# How to find basis of intersection of 2 vector spaces with given basis?

Suppose we have vector spaces $\mathbb W$ and $\mathbb V$.

A given basis for $\mathbb W$ is $\{(1, 1, 0, -1), (0, 1, 3, 1)\}$, and a given basis for $\mathbb V$ is $\{(1, 2, 2, -2), (0, 1, 2, -1)\}$.

How to find a basis for $\mathbb W\cap\mathbb V$?

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To get the best possible answers, you should explain what your thoughts on the problem are so far. That way, people won't tell you things you already know, and they can write answers at an appropriate level; also, people tend to be more willing to help you if you show that you've tried the problem yourself. – Zev Chonoles Mar 7 '13 at 21:58
@ZevChonoles Thank you for this comment. in fact, I have no idea how to try to solve this, thus, it didn't tried anything. moreover, I would like to get tutorial how to do o. – Billie Mar 7 '13 at 22:00
@user1798362 Can you find $\Bbb W \cap \Bbb V$? I mean the actual set, not a basis. – Git Gud Mar 7 '13 at 22:01
@GitGud I guess I do. – Billie Mar 7 '13 at 22:02
@user1798362 Start by findint, simplyfing the conditions that define the set and post your work on the question. Or just read the answer below. – Git Gud Mar 7 '13 at 22:03

Take any $\vec x\in\Bbb V\cap\Bbb W.$ Since $\vec x\in\Bbb V,$ then there exist unique scalars $\alpha,\beta$ such that $$\vec x=(\alpha,2\alpha+\beta,2\alpha+2\beta,-2\alpha-\beta).\tag{\clubsuit}$$ Similarly, $\vec x\in\Bbb W,$ so there exist unique scalars $\gamma,\delta$ such that $$\vec x=(\gamma,\gamma+\delta,3\delta,-\gamma+\delta).\tag{\diamondsuit}$$ By $(\clubsuit)$ and $(\diamondsuit),$ we have \begin{align}\vec 0 &= \vec x-\vec x\\ &= (\alpha-\gamma,2\alpha+\beta-\gamma-\delta,2\alpha+2\beta-3\delta,-2\alpha-\beta+\gamma-\delta).\end{align}\tag{\heartsuit} Since $0=\alpha-\gamma$ (why?) then $\gamma=\alpha,$ so replacing $\gamma$ with $\alpha$ in $(\heartsuit)$ gives us \begin{align}\vec 0 &= (\alpha-\alpha,2\alpha+\beta-\alpha-\delta,2\alpha+2\beta-3\delta,-2\alpha-\beta+\alpha-\delta)\\ &= (0,\alpha+\beta-\delta,2\alpha+2\beta-3\delta,-\alpha-\beta-\delta).\end{align}\tag{\spadesuit} Since $0=\alpha+\beta-\delta$ (why?), then $\delta=\alpha+\beta,$ so replacing $\delta$ with $\alpha+\beta$ in $(\spadesuit)$ gives us \begin{align}\vec 0 &= \bigl(0,\alpha+\beta-[\alpha+\beta],2\alpha+2\beta-3[\alpha+\beta],-\alpha-\beta-[\alpha+\beta]\bigr)\\ &= (0,0,-\alpha-\beta,-2\alpha-2\beta).\end{align}\tag{*} Since $0=-\alpha-\beta$ (why?), then $\beta=-\alpha,$ and so replacing $\beta$ with $-\alpha$ in $(\clubsuit),$ we have \begin{align}\vec x &= \bigl(\alpha,2\alpha+[-\alpha],2\alpha+2[-\alpha],-2\alpha-[-\alpha]\bigr)\\ &= (\alpha,\alpha,0,-\alpha).\end{align}

Hence, every vector in $\Bbb V\cap\Bbb W$ is a scalar multiple of $(1,1,0,-1).$

On the other hand, every scalar multiple of $(1,1,0,-1)$ is immediately in $\Bbb W$ since $(1,1,0,-1)$ is, and noting that $(1,1,0,-1)=(1,2,2,-2)-(0,1,2,-1)$, then $(1,1,0,-1)$ is in $\Bbb V$, as well, and so every scalar multiple of $(1,1,0,-1)$ is in $\Bbb V.$ Thus, every scalar multiple of $(1,1,0,-1)$ is a vector in $\Bbb V\cap\Bbb W.$

By double inclusion, $\Bbb V\cap\Bbb W$ is precisely the set of scalar multiples of $(1,1,0,-1),$ and so a convenient basis for $\Bbb V\cap\Bbb W$ would be $$\bigl\{(1,1,0,-1)\bigr\}.$$

Let me sum up the general idea, here, so you can hopefully apply it in other cases. Let's suppose that we've got spaces $\Bbb V$ and $\Bbb W$ of dimension $m$ and $n,$ respectively--assume that $m\leq n$ without loss of generality--with given bases $\{\vec v_1,...,\vec v_m\}$ and $\{\vec w_1,...,\vec w_n\}$ (respectively).

$\Bbb V\cap\Bbb W$ is necessarily a subspace of $\Bbb V$ and of $\Bbb W.$ The idea, then, is to start with a general vector $\vec x$ in $\Bbb V\cap\Bbb W,$ then write this as a general vector in $\Bbb V$ and as a general vector in $\Bbb W$--that is write $$\vec x=\alpha_1\vec v_1+\cdots\alpha_m\vec v_m\quad\text{ and }\quad \vec x=\beta_1\vec w_1+\cdots+\beta_n\vec w_n$$ for some unique scalar $\alpha_i$s and $\beta_j$s.

Now, since $\Bbb V\cap\Bbb W$ is a subspace of $\Bbb V$, then it has dimension of at most $m$, so we'll be able to eliminate most of the scalars by setting $$\vec0=\left(\alpha_1\vec v_1+\cdots\alpha_m\vec v_m\right)-\left(\beta_1\vec w_1+\cdots+\beta_n\vec w_n\right),\tag{#}$$ then repeatedly back-substituting as above. In particular, we'll be able to eliminate all of the $\beta_j$s (since we assumed $m\leq n$), and once we've done that, we'll want to eliminate as many of the $\alpha_i$s as possible.

At some point, if we keep substituting into the chain of vector equations that started with $(\#)$, then we may end up with the equation $\vec 0=\vec 0$. For example, if we'd substituted $\beta=-\alpha$ into $(*)$ above, that's what we'd get. If that happens, we'll know we've reduced as far as we can, and we'll want to use the relationships we've determined between the various $\alpha_i$s along the way to rewrite $x$ in terms of as few $\alpha_i$s as possible--that's what we did above when we subbed back into $(\clubsuit)$--and that will show that every vector in $\Bbb V\cap\Bbb W$ can be written as a linear combination of some vectors, which should comprise the basis for $\Bbb V\cap\Bbb W$. (It's worth checking, of course.) Consider the spaces $\Bbb V$ and $\Bbb W$ with respective bases $$\{(1,0,0,0),(1,2,2,-2),(0,1,2,-1)\}$$ and $$\{(0,1,0,-1),(1,1,0,-1),(0,1,3,1)\}$$ for an example where we still have more than one $\alpha_i$ left at the end, and see if you can pick out appropriate basis vectors in that case.

Now, it might be that we end up discovering that all the $\alpha_i$s are $0$. In that case, $\Bbb V\cap\Bbb W=\{\vec 0\}$, and the basis is the empty set. Consider for example the spaces $\Bbb V$ and $\Bbb W$ with respective bases $\{(1,1)\}$ and $\{(1,-1)\}$ to see how this can happen.

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Thank you but I have question - "Since 0=−α−β (why?), then β=−α, and so replacing β with −α, we have" - How did you got from "(0,0,−α−β,−2α−2β)" to "(α,α,0,−α)"? the 2 first vector wasn't being 0? – Billie Mar 9 '13 at 15:26
Notice which vector I made that substitution in. The first two I made the substitution in $0=x-x$, but not that one. – Cameron Buie Mar 9 '13 at 15:29
Sorry, I still don't get it. – Billie Mar 9 '13 at 15:37
I made the last substitution into $x$, not $0=x-x$. We had $$x=(\alpha,2\alpha+\beta,2\alpha+2\beta,-2\alpha-\beta),\tag{\clubsuit}$$ yes? Once we found out that we needed $\beta=-\alpha,$ we were able to rewrite $x$ in terms of $\alpha$ only by substituting $-\alpha$ for $\beta$ in $(\clubsuit).$ Do you see it now? – Cameron Buie Mar 9 '13 at 15:54
You'd better highlight it in your answer .. It wasn't clear. But now it is. Great. thank you very much. – Billie Mar 9 '13 at 16:44

$\mathbb W \cap \mathbb V$ is either $\mathbb W (=\mathbb V)$, a line that is common to the two, or just the origin. If the two spaces are the same, you can express both basis vectors of one in terms of the vectors of the other. So write $(1,1,0,-1)=a(1,2,2,-2)+b(0,1,2,-1)$, then for the other. If you can find $a$ and $b$ the spaces are the same and either of the bases you have will work. If not, try to find a vector in the intersection by writing $c(1,1,0,-1)+d(0,1,3,1)=a(1,2,2,-2)+b(0,1,2,-1)$ and see if you can find a solution. If so, the vector that is the solution is a basis. Else the intersection is only the origin, which has no basis.

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