(Theorem) Rank-nullity

The rank-nullity theorem says that a vector space can be thought of as a combination of vectors that “die” (the vectors in the kernel) and the vectors that “survive” (the vectors in the image) after a linear transformation.

Theorem: Rank-nullity

If $A:V\to W$ is a linear map between finite-dimensional vector spaces, we have

$$\text{rank}(A)+\text{null}(A)=\text{dim}(V).$$

As a corollary of the theorem, we can use facts about $A$ to determine the relative dimensions of $V$ and $W$.

Corollary

• $A\text{ is injective}⟹\text{dim}(V)\le \text{dim}(W)$
• $A\text{ is surjective}⟹\text{dim}(V)\ge \text{dim}(W)$
• $A\text{ is bijective}⟹\text{dim}(V)=\text{dim}(W)$

Proof from Honors Mathematics A.

To prove the rank-nullity theorem, our goal is to:

• Use the bases of image and kernel to produce a basis for all of $V$ $\{w_1,\cdots ,w_m,u_1,\cdots ,u_k\}$.
• Show $\text{dim}V=m+k=\text{rank}(A)+\text{null}(A)$.

We start with the following fact:

• Pick a basis $\{w_1,\cdots ,w_m\}$ for $\text{im}(A)$, so $\text{rank}(A)=m$.
  • By definition of image, $\exists$ vectors $v_1,\cdots ,v_n\in V$ so that $A{v}_i=w_i$ for all $1\le i\le m$.
• Pick a basis $\{u_1,\cdots ,u_k\}$ for $\text{ker}(A)$, so $\text{null}(A)=k$.
• The first set spans the part of $V$ which maps identically onto $\text{im}(A)$, and the second set spans the part of $V$ which is sent to 0.

The following is a sketch of the proof:

1. Show $\{w_1,\cdots ,w_m,u_1,\cdots ,u_k\}$ forms a linearly independent set.
   1. Show second sequence has only the trivial relation.
   2. Show first sequence has only the trivial relation.
2. Verify $\{w_1,\cdots ,w_m,u_1,\cdots ,u_k\}$ spans $V$ by picking an arbitrary $v\in V$.