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

Apply the preceding rank bounds for a sum of linear maps to $\alpha =(\alpha +\beta )+(-\beta )$. Since multiplying a linear map by $-1$ does not change its image, $\mathrm{rank}\alpha \le \mathrm{rank}(\alpha +\beta )+\mathrm{rank}\beta$. Thus $\mathrm{rank}(\alpha +\beta )\ge \mathrm{rank}\alpha -\mathrm{rank}\beta$. Interchanging the maps gives the other difference, proving