Rank-nullity theorem

Let $𝑇 \in 𝖵 𝖾 𝖼 𝗍 𝕂 (𝑈, 𝑉)$ be a linear map. Then any complement of the kernel is isomorphic to the image #m/thm/linalg

(k e r 𝑇) 𝑐 ≅ i m 𝑇

and thus the sum of the rank and the nullity equals the dimension of $𝑈$ ¹

r a n k 𝑇 + n u l l i t y 𝑇 = d i m 𝑈 .

In full generality, this is downstream of AC.

Proof

By ^Existence we have $𝑈 = k e r 𝑇 \oplus (k e r 𝑇) 𝑐$ whence $d i m 𝑈 = n u l l i t y 𝑇 + d i m ((k e r 𝑇) 𝑐)$ . Let $𝑇 𝑐 = 𝑇 ↾ (k e r 𝑇) 𝑐$ . Note $𝑇 𝑐$ is monic since $k e r 𝑇 𝑐 = (k e r 𝑇) \cap (k e r 𝑇) 𝑐 = {0}$ . Let $𝑇 𝑣 \in i m 𝑇$ . Since $𝑣 = 𝑢 + 𝑢 𝑐$ for $𝑢 \in k e r 𝑇$ and $𝑢 \in (k e r 𝑇) 𝑐$ we have

𝑇 𝑣 = 𝑇 𝑢 + 𝑇 𝑢 𝑐 = 𝑇 𝑢 𝑐 = 𝑇 𝑐 𝑢 𝑐 \in i m (𝑇 𝑐)

hence $i m (𝑇) \subseteq i m (𝑇 𝑐)$ so $𝑇 𝑐 : (k e r 𝑇) 𝑐 \to i m 𝑇$ is an isomorphism It follows immediately that $r a n k 𝑇 + n u l l i t y 𝑇 = d i m 𝑈$ .

Corollaries

It follows that an endomorphism $𝑇$ on a finite-dimensional vector space is monic iff it is epic.

#state/tidy | #lang/en | #SemBr

2008. Advanced Linear Algebra, p. 63 ↩

Rank-nullity theorem

Corollaries

Footnotes