Proof that an injective endomorphism of a finite-dimensional space is surjective

Proof

We use the rank–nullity theorem to compare the kernel and the image.

1. Let $n = \dim V$. By the rank–nullity theorem, $\dim \im T + \dim \ker T = n$, where $\im T$ is the image and $\ker T$ the kernel.
2. A subspace of dimension $0$ has the empty family as basis, so it equals $\{0\}$. Conversely $\{0\}$ has dimension $0$.
3. (1 implies 2.) Suppose $T$ is injective. By the kernel criterion, $\ker T = \{0\}$, so $\dim \ker T = 0$.
4. By step 1, $\dim \im T = n = \dim V$. Since $\im T$ is a subspace of $V$, a subspace of full dimension is the whole space gives $\im T = V$. So $T$ is surjective.
5. (2 implies 1.) Suppose $T$ is surjective. Then $\im T = V$, so $\dim \im T = n$.
6. By step 1, $\dim \ker T = 0$, so $\ker T = \{0\}$ by step 2. By the kernel criterion, $T$ is injective.
7. (3 is equivalent to 1 and 2.) A map is bijective exactly when it is injective and surjective.
8. If $T$ is bijective, it is injective, so condition 1 holds. If condition 1 holds, then condition 2 holds by step 4, so $T$ is bijective.
9. By steps 3 to 8, the three conditions are equivalent.