Rank Nullity Theorem

• If $V,W$ are finite dimensional vector spaces and $T:V\to W$ is linear
• Then: $dim(ker(T))+dim(Image(T))=dim(V)$
• or, $dim(V)=nullity(T)+rank(T)$

Proof

1. Refer to the proof of Bases for Images and Kernel Procedure 2
2. Then, we need to show $\{T(v_{k+1}),\dots ,T(v_n)\}$ is a basis
3. Checking linear independence:
   1. Suppose for sake of contradiction that $0=a_{k+1}T(v_{k+1})+\cdots +a_{n}T(v_n)$ has some non-zero coefficient $a_z\ne \hat{0}$
   2. This gives: $\overrightarrow{0}=T(\overrightarrow{0})=T(a_{k+1}{v}_{k+1}+\cdots +a_n{v}_n)$
   3. Therefore $a_{k+1}{v}_{k+1}+\cdots +a_n{v}_n\in ker(T)$u
   4. But we already have a basis for $ker(T)=\{b_1{u}_1,\dots ,b_k{u}_k\}$
   5. $⟹a_{k+1}{v}_{k+1}+\cdots +a_n{v}_n=b_1{u}_1+\cdots +b_k{u}_k+0v_{k+1}+\cdots +0v_n$
   6. Note that this gives us a non-unique representation in the basis $\{u_1,u_k,\dots ,v_{k+1},\dots ,v_n\}$
   7. $\{T(v_{k+1}),\dots ,T(v_n)\}$ is a indep

Alternate Proof

1. With $dim(V)=n$
2. Then, basis $\beta$ of $V$, then $|\beta |=n$
3. $ker(T)\subset V$ by defn of kernel as $ker(T)$ is a subspace of $V$
4. Choosing ${\beta }_1$ as the basis of $ker(T)$
5. Then, basis extended to form a basis of vector space by (some theorem)
6. $\beta =\{b_1,\dots ,b_k,v_{k+1},\dots ,v_n\}$
7. Applying $T$ to the span of this basis will get us the image
8. $T(a_1{b}_1+\cdots +a_n{v}_n)$
9. $a_{1}T(b_1)+\cdots +a_{k}T(b_k)+a_{k+1}T(v_{k+1})+a_{n}T(v_n)$
10. $=0+\cdots +0+a_{k+1}T(v_{k+1})+a_{n}T(v_n)$
11. Then, $\{v_{k+1},\dots ,v_n\}$ is the basis for the image
12. Thus, $dim(ker(T))=dim(Image(T))$

Use Cases