 Solution

For a linear map $T:V\to V$, its kernel of a linear map and image of a linear map areIts nullity of a linear map is $n(T)=\dim \ker T$, and its rank of a linear map is $r(T)=\dim \mathrm{im}T$.

Choose a basis $u_1,\dots ,u_k$ of $\ker T$ and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_m$ of $V$. The vectors $T(v_1),\dots ,T(v_m)$ span $\mathrm{im}T$, because applying $T$ to an arbitrary linear combination of the full basis eliminates the $u_i$ terms. They are also linearly independent: if ${\sum }_j{b}_{j}T(v_j)=0$, then ${\sum }_j{b}_j{v}_j\in \ker T$, so ${\sum }_j{b}_j{v}_j={\sum }_i{a}_i{u}_i$ for some coefficients. Independence of the extended basis forces all $a_i,b_j$ to vanish.

Consequently $r(T)=m$, $n(T)=k$ and $\dim V=k+m$. This proves the rank-nullity theorem: