We have the following vectors:
$v_1 =\begin{bmatrix}1 \\ 1 \\ 2 \\ 2\end{bmatrix}$, $v_2 =\begin{bmatrix}0 \\ 1 \\ 0 \\ -1\end{bmatrix}$, $v_3 =\begin{bmatrix}1 \\ 0 \\ 2 \\ 3\end{bmatrix}$, $v_4 =\begin{bmatrix}-1 \\ 1 \\ 1 \\ 0\end{bmatrix}$, $v_5 =\begin{bmatrix}-1 \\ 1 \\ 4 \\ 4\end{bmatrix}$
Define the subspaces $V$ and $W$ as $V=span\{v_1,v_2,v_3\}$ and $W=span\{v_4,v_5\}$
The question is to find a nonzero vector $u$ that belongs to both $V$ and $W$.
I've done the calculations by finding that $span\{v_1, v_2\}$ and $span\{v_4, v_5\}$ are bases for $V$ and $W$ respectively and by setting $u=\alpha_1v_1 +\alpha_2v_2$, $u=\beta_1v_4 +\beta_2v_5$ and setting those equal to find $u$ is any vector of the form $u=\lambda\begin{bmatrix}1 \\ -1 \\ 2 \\ 4\end{bmatrix}$
My question is whether there is a simpler way to do this that I may be missing. We haven't covered this in class, and previous parts of the question focus on the column space of the matrix formed by the 5 vectors. This method seems perhaps a bit too "brute force", is there any intuition I'm missing?
Best Answer
Consider the set $\{v_1, v_2, v_4, v_5\}$ and try to express each vector as a linear combination of its predecessors. $v_1$ and $v_2$ are linearly independent so the first vector for which this might be possible is $v_4$. However, by inspection we see that it isn't possible to express $v_4$ as a linear combination $v_1$ and $v_2$.
Next we try to express $v_5$ as a linear combination of $v_1, v_2, v_4$. We get that $$v_5 = v_1-2v_2+2v_4$$
so $$\underbrace{v_1-2v_2}_{\in V} = \underbrace{-2v_4+v_5}_{\in W}$$ is your vector.