Proof of Two Subspaces Whose Dimensions Add Up to More Than the Whole Space Meet

Routine facts about concatenating tuples, splitting their linear combinations, and negatives in a vector space are used without comment (Toolkit for AL3).

Assume $\dim U+\dim W>\dim V$, and let $p=\dim U$ and $q=\dim W$. The subspaces $U$ and $W$ are vector spaces with the operations of $V$ (Linear Subspaces Are Vector Spaces; Intersections (i)), so by Finite-Dimensional Vector Spaces and Their Dimension (ii) we may choose a basis $(u_{1},\dots,u_{p})$ of $U$ and a basis $(w_{1},\dots,w_{q})$ of $W$. Each of these families is linearly independent in its subspace, and therefore also as a family in $V$, by Linear Subspaces Are Vector Spaces; Intersections (iv).

Consider the $(p+q)$-tuple $(u_{1},\dots,u_{p},w_{1},\dots,w_{q})$ in $V$. Its length $p+q$ exceeds $\dim V$, so by Properties of Dimension (ii) it is not linearly independent, that is, it is linearly dependent. Hence there are scalars $\alpha_{1},\dots,\alpha_{p},\beta_{1},\dots,\beta_{q}\in K$, not all $0$, with

$$\sum_{i=1}^{p}\alpha_{i}u_{i}+\sum_{j=1}^{q}\beta_{j}w_{j}=0.$$

Put $v=\sum_{i=1}^{p}\alpha_{i}u_{i}$; by the display, also $v=\sum_{j=1}^{q}(-\beta_{j})w_{j}$. A linear combination of vectors of a subspace lies in that subspace (Linear Subspaces Are Vector Spaces; Intersections (iv)), so the first expression shows $v\in U$ and the second $v\in W$. Thus $v\in U\cap W$.

It remains to see that $v\neq0$. Suppose $v=0$. Then $\sum_{i=1}^{p}\alpha_{i}u_{i}=0$, and the independence of $(u_{1},\dots,u_{p})$ forces every $\alpha_{i}$ to be $0$. The display then reads $\sum_{j=1}^{q}\beta_{j}w_{j}=0$, and the independence of $(w_{1},\dots,w_{q})$ forces every $\beta_{j}$ to be $0$. So all the scalars vanish, contrary to their choice. Hence $v$ is a nonzero vector of $U\cap W$, which proves (i).