Let $V$ and $W$ be finite-dimensional vector spaces and $T: V\rightarrow W$ be linear. Prove that if $\dim (V) < \dim (W)$, then $T$ cannot be onto. I know that this proof is related with the dimension theorem, but I don't know how. Any help is appreciated.
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$\begingroup$ 1. prove that the images of a basis of V span the image set of T. 2. if the map was onto, then that set, say of n, vectors would span W... $\endgroup$– RougeSegwayUserSep 8, 2014 at 3:34
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2 Answers
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By Rank-Nullity Theorem,
$$\dim V = \dim N(T) + \dim R(T).$$
Thus, $$\dim R(T) = \dim V - \dim N(T) < \dim W.$$
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Suppose that $\dim (V) = n < \dim (W)$. Then $V$ has some basis $v_1,v_2,\ldots,v_n$. If $T$ is onto, then $T(V) = W$, so that $T(v_1),T(v_2),\ldots,T(v_n)$ is a spanning set for $W$. But this means that $\dim(W) \leq n$, which is a contradiction.
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$\begingroup$ Why doesn't a linear transformation create more basis? $\endgroup$ Mar 21, 2018 at 17:16