= Solution
For a <linear map> $\alpha:U\to V$ with $U$ a <finite-dimensional vector space>, the <rank-nullity theorem> is
$$
\boxed{\dim U=\dim\ker\alpha+\dim\operatorname{im}\alpha.}
$$
Choose a <basis> $u_1,\ldots,u_k$ of the <kernel of a linear map> and extend it to a basis $u_1,\ldots,u_k,v_1,\ldots,v_r$ 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 $\sum_j c_jv_j$ 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 $\mathbb Re_3$ and its image is $\operatorname{span}(e_1,e_2)$. It fails for $\beta(x,y,z)=(y,z,0)$: the kernel is $\mathbb Re_1$, contained in the two-dimensional image $\operatorname{span}(e_1,e_2)$. The <rank-nullity theorem> fixes the sum of the dimensions, but does not imply zero intersection.
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