# Finding a vector that belong to two subspaces spanned by vectors.

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?

• Welcome to Mathematics Stack Exchange. Can you see that both $v_4$ and $v_5$, and therefore all elements of $W$, have their first component the additive inverse of the second component? Can you see that $v_1-2v_2$ also has that property? $v_1-2v_2$ is your vector! Commented Nov 26, 2019 at 4:13

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.