= Solution
Use the standard <basis> $e_1,\ldots,e_n$. For the smallest possible rank, set $\alpha e_j=e_j$ for $j\leq r$ and zero otherwise, and set $\beta e_j=-e_j$ for $j\leq s$ and zero otherwise. Then the maps have ranks $r,s$, while their sum has nonzero diagonal entries only for $r<j\leq s$. Hence
$$
\boxed{\operatorname{rank}(\alpha+\beta)=s-r.}
$$
For full rank when $r+s\geq n$, instead take $\alpha e_j=e_j$ for $j\leq r$ and zero otherwise, and $\beta e_j=e_j$ for $j\geq n-s+1$ and zero otherwise. Their supports cover every basis index because $n-s+1\leq r+1$. The sum has diagonal entries one where just one map is nonzero and two where both are nonzero. None vanish over $\mathbb R$, so these <linear maps> have ranks $r,s$ and
$$
\boxed{\operatorname{rank}(\alpha+\beta)=n.}
$$
Back to article page
