Find Intersection Of 2 Sub-spaces

Question

$U=span((1,1 ,0),(2 ,0 ,1))$ $W=span((1,1,1),(5,3,1))$

what is $U \cap W$?

can I find the rref of them both and to find the intersections? that mean

$U= \begin{pmatrix} 1 & 0 & 0.5 \\ 0 & 1 & -0.5 \end{pmatrix}$

$W= \begin{pmatrix} 1 & 0 & -1 \\ 0 & 1 & 2 \end{pmatrix}$

so is $U \cap W$=$Span \begin{pmatrix} 1 & 0 & 0 \\ 0 & 1 & 0 \end{pmatrix}$

See also: http://math.stackexchange.com/questions/598880/calculating-a-basis-of-vector-space-u-cap-v, http://math.stackexchange.com/questions/25371/how-to-find-basis-for-intersection-of-two-vector-spaces, http://math.stackexchange.com/questions/165361/linear-algebra-linear-transformation, http://math.stackexchange.com/questions/97880/, http://math.stackexchange.com/questions/139420/, http://math.stackexchange.com/questions/849433/ — Martin Sleziak, Aug 08 '14 at 06:08

score 2 · Accepted Answer · answered Aug 07 '14 at 09:46

2

Well, $U \cap W$ is the space consisting of all vectors that can be formed both as a linear combination of the elements of U and V. So, for any $\vec v \in U \cap W$: $$ \vec v = c_1 (1, 1, 0) + c_2 (2, 0, 1) $$ and $$ \vec v = d_1(1, 1, 1) + d_2(5, 3, 1)$$ where $c_1, c_2, d_1, d_2 \in \mathbb{R}$. Does that help?

answered Aug 07 '14 at 09:46

MathMajor

6,478

Or maybe it might be easier to use the vectors from the reduced matrix; the computations will be easier. (We have more zeroes.) – Martin Sleziak Aug 08 '14 at 05:56
@Toxicz I have 3 equations with 4 unknowns – gbox Sep 17 '14 at 08:33

score 1 · Answer 2 · edited Apr 13 '17 at 12:21

From the rrefs you calculated, you know that $\dim U=2$ and $\dim V=2$. (And this can be also seen without calculating rref: We have only two vectors generating a subspace, and neither of these vectors is a multiple of the other one, so they are linearly independent.)

If you make rref from the matrix containing all four vectors, you will find out, that $\dim(U+V)=3$. $$\begin{pmatrix} 1 & 0 & \frac12\\ 0 & 1 & -\frac12\\ 1 & 0 & -1\\ 0 & 1 & 2 \end{pmatrix}\sim \begin{pmatrix} 1 & 0 & 0\\ 0 & 1 & 0\\ 0 & 0 & 1\\ 0 & 0 & 0 \end{pmatrix} $$

So from the formula $\dim(U)+\dim(V)=\dim(U+V)-\dim(U\cap V)$ we can calculate $$\dim(U\cap V)=1.$$ So the intersection is generated by single vector.

If we notice that $$(5,3,1)=5(1,0,\frac12)+3(0,1,-\frac12) = 3(1,1,0)+(2,0,1)$$ then we see that the vector $(5,3,1)$ belongs to both $U$ and $V$, i.e., $(5,3,1)\in U\cap V$.

So now, without any other calculations, we can say that $$U\cap V = \operatorname{span}\{(5,3,1)\}$$

However, in this case we were simply lucky that we only needed one vector and we were able to guess a vector which belongs to both given subspaces.

If we want to avoid guesswork, a reasonable way to go is to solve a system of equations as suggested in Toxicz's answer.

Find Intersection Of 2 Sub-spaces

2 Answers2