Problem
Suppose $U$ and $W$ are both 4-dimensional subspaces of $\mathbb{C}^6$. Prove that there exist two vectors in $U \cap W$ such that neither of these vectors is a scalar multiple of the other
Solution
By theorem 2.43, 2.38 and the fact that $U + W$ is a subspace of $\mathbb{C}^6$ by theorem 1.39: $$6 = \dim \mathbb{C}^6 \geq \dim (U + W) = \dim U + \dim W - \dim U \cap W$$ $$= 4 + 4 - \dim U \cap W$$ Implying that $\dim U \cap W \geq 2$, therefore by the definition of dimension. there must exist two linearly independent vectors, a basis for $U \cap W$ which is a vector space as per P1C10. Therefore we can let $v_1,v_2 \in U \cap W$ be a basis for $U \cap W$ implying that the only solution to $a_1v_1 = a_2v_2$, for $a_1,a_2 \in \mathbb{C}$ is the trivial one.