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Let $$V$$ and $$W$$ be finite-dimensional vector spaces and $$T: V \rightarrow W$$ be linear.

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

(b) Prove that if $$\operatorname{dim}(V)>\operatorname{dim}(W)$$, then $$T$$ cannot be one-to-one.
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(a) Prove that if $$\operatorname{dim}(V)<\operatorname{dim}(W)$$, then $$T$$ cannot be onto.

Solution: Suppose $$\operatorname{dim}(V)<\operatorname{dim}(W)$$, and assume (by means of contradiction) that $$T$$ is onto. Then image $$(T)=W$$, and thus $$\operatorname{rank}(T)=\operatorname{dim}(W)$$. By the dimensions theorem, we know
$\begin{gathered} \operatorname{dim}(V)=\operatorname{rank}(T)+\text { nullity }(T) \\ =\operatorname{dim}(W)+\text { nullity }(T) \end{gathered}$
Since $$\operatorname{dim}(V)<\operatorname{dim}(W)$$, this implies nullity $$(T)=\operatorname{dim}(V)-\operatorname{dim}(W)<0$$, which is a contradiction since nullity can not be negative. Thus $$T$$ is NOT onto.

(b) Prove that if $$\operatorname{dim}(V)>\operatorname{dim}(W)$$, then $$T$$ cannot be one-to-one.
Solution: Similar argument to (a). See if you can get it.

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