# The basis for intersection of two subspaces

I know there are other posts on how to do this, but I was wondering if there is a slicker method.

$$X = Span\{(1,1,0,-1),(1,2,3,0),(2,3,3,-1)\}$$ $$Y = Span\{(1,2,2,-2),(2,3,2,-3),(1,3,4,-3)\}$$

I want to find a basis for $$X \cap Y$$

The method I have been trying to use is by letting $$v \in X$$ and $$v \in Y$$ and attempting to solve the equation $$v - v = 0$$. However it seems very long and tedious. I was thinking there may be a better method.

Well, for this particular problem, there are some shortcuts that make life a bit easier. Let $$X = \operatorname{span} \left\{v_1,v_2,v_3 \right\}$$ and $$Y = \operatorname{span} \left\{w_1,w_2,w_3 \right\}$$, where $$v_1=(1,1,0,-1)$$, $$v_2=(1,2,3,0)$$ etc.

You could firstly note that $$(1,1,0,-1)+(1,2,3,0)=(2,3,3,-1) \implies v_3 \in \operatorname{span}\left\{v_1, v_2\right\},$$ and $$3(1,2,2,-2)-(2,3,2,-3)= (1,3,4,-3) \implies w_3 \in \operatorname{span} \left\{w_1, w_2 \right\},$$ and so your problem reduces to finding a basis for $$X \cap Y$$, where $$X = \operatorname{span} \left\{v_1,v_2 \right\}$$ and $$Y = \operatorname{span} \left\{w_1,w_2 \right\}$$.

Now, taking a similar approach with the resulting sets $$X$$ and $$Y$$, we could note that $$(2,3,2,-3) - (1,2,2,-2)=(1,1,0,-1) \implies v_1 \in \operatorname{span} \left\{w_1, w_2 \right\},$$ and so on.

• I see, so in the end our basis for X $\cap$ Y is $\{v_2,w_1,w_2\}$ since $v_2$ is not in the span of $w_1, w_2$ Dec 1, 2020 at 10:08
• You're right that $v_2 \notin \operatorname{span}\left\{w_1, w_2 \right\}$, but be careful when it comes to the answer. A vector $v$ is in $X \cap Y$ if $v \in X$ and $v \in Y$. That is, $v \in \operatorname{span}\left\{v_1, v_2 \right\}$ and $v \in \operatorname{span}\left\{w_1, w_2 \right\}$. So we know that $v_1 \in X \cap Y$ but $v_2 \notin X \cap Y$. Dec 1, 2020 at 10:37
• So surely we have that the basis is just the one vector $v_1$ as that is the only vector that we know is in both subspaces. Dec 1, 2020 at 11:23
• Right! Well done! Dec 1, 2020 at 23:42

This is a perfect situation where changing the point of view can provide some good insight. You are trying to find the intersection of two vectors subspaces of $$\mathbb{R}^4$$ and you are given those spaces as the spans of sets of vectors. In other words you are given X and Y as the column spaces of matrix $$V=[v_i]$$ an $$W=[w_i]$$ (where the columns are the vectors you gave above). At first glance it's not clear how to find the intersection. Indeed, what do we do? Write a matrix / take echelon form? But there are two matrices... Hmm... However...

In general - there are straightforward efficient ways to find the null space or column space of a matrix. In this case, thinking about null spaces turns out to help a lot!

Indeed, suppose instead we were give these (same) spaces not as the column space of matrices, but instead as the null space of a matrix, say $$X = \mathrm{Nul }A$$ and $$Y = \mathrm{Nul }B$$, then it would be easy to find the intersection of $$X$$ and $$Y$$. Indeed, we can think of $$X$$ as solutions to $$Ax=0$$ and $$Y$$ as solutions to $$Bx=0$$, and so to find $$X\cap Y$$ we just need to combine all of these linear equations into one big system, namely the matrix with $$A$$ and $$B$$ stacked on top of each other. Then if we solve: $$\begin{bmatrix}A \\ B \end{bmatrix} x = 0$$ then the set of solutions will be $$X\cap Y$$ so we just need to row-reduce this bigger matrix and then we can read off the null space from that.

Ok, I know what you're thinking - how do we find the matrices $$A$$ and $$B$$. Well this is where some more linear algebra magic can come to the rescue. So the question is: Given that $$X = Col(V)$$ how can you find an $$A$$ so that $$X = Nul(V)$$. Well here's a sneaky trick, the answer is Find a basis for the null space of $$V^T$$ (a null space calculation) and then write these basis vectors as columns of a matrix, call it $$P$$. Then by construction we have that $$V^TP = 0$$. Now if we take the transpose we get $$P^TV = 0$$, which means that the columns of $$V$$ (i.e. our original vectors in your problem) are in the null space of $$P^T$$. So $$P^T$$ is the matrix that you want.''

I don't know how much linear algebra you've seen, and this is surely not the fastest way to do this, but you sound like you'd be interested in learning a bit more and I'd be happy to fill in more details.

In pseudocode if you want to flip this from a "span" problem to a "null space" problem the ideas you want are:

X = col(V); Y = col(W);

A = transpose ( null transpose( V ) );

B = transpose ( null transpose( W ) );

Then $$X\cap Y$$ = null space of $$\begin{bmatrix} A\\ B\end{bmatrix}$$