Rank–Nullity Theorem

Definition

Linear Transformation

$\dim (V)=\dim (\ker L)+\dim (\mathrm{im}L)$ where $V$ is a finite-dimensional Vector Space and $L:V\to W$ is a Linear Map

The dimension of vector space $V$ is equal to the sum of the dimension of kernel of $L$ and the dimension of image of $L$.

Proof

We want to show that $\{L({\mathbf{v}}_{k+1}),\dots ,L({\mathbf{v}}_n)\}$ is the basis of $\mathrm{im}L$, or $\dim (V)=\dim (\mathrm{im}L)$

Let $B_V=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_n\}$ be the basis of Vector Space $V$, and $B_{\ker L}=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_k\}$ where $k\le n$ be the basis of the $\ker L\subset V$

Span

$\forall L(\mathbf{v})\in \mathrm{im}L:\mathbf{v}=c_1{\mathbf{v}}_1+c_2{\mathbf{v}}_2+\cdots +c_k{\mathbf{v}}_k+c_{k+1}{\mathbf{v}}_{k+1}+\cdots +c_n{\mathbf{v}}_n$ Since $L({\mathbf{v}}_1)+L({\mathbf{v}}_2)+\cdots +L({\mathbf{v}}_k)=0$ by the definition of $B_{\ker L}$ $L(\mathbf{v})=L(c_{k+1}{\mathbf{v}}_{k+1}+\cdots +c_n{\mathbf{v}}_n)=c_{k+1}L({\mathbf{v}}_{k+1})+\cdots +c_n({\mathbf{v}}_n)\in \mathrm{im}L$ $\therefore \mathrm{span}\{L({\mathbf{v}}_{k+1}),\dots ,({\mathbf{v}}_n)\}=\mathrm{im}L$

Linear Independence

Let $c_{k+1}L({\mathbf{v}}_{k+1})+\cdots +c_{n}L({\mathbf{v}}_n)=0$ We can transform the equation$c_{k+1}L({\mathbf{v}}_{k+1})+\cdots +c_{n}L({\mathbf{v}}_n)=L(c_{k+1}{\mathbf{v}}_{k+1}+\cdots +c_n{\mathbf{v}}_n)=0$ by the Linearity of $L$ Then, $c_{k+1}{\mathbf{v}}_{k+1}+\cdots +c_n{\mathbf{v}}_n\in \ker L$

The expression can be expressed using only the basis vectors. So, $\exists c_1,\dots ,c_k,(c_1{\mathbf{v}}_1+\cdots +c_k{\mathbf{v}}_k=c_{k+1}{\mathbf{v}}_{k+1}+\cdots +c_n{\mathbf{v}}_n)$ $\iff c_1{\mathbf{v}}_1+\cdots +c_k{\mathbf{v}}_k-c_{k+1}{\mathbf{v}}_{k+1}-\cdots -c_n{\mathbf{v}}_n=0$ by subtracting LHS - RHS

Since $B_V=\{{\mathbf{v}}_1,\dots ,{\mathbf{v}}_n\}$ is a basis of the vector space, $c_1=c_2=\cdots =c_n=0$ $\Rightarrow c_{k+1}+\cdots +c_n=0$ $\therefore \{L({\mathbf{v}}_{k+1}),\dots ,L({\mathbf{v}}_n)\}$ is linearly independent.

Since $\{L({\mathbf{v}}_{k+1}),\dots ,L({\mathbf{v}}_n)\}$ satisfies Span and Linear Independence, is the basis of $\mathrm{im}L$ $B_{\mathrm{im}L}=\{L({\mathbf{v}}_{k+1}),L({\mathbf{v}}_{k+2}),\dots ,L({\mathbf{v}}_n)\}$

Matrices

$n=\mathrm{rank}(M)+\mathrm{nullity}(M)$ where $M\in {\mathcal{M}}_{m\times n}(F)$

Proof

Let a matrix $A$ be the echelon form of the matrix $M$, and $\mathrm{rank}(M)=r\le n$ Then, the number of pivots of $A$ is $r$. Also the number of free variables of $M\mathbf{x}=0$ become $n-r$ $\therefore \mathrm{nullity}(M)=n-r\Rightarrow \mathrm{rank}(M)+\mathrm{nullity}(M)=n$

Visualization

$\dim (\mathrm{Ker}(A))=2,\dim (\mathrm{Im}(A))=1$