Let \(\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3}\) and \(\set{\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) be bases for \(U\) and \(V\) (respectively). Then, the set \(\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3,\,\vect{v}_1,\,\vect{v}_2,\,\vect{v}_3}\) is linearly dependent, since Theorem G says we cannot have 6 linearly independent vectors in a vector space of dimension 5. So we can assert that there is a nontrivial relation of linear dependence,
\begin{equation*}
a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3+b_1\vect{v}_1+b_2\vect{v}_2+b_3\vect{v}_3=\zerovector
\end{equation*}
where \(a_1,\,a_2,\,a_3\) and \(b_1,\,b_2,\,b_3\) are not all zero.

We can rearrange this equation as
\begin{equation*}
a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3
\end{equation*}
This is an equality of two vectors, so we can give this common vector a name, say \(\vect{w}\text{,}\)
\begin{equation*}
\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3
\end{equation*}
This is the desired nonzero vector, as we will now show.

First, since \(\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3\text{,}\) we can see that \(\vect{w}\in U\text{.}\) Similarly, \(\vect{w}=-b_1\vect{v}_1-b_2\vect{v}_2-b_3\vect{v}_3\text{,}\) so \(\vect{w}\in V\text{.}\) This establishes that \(\vect{w}\in U\cap V\) (Definition SI).

Is \(\vect{w}\neq\zerovector\text{?}\) Suppose not, in other words, suppose \(\vect{w}=\zerovector\text{.}\) Then
\begin{equation*}
\zerovector=\vect{w}=a_1\vect{u}_1+a_2\vect{u}_2+a_3\vect{u}_3
\end{equation*}
Because \(\set{\vect{u}_1,\,\vect{u}_2,\,\vect{u}_3}\) is a basis for \(U\text{,}\) it is a linearly independent set and the relation of linear dependence above means we must conclude that \(a_1=a_2=a_3=0\text{.}\) By a similar process, we would conclude that \(b_1=b_2=b_3=0\text{.}\) But this is a contradiction since \(a_1,\,a_2,\,a_3,\,b_1,\,b_2,\,b_3\) were chosen so that some were nonzero. So \(\vect{w}\neq\zerovector\text{.}\)

How does this generalize? All we really needed was the original relation of linear dependence that resulted because we had “too many” vectors in \(W\text{.}\) A more general statement would be: Suppose that \(W\) is a vector space with dimension \(n\text{,}\) \(U\) is a subspace of dimension \(p\) and \(V\) is a subspace of dimension \(q\text{.}\) If \(p+q\gt n\text{,}\) then \(U\cap V\) contains a nonzero vector.