Past exam of the mathematics course of the University of Cambridge / 2003 / ib / Paper 1 / 14E (Linear Mathematics)

Choose a basis $e_1,\dots ,e_k$ for $U\cap U^{\prime }$. Extend it to a basis $e_1,\dots ,e_k,u_1,\dots ,u_r$ of $U$, and separately to a basis $e_1,\dots ,e_k,v_1,\dots ,v_s$ of $U^{\prime }$. The combined list spans $U+U^{\prime }$.

To prove linear independence, suppose a linear combination of the combined list is zero. Then the combination of the $v_j$ lies in $U\cap U^{\prime }$, so it can be expressed using the $e_i$. Independence of the chosen basis of $U^{\prime }$ forces every $v_j$ coefficient to vanish. Independence of the chosen basis of $U$ then forces the remaining coefficients to vanish. Thus the combined list is a basis, with $k+r+s$ elements. This proves the dimension formula for a sum of subspaces:

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

Use the standard basis $e_1,\dots ,e_n$. For the smallest possible rank, set $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and set $\beta e_j=-e_j$ for $j\le s$ and zero otherwise. Then the maps have ranks $r,s$, while their sum has nonzero diagonal entries only for $r<j\le s$. HenceFor full rank when $r+s\ge n$, instead take $\alpha e_j=e_j$ for $j\le r$ and zero otherwise, and $\beta e_j=e_j$ for $j\ge n-s+1$ and zero otherwise. Their supports cover every basis index because $n-s+1\le r+1$. 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