 Solution

The rank-nullity theorem says that for a linear map $T:V\to W$ with finite-dimensional domain of a function $V$,Take a basis $k_1,\dots ,k_r$ of the kernel of a linear map and use basis extension to obtain a basis $k_1,\dots ,k_r,u_1,\dots ,u_s$ of $V$. We show that $T(u_1),\dots ,T(u_s)$ is a basis of the image of a linear map.

Every $v\in V$ is a linear combination of the extended basis, and its $k_i$ terms vanish under $T$. Therefore $T(v)$ is a linear combination of the $T(u_j)$, proving the required linear span property. For linear independence, suppose ${\sum }_j{b}_{j}T(u_j)=0$. Then ${\sum }_j{b}_j{u}_j\in \ker T$, so ${\sum }_j{b}_j{u}_j={\sum }_i{a}_i{k}_i$ for some real scalars $a_i$. The extended basis is linearly independent, forcing every $a_i$ and $b_j$ to be zero. Thus the image of a linear map has dimension $s$, the kernel of a linear map has dimension $r$, and $\dim V=r+s$ proves the result.