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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.

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  • $\begingroup$ 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. $\endgroup$ – Gerry Myerson Dec 6 '18 at 6:04
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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 !

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  • $\begingroup$ 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? $\endgroup$ – FundementalJTheorem Dec 6 '18 at 6:29
  • $\begingroup$ 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 $T(x,y)=x$. $\endgroup$ – Fred Dec 6 '18 at 11:08
  • $\begingroup$ If $T$ is onto, then $Im(T)=W$, hence $\dim Im(T)=n$, thus $m= \dim ker(T)+ \dim Im(T)= \dim ker(T)+n \ge n.$ $\endgroup$ – Fred Dec 6 '18 at 11:10

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