Calculating a basis of vector space $U \cap V$

Question

So I have two vector spaces:
$ U := \langle(1,2,1,2), (1,2,3,3), (1,2,2,3)\rangle $ and
$ V := \langle(2,0,2,1), (3,2,3,2), (0,4,0,1)\rangle $

I was able to calculate the base of both $U$ and $V$:
$ B_U = \langle(1,2,1,2), (1,2,3,3), (1,2,2,3)\rangle $ since the vectors linearly independent.
$ B_V = \langle(2,0,2,1), (3,2,3,2))\rangle $ since you can write $(0,4,0,1)$ as $2*(3,2,3,2) - 3*(2,0,2,1)$.

However, I have no clue for to do it for $U \cap V$. Could you please point out how to go about doing that and/or giving me an example? Thanks in advance.

This also helped me – Christian Schnorr Dec 09 '13 at 10:42 — Christian Schnorr, Dec 09 '13 at 10:42

score 3 · Accepted Answer · answered Dec 08 '13 at 23:31

3

Note that there must be some overlap, since $U$ is $3$-dimensional and $V$ is $2$-dimensional. So, either $V\subseteq U$, or they intersect in a one dimensional subspace. Can you write both basis vectors of $V$ in terms of the basis for $U$? If not, then the intersection is one-dimensional. I'd just look at what vectors are in $U$ and $V$. For instance $(0,0,2,1)$ and $(0,0,1,0)$ are in $U$. Thus $(0,0,0,1)$ is also in $U$.

answered Dec 08 '13 at 23:31

J126

17,451

I found the basis of $U \cap V$ to be $(1,2,-1,1)$ - would that be correct? – Christian Schnorr Dec 09 '13 at 10:44
@Jenox I get $(1,2,1,1)$. – J126 Dec 09 '13 at 13:02

Calculating a basis of vector space $U \cap V$

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