I'm aware that the rank-nullity theorem states that $\dim V = \operatorname{dim null}(T) + \operatorname{dim range}(T)$, but I'm unable to see how I can apply the theorem to get that a one-to-one linear map from a finite-dimensional vector space to itself is onto.
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1$\begingroup$ $T$ is one-to-one iff $null(T)=\left\{0\right\}$, which is $0$-dimensional. What is $\dim(range(T))$? $\endgroup$– Luiz CordeiroAug 6, 2018 at 3:57
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$\begingroup$ dim(range(T)) should be the dimension of the vector space, correct? $\endgroup$– K.MAug 6, 2018 at 3:59
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$\begingroup$ Yes. Then what happens if $W\subseteq V$ and $\dim W=\dim V$? $\endgroup$– Luiz CordeiroAug 6, 2018 at 4:00
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$\begingroup$ Then $W$ = $V$? $\endgroup$– K.MAug 6, 2018 at 4:01
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$\begingroup$ Yes. You need to think of dimension as a sort of ``size'' of a space. An analogy would be the following: If a subset $X$ of a finite set $Y$ has the same number of elements as $Y$, then $X=Y$. $\endgroup$– Luiz CordeiroAug 6, 2018 at 4:07
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2 Answers
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The range of $T$ is a subspace of the finite dimensional space $V$. It equals $V$ if and only if it has the same dimension as $V$. According to the rank-nullity theorem, it has the same dimension as $V$ if and only if the kernel of $T$ has dimension $0$, that is is $0$.
Hence, you may see the following equivalence thanks to the rank-nullity theorem:
A linear map from a finite-dimensional vector space to itself is onto in and only if it is injective (and so if and only if it is an isomorphism).
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1$\begingroup$ I meant injective, the property that to any element in the image corresponds only one preimage. (English is not my main language - I may edit my answer to make it clearer). $\endgroup$– SuzetAug 6, 2018 at 4:10
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Suppose $f$ is not onto then there must exist a vector $x$ in $V$ having no pre-image in $V$, so $\vert f(V)\vert <\vert V\vert $. It suggests that there must exist distinct $x_1$, $x_2\in V$ s.t. $f(x_1)=f(x_2)$ i.e. $f$ is not one-one ($\mathbb\ contradiction$).
Hence $f$ is onto as well.
answered Aug 6, 2018 at 4:24