A linear transformation between spaces of equal dimension is injective iff it's surjective iff it's an isomorphism

up:: Linear Transformation

Let $T:U\to V$ a linear transformation between Vector Space with equal Dimension $\dim U=\dim V=n$.

We seek to prove that

• A) Bijectiveness $⟺$
• B) Injectiveness $⟺$
• C) Surjectiveness

It is trivial that $A)⟹B)$ and that $A)⟹C)$.

Injective $⟹$ Bijective

By the Kernel-Image Theorem, we have that

$$\dim U(=\dim V)=\dim ImT$$

Thus, the dimension of the codomain is equal to the dimension of $T$‘s image ─ which implies that $T$ is Surjective, and thus Bijective.

Surjective $⟹$ Bijective

If $T$ is surjective, we have that $\dim V=\dim ImT$, for which we have that

$$\dim \ker T=0$$

for which $T$ is Injective, since A linear transformation is injective iff its kernel is trivial.

Corollaries

• An endomorphism is injective iff it’s surjective

---

References

• Um Curso de Álgebra Linear, Flávio Ulhoa Coelho & Mary Lilian Lourenço. Editora EDUSP.