Definition. A vector space $V$ is finite dimensional if it has a basis consisting of a finite number of vectors. The unique number of vectors in each basis for $V$ is called the dimension of $V$ and is denoted by $\text{dim}(V)$.

Definition. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. The null space (or kernel) $N(T)$ of $T$ is the set of all vectors $x\in V$ such that $T(x)=0$; that is, $N(T)=\{x\in V:T(x)=0\}$. The range (or image) $R(T)$ of $T$ is the subset of $W$ consisting of all images, under $T$, of vectors in $V$; that is, $R(T)=\{T(x):x\in V\}$.

Definition. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. The nullity of $T$ is the dimension of $N(T)$ denoted by $\text{nullity}(T)$. The rank of $T$ is the dimension of $R(T)$ denoted by $\text{rank}(T)$.

Rank Nullity Theorem. Let $V$ and $W$ be vector spaces and let $T:V\to W$ be linear. If $V$ is finite dimensional, then $\text{dim}(V)=\text{rank}(T)+\text{nullity}(T)$.

Proof. Suppose that $\{u_1,\ldots,u_m\}$ forms a basis for $N(T)$. We can extend this to form a basis for $V$, $\{u_1,\ldots,u_m,w_1,\ldots,w_n\}$. Since $\text{nullity}(T)=m$ and $\text{dim}(V)=m+n$, it suffices to show that $\text{rank}(T)=n$. I claim that $S=\{Tw_1,\ldots,Tw_n\}$ forms a basis for $R(T)$.

First we show that $S$ spans $R(T)$. Let $v\in V$. Then there exist unique scalars $a_1,\ldots,a_m$ and $b_1,\ldots,b_n$ such that

$v=a_1u_1+\ldots+a_mu_m+b_1w_1+\ldots+b_nw_n$.

Thus,

$Tv=a_1Tu_1+\ldots+a_mTu_m+b_1Tw_1+\ldots+b_nTw_n$.

Since each $u_i$ is in $N(T)$, we have that $Tu_i=0$ and so $Tv=b_1Tw_1+\ldots+b_nTw_n$.  This shows that $S$ spans $R(T)$.

Next we show that $S$ is linearly independent. Suppose that

$c_1Tw_1+\ldots+c_nTw_n=0$

for scalars $c_i$. Since $T$ is linear, we have that

$T(c_1w_1+\ldots+c_nw_n)=0$.

Thus, $c_1w_1+\ldots+c_nw_n$ is in $N(T)$. Since $u_i$ span $N(T)$, there exist scalars $d_i$ such that

$c_1w_1+\ldots+c_nw_n=d_1u_1+\ldots+d_mu_m$.

Which implies that

$c_1w_1+\ldots+c_nw_n-(d_1u_1+\ldots+d_mu_m)=0$.

But, $\{u_1,\ldots,u_m,w_1,\ldots,w_n\}$ is a basis for $V$ which means that all the $c_i$ and $d_i$ must be zero. Thus, $S$ is linearly independent and is indeed a basis for $R(T)$. This proves that $\text{rank}(T)=n$ as desired. $\Box$