 Solution

For a linear map $\alpha :U\to V$ with $U$ a finite-dimensional vector space, the rank-nullity theorem isChoose a basis $u_1,\dots ,u_k$ of the kernel of a linear map and extend it to a basis $u_1,\dots ,u_k,v_1,\dots ,v_r$ of $U$. The vectors $\alpha (v_j)$ span the image of a linear map, because the kernel basis contributes zero. If ${\sum }_j{c}_j\alpha (v_j)=0$, then ${\sum }_j{c}_j{v}_j$ belongs to the kernel and is a combination of the $u_i$; independence of the extended basis makes every $c_j=0$. Thus the image basis has $r$ elements and $\dim U=k+r$, proving the theorem.

For a rank-two map on ${\mathbb{R}}^3$, the direct sum occurs for $\alpha (x,y,z)=(x,y,0)$: its kernel is $\mathbb{R}{e}_3$ and its image is $\mathrm{span}(e_1,e_2)$. It fails for $\beta (x,y,z)=(y,z,0)$: the kernel is $\mathbb{R}{e}_1$, contained in the two-dimensional image $\mathrm{span}(e_1,e_2)$. The rank-nullity theorem fixes the sum of the dimensions, but does not imply zero intersection.