + \dim N(T) = \dim(V)$, this makes $\dim R(T) = \dim(W)$, and thus $V$ maps onto $W$, which contradicts the fact that $\dim(V) > \dim(W)$ and hence proving the statement.

Is my prove correct? Could anyone explain? Also could anyone show me how to do the prove directly instead of using contradiction? Thanks

• For $(a)$, there are many linear transformations that are onto but not one-to-one. For example $T:\mathbb{R}^2 \rightarrow \mathbb{R}$, $T(x,y) = x$. You also have your formula wrong. You should have $$\dim R(T) + \dim N(T) = \dim (V).$$
– J126
Feb 12 '16 at 13:05
• Probably you can take advantage reading here: math.stackexchange.com/a/1651855/309917 Feb 12 '16 at 13:08

Suppose $\dim V<\dim W$; then $$\dim R(T)=\dim V-\dim N(T)<\dim W-\dim N(T)<\dim W$$ so $\dim R(T)<\dim W$ and $T$ is not onto.
Suppose $\dim V>\dim W$; then $$\dim N(T)=\dim V-\dim R(T)>\dim W-\dim R(T)\ge0$$ so $\dim N(T)>0$ and $T$ is not one-to-one.
What about your proofs? The fact that an onto linear map is also one-to-one is valid only if domain and codomain have the same dimension. So you can't use the fact that $T$ is one-to-one in the first attempt.