= Solution
For every $v\in V$, $(\alpha+\beta)v=\alpha v+\beta v$, so $\operatorname{im}(\alpha+\beta)\subseteq\operatorname{im}\alpha+\operatorname{im}\beta$. Taking <dimensions> and applying the preceding <dimension formula for a sum of subspaces> gives
$$
\operatorname{rank}(\alpha+\beta)\leq\dim(\operatorname{im}\alpha+\operatorname{im}\beta)\leq\operatorname{rank}\alpha+\operatorname{rank}\beta.
$$
This proves the upper <rank bounds for a sum of linear maps>, without assuming the two images have zero intersection.
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