Finding a basis for the intersection of two vector subspaces.

Question

Suppose:

$V_1$ is the subspace of $\mathbb{R}^3$ given by $V_1 = \{(2t-s,t,t+s)|t,s\in\mathbb{R}\}$

and

$V_2$ is the subspace of $\mathbb{R}^3$ given by $V_2 = \{(s,t,s)|t,s\in\mathbb{R}\}$.

How could a basis and dimension for $V_1 \cap V_2$ be determined?

I thought of maybe setting the respective components of the two spaces equal to each other such that $2t-s = s$, $t=t$, and $t+s=s$.

Solving each of these equations it can be found that $2t=2s$, $t=t$, and $t=0 \implies s=0 \implies V_1 \cap V_2 = \{(0,0,0)\} \implies \text{dim}(V_1 \cap V_2)=0$.

This seems like it isn't right though, but I'm really uncertain.

Answer 1

Another possibility:

Let us first consider the set of points in $V_1\cap V_2$. Suppose $v\in V_1\cap V_2$. Then $v$ must be in the form of $(a,b,a)$ for some $a,b\in \mathbb{R}$. In addition. $(a,b,a)=(2t-s,t,t+s)$ for some $t,s\in \mathbb{R}$. Then we obtain the relation $2a=3b$. This shows that $V_1\cap V_2\subseteq \{ (a,2a/3,a) | a\in \mathbb{R} \}$.

Now we will show that $\{ (a,2a/3,a) | a\in \mathbb{R} \}\subseteq V_1\cap V_2$. Note that for $(t,s)=(2/3,1/3)$, we have that $(2t-s,t,t+s)=(1,2/3,1)$. Thus $(1,2/3,1)\in V_1\cap V_2$. Since $V_1\cap V_2$ is a subspace, we have that $\{ (a,2a/3,a) | a\in \mathbb{R} \}\subseteq V_1\cap V_2$

Hence $V_1\cap V_2= \{ (a,2a/3,a) | a\in \mathbb{R} \}$. This gives us the basis and dimension for $V_1\cap V_2$.

Answer 2

Here is one way:

The idea is to write $V_k$ as $\ker u_k^T$ for some $u_k$, then $v \in V_1 \cap V_2$ iff $v \in \ker \begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix}$. If $u_k$ are linearly independent, then if $u_3\neq 0$ is orthogonal to $u_1,u_2$, we see that $V_1 \cap V_2 = \operatorname{sp} \{ u_3 \}$.

In the following we use the fact that $(\ker C)^\bot = {\cal R} C^T$.

It should be straightforward to show that $\dim V_1 = \dim V_2 = 2$,

Hence $\dim V_1^\bot = \dim V_2^\bot = 1$.

Suppose $(x,y,z)^T \in V_1^\bot$, then we must have $t(2x+y+z) + s(z-x) = 0$ for all $s,t$. Hence $z=x$ and $y=-3x$. Letting $x=1$ gives $u_1=(1,-3,1)^T$ and hence $V_1^\bot = \operatorname{sp} \{u_1\}$, and so $V_1 = \ker u_1^T$.

Suppose $(x,y,z)^T \in V_2^\bot$, then we must have $s(x+z)+ty=0$ for all $s,t$. Hence $y=0, z=-x$. Letting $x=1$ gives $u_2=(1, 0 ,-1)^T$ and hence $V_1^\bot = \operatorname{sp} \{u_2\}$, and so $V_2 = \ker u_2^T$.

Hence $v \in V_1 \cap V_2$ iff $v \in \ker u_1^T$ and $v \in \ker u_2^T$ iff $u \in \ker \begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix}$. Solving $\begin{bmatrix} u_1^T \\ u_2^T \end{bmatrix} (x,y,z)^T = 0$ gives $z=x, y = {1 \over 3} 2x$. Choosing $x=3$ gives $u_3 = (3,2,3)^T$.

Hence $v \in V_1 \cap V_2$ iff $v \in \operatorname{sp} \{ u_3 \}$.

Answer 3

In general, to find the intersection of two subspaces, you express a vector in the intersection in two different ways and then solve the system of linear equations that result, a la copper.hat’s answer.

In the special case of $\mathbb R^3$, there’s a quick and dirty way to find the intersection of two different planes through the origin. Since their intersection will be orthogonal to both normals, you can get the answer by computing three cross products—two to compute the normals and another to find a vector orthogonal to them. With only 3 dimensions, this last vector must lie on the line of intersection.

To illustrate, the plane $V_1$ is spanned by $(-1,0,1)^T$ and $(2,1,1)^T$, so a normal is $(-1,0,1)^T\times(2,1,1)^T=(-1,3,-1)^T$, while $V_2$ is spanned by $(1,0,1)^T$ and $(0,1,0)^T$ with normal $(1,0,1)^T\times(0,1,0)^T=(-1,0,1)^T$. The cross product of these normals is $(-1,3,-1)^T\times(-1,0,1)^T=(3,2,3)^T$, just as was found by the general method.

This also gives a way to find the intersection of two lines in the plane or the line through two points in the plane: use homogeneous coordinates and take their cross product.