I'm trying to find $V \cap W$ if:
$V = \text{span}( \{ (2,1,3,-1),(3,2,-1,2),(9,5,8,-1),(7,4,5,0) \} )$ and $W = \text{span}( \{(5,3,2,1),(6,4,-2,4),(1,1,-4,3),(4,2,6,-2) \})$.
Based on the given information, I've found the following basis of $V$ and $W$:
$B_V= \{ (2,1,3,-1),(3,2,-1,2) \} $ and $B_W= \{(5,3,2,1),(1,1,-4,3) \}$.
Because $V \cap W = \{x | x \in V \land x \in W \}$, vector $x$ can be written as:
$x= \alpha \cdot (2,1,3,-1) + \beta \cdot (3,2,-1,2) $ and
$x= \gamma\cdot (5,3,2,1) + \delta\cdot (1,1,-4,3) $.
Then we can write:
$ \alpha \cdot (2,1,3,-1) + \beta \cdot (3,2,-1,2) = \gamma\cdot (5,3,2,1) + \delta\cdot (1,1,-4,3)$ which yields in a system of four equations:
$$2 \alpha + 3 \beta = 5 \gamma + \delta $$
$$ \alpha + 2 \beta = 3 \gamma + \delta $$
$$ 3 \alpha – \beta = 2 \gamma – 4 \delta $$
$$ – \alpha + 2 \beta = \gamma + 3 \delta $$
Is this approach correct? Any hint to how to solve this system and in turn solve $V \cap W$?
Best Answer
No, you did not find the bases of $V$ and of $W$. What you did find was a basis of each of those spaces.
And your approach is fine. But it's easier to see that$$(5,3,2,1)=(2,1,3,−1)+(3,2,−1,2)$$and that$$(1,1,−4,3)=-(2,1,3,−1)+(3,2,−1,2).$$Therefore, $W\subset V$. It follows from this and from the fact that $\dim V=\dim W=2$ that $V=W$. So, $V\cap W=V=W$.