# Using the dimension formula to prove isomorphism

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.

• $V$ must be finite-dimensional, otherwise the assertions don't follow. – Daniel Fischer Oct 6 '15 at 18:54
• What did you try? – Bernard Oct 6 '15 at 18:55
• 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. – Alex Jeon Oct 6 '15 at 19:00
• Injective means that every vector in V maps to a unique vector in V. – Yunus Syed Oct 6 '15 at 19:02

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?
We have these equivalences (using $\dim \ker T+\operatorname{rank}T=\dim V$)
$$T\;\text{is injective}\iff\ker T=\{0\}\iff\dim\ker T=0\\\iff\operatorname{rank}T=\dim V\iff T \;\text{is surjective}$$