# To determine basis for $V \cap W$

Let $V$ be vector subspace of$R^{4}$ spanned by vectors $( 1,1,1,-1)$ and $(1,-1,0,1)$. Let $W$ be another vector subspace of $R^{4}$ spanned by $(1,1,-1,1)$ and $(1,3,4,-5)$. Determine basis of $V \cap W.$

What I did is that I found out dimension of$V \cap W$ which is 1. Now I need a vector which is spanned by both of basis .I tried subtracting basis of first subspace and checking whether it can be written as linear combination of other basis but I am not geting it. Thanks for help

• Have you tried setting a general linear combination of the spanning vectors in V equal to a general linear combination of spanning vectors in W and then algebraically deriving some requirements for your needed vector? – Jonathan Hebert Jan 7 '15 at 16:41
• @Doop yes i did and i got 4 equations in 2 unknowns .but they are not giving consistent solution – godonichia Jan 7 '15 at 16:43

If the dimension of $V \cap W$ is $1$, then any singleton $\{v\} \subset V \cap W$ is a basis, except for the singleton containing the zero vector.

To find such a vector, we find one that can be expressed as a linear combination of the vectors we know to span $V$ and $W$. Thus, let $a, b, c, d \in R$.

$a(1,1,1,-1) + b(1,-1,0,1) = c(1,1,-1,1) + d(1,3,4,-5)$

$\Rightarrow (a+b, a-b, a, b-a) = (c+d, c+3d, -c+4d, c-5d)$

Leaving us with four equations:

$a + b = c + d, a - b = c + 3d, a = 4d - c, a + b = c - 5d$

From which we can deduce $a = 3c, b = -c, c = d$ as a sufficient restriction. With that restriction, let $a = 1$ so that $a = 1, b = -\frac{1}{3}, c = \frac{1}{3}, d = \frac{1}{3}$.

This gives $\{(\frac{2}{3}, \frac{4}{3}, 1, -\frac{4}{3})\}$ a one of many possible bases.