# Proving a Linear Transformation if onto/one-to-one for vector spaces

Given two vector spaces $$V$$ and $$W$$ of finite dimensions and a linear transformation $$T: V \rightarrow W$$, let $$m = dim(V)$$ and $$n = dim(W)$$. Show that if $$T$$ is one to one, then $$m \leq n$$ and if $$T$$ is onto then $$m \geq n$$.

So I am thinking this is best approached by contradiction. Can I say that the number of vectors in the basis of $$V$$ is $$m$$ and the number of vectors in the basis of $$W$$ is $$n$$ and to assume $$m \geq n$$. Let $$V = \{v_1,v_2,...,v_m\}$$ and $$W = \{w_1,w_2,...,w_n\}$$, if we consider the function $$T: V \rightarrow W$$ then the number of vectors in the $$span(V) \leq span(W)$$ so then you wouldn't have unique mapping and $$T$$ isn't one to one.

I'm not sure where to even begin for the onto part of the question.

• If we are talking about vector spaces over the real numbers, then the number of vectors in the span of a nonempty subset $S$ is the same no matter how many elements $S$ has. I think you are very confused about vector space, basis, span, and dimension. – Gerry Myerson Dec 6 '18 at 6:04

We have $$m= \dim V = \dim ker(T)+ \dim Im(T).$$
If $$T$$ is one-to-one, then $$\dim ker(T)=0$$, thus $$m = \dim Im(T) \le \dim W =n.$$
If $$T$$ is onto, then...... your turn !
• Then it's an equivalent statement but instead starting with $n=dim(W)$? I know that $T$ is onto if $dim(V)=dim(W)$ so does it hold for $\leq$ relationships? – FundementalJTheorem Dec 6 '18 at 6:29
• It is not correct, that if $T$ is onto, then $\dim(V)=\dim(W)$. Example : let $T \mathbb R^2 \to \mathbb R$ be defined by $T(x,y)=x$. – Fred Dec 6 '18 at 11:08
• If $T$ is onto, then $Im(T)=W$, hence $\dim Im(T)=n$, thus $m= \dim ker(T)+ \dim Im(T)= \dim ker(T)+n \ge n.$ – Fred Dec 6 '18 at 11:10