The official question reads:
Suppose U and W are each 5-dimensional subspaces of $\mathbb{C}^8$.
Prove that there are two vectors $u,w ∈ U∩W$ $\hspace{1mm}$ s.t. $\hspace{1mm}$ $u≠zw$ $\hspace{1mm}$ ∀ z ∈ $\mathbb{C}$
First I considered one subspace being in $\mathbb{R}^5$ and the other in $\mathbb{C}^5$, but I don't think that would even work here because it's asking for a generalized proof given two subspaces of $\mathbb{C}^8$.
But here's where I'm really stuck. The intersection of two subspaces is itself a subspace, right? So then the three subspace properties hold (Additive Identity, Closure under Addition, Closure under Scalar Multiplication).
Since the question reads for any z ∈ $\mathbb{C}$, I'm thinking well if we let $z=a+bi$, then it's possible that $b=0$, at which point all we're dealing with is scalar multiplication, $\hspace{1mm}$ i.e. $\hspace{1mm}$ $u=aw$ $\hspace{2mm} ∀ a ∈ \mathbb{R}$.
But this is totally allowed since we're working within a subspace.
Perhaps I need $u$ & $w$ to both be linearly independent, which would mean I should look at the bases of $U$ and $W$? But then what would the intersection look like? Thank you for taking the time to assist.
You're right with your second line of inquiry. Indeed, linear independence implies this condition (certainly if $u = zw$, then $u - zw = 0$ is a non-trivial linear combination of $u$ and $w$ that produces the zero vector). So, it suffices to show that $\dim(U \cap W) \ge 2$; if this is true, we can simply take the first two vectors of the basis to get two linearly independent vectors $u$ and $w$.
Now, the key result here is to use the following formula: $$\dim(U) + \dim(V) - \dim(U \cap V) = \dim(U + V).$$ Note that $U + V$ is a subspace of $\Bbb{C}^8$, and hence has dimension at most $8$. Also, $\dim(U) + \dim(V) = 5 + 5 = 10$ by assumption. Thus $$\dim(U \cap V) = 10 - \dim(U + V) \ge 10 - 8 = 2,$$ as required.