Let \(\{u_1,\ldots,u_r\}\) be a basis for \(\ker(T)\text{.}\) We extend this basis to a basis of \(V\text{,}\) say \(\{u_1,\ldots,u_r,v_1,\ldots,v_s\}\text{.}\) We claim that \(\{T(v_1),\ldots,T(v_s)\}\) is a basis for \(\Im(T)\text{.}\) We first check the independence. Suppose that
\begin{align*}
0\amp=\sum\alpha_iT(v_i)\\
\amp= T\big(\sum\alpha_iv_i\big)
\end{align*}
Thus, \(\sum\alpha_iv_i\in\ker(T)\) and hence there exists \(\beta_j\in F\) such that
\begin{equation*}
\sum\alpha_iv_i=\sum\beta_ju_j\quad\text{i.e.,}\quad\sum_i\alpha_iv_i-\sum_j\beta_ju_j=0.
\end{equation*}
Since \(\{u_1,\ldots,u_r,v_1,\ldots,v_n\}\) is a basis we have \(\alpha_i=0\) and \(\beta_j=0\) for each \(i\) and \(j\text{.}\) Therefore, \(\{T(v_j)\}\) is linearly independent.
Given any \(w\in\Im(T)\) there exists \(v\in V\) such that \(w=T(v)\text{.}\) Let
\begin{equation*}
v=\sum\gamma_iu_i+\sum\delta_jv_j.
\end{equation*}
Using \(\{u_i\}\) is a basis of \(\ker(T)\) we have the following.
\begin{align*}
w\amp=T(v)\\
\amp=\sum\gamma_iT(u_i)+\sum\delta_jT(v_j)\\
\amp=\sum\delta_jT(v_j)
\end{align*}
Hence \(\{T(v_j)\}\) spans \(\Im(T)\text{.}\) Since \(\{T(v_j)\}\) is linearly independent and spans the \(\Im(T)\) it is a basis of \(\Im(T)\text{.}\) We thus have
\begin{equation*}
\dim_F\Im(T)=s\quad\text{and}\quad\dim_F\ker(T)=r.
\end{equation*}
Therefore,
\begin{equation*}
\dim_FV={\rm nullity}(T)+{\rm rank}(T).
\end{equation*}
