An endomorphism is injective iff it's surjective

up:: Linear Transformation

Given an Endomorphism $T:V\to V$, we seek to prove that it is injective if and only if it is surjective.

By the Kernel-Image Theorem, we have that

$$\dim V=\dim \ker T+\dim ImT$$

For clarity, I’ll denote $V_D$ the vector space of $T$‘s domain, and $V_{CD}$ the vector space of its codomain.

Injective $⟹$ Surjective

Since A linear transformation is injective iff its kernel is trivial, we have that

$$\dim V_D=\dim ImT$$

But since $V_D=V_{CD}$, we have that $\dim ImT=\dim V_{CD}$, and thus $T$ is Surjective¹.

Surjective $⟹$ Injective

If $T$ is surjective, we have that $\dim ImT=\dim V_{CD}=\dim V_D$, for which we have

$$\dim \ker T=0$$

for which we conclude that $T$ is Injective.

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References

• Notas sobre Álgebra Linear

Footnotes

1. Since $ImT\subseteq V_{CD}$ as a subspace, if both have the same Dimension, then they must be the same space. ↩