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Let $V$ be a finite-dimensional vector space and $T: V\rightarrow V $. $T$ is a linear transformation. Use the dimension formula to prove that if $T$ is injective, it must also be surjective; if T is surjective, it must also be injective.

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    $\begingroup$ $V$ must be finite-dimensional, otherwise the assertions don't follow. $\endgroup$ – Daniel Fischer Oct 6 '15 at 18:54
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    $\begingroup$ What did you try? $\endgroup$ – Bernard Oct 6 '15 at 18:55
  • $\begingroup$ I tried using the fact that the kernel/nullspace of V will only contain the zero vector if T is 1-1, but I wasn't sure if the nullspace would have 1 dimension or 0 dimensions. I think the nullspace would be 1-dimensional, because the zero vector is the one and only basis vector of the nullspace. $\endgroup$ – Alex Jeon Oct 6 '15 at 19:00
  • $\begingroup$ Injective means that every vector in V maps to a unique vector in V. $\endgroup$ – Yunus Syed Oct 6 '15 at 19:02
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Let $T:V\to V$ be a linear map on a finite-dimensional vector space $V$. Recall that

  • $T$ is injective if and only if $\dim\ker T=0$
  • $T$ is surjective if and only if $\dim\DeclareMathOperator{image}{image}\image T=\dim V$

The Rank-Nullity Theorem states that the equality $$ \dim\ker T+\dim\image T=\dim V $$ always holds. Can you combine the above to prove your result?

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We have these equivalences (using $\dim \ker T+\operatorname{rank}T=\dim V$)

$$T\;\text{is injective}\iff\ker T=\{0\}\iff\dim\ker T=0\\\iff\operatorname{rank}T=\dim V\iff T \;\text{is surjective}$$

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