# How to find basis for intersection of two vector spaces

What is the general way of finding the basis for intersection of two vector spaces?

Suppose I'm given the bases of two vector spaces U and W: $$\mathrm{Base}(U)= \left\{ \left(1,1,0,-1\right), \left(0,1,3,1\right) \right\}$$ $$\mathrm{Base}(W) =\left\{ \left(0,-1,-2,1\right), \left(1,2,2,-2\right) \right\}$$

I already calculated $U+W$, and the dimension is $3$ meaning the dimension of $U \cap W$ is $1$.

The answer is supposedly obvious, one vector is the basis of $U \cap W$ but how do I calculate it?

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The procedure can be described as: (i) put down a list of the basis vectors of $U$ and $W$ (ii) Perform base changes on both list until a common vector appears. Keep that one. Repeat for the other vectors, until no common vectors can be found anymore. – shuhalo Mar 7 '11 at 0:16
A = (Base(U) | Base(W)) as columns. The answer is the nullspace of A? – Annan Jun 25 '12 at 10:34
@Annan I think what it ends up meaning is that the basis for the intersection will be basis vectors for example from U which are linear combinations of basis vectors from W, or the other way around. Another way of thinking about it is that you're looking for vectors which are in the column space / span of both sets which I think can only happen when some basis vector of one set is in the column space of the other set. – Robert S. Barnes Nov 5 '12 at 7:28

## 4 Answers

Assume $\textbf{v} \in U \cap W$. Then $\textbf{v} = a(1,1,0,-1)+b(0,1,3,1)$ and $\textbf{v} = x(0,-1,-2,1)+y(1,2,2,-2)$.

Since $\textbf{v}-\textbf{v}=0$, then $a(1,1,0,-1)+b(0,1,3,1)-x(0,-1,-2,1)-y(1,2,2,-2)=0$. If we solve for $a, b, x$ and $y$, we obtain the solution as $x=1$, $y=1$, $a=1$, $b=0$.

so $\textbf{v}=(1,1,0,-1)$

You can validate the result by simply adding $(0,-1,-2,1)$ and $(1,2,2,-2)$

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What would you do if U had dimension 3? So you would have 3+2=5 variables and 4 equations. – Martin Thoma Jan 7 '12 at 14:15

Parameterize both vector spaces (using different variables!) and set them equal to each other. Then you will get a system of 4 equations and 4 unknowns, which you can solve. Your solutions will be in both vector spaces.

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It is a one dimensional vector space, so find any non-zero vector which is in both spaces and it will be a basis.

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any nonzero vector that is in both... – Arturo Magidin Mar 6 '11 at 22:45
Duly noted, thanks! – Jon Beardsley Mar 7 '11 at 20:44

The comment of Annan with slight correction is one possibility of finding basis for the intersection space $U \cap W$, the steps are as follow:

1) Construct the matrix $A=\begin{pmatrix}\mathrm{Base}(U) & | & -\mathrm{Base}(W)\end{pmatrix}$ and find the basis vectors $\textbf{s}_i=\begin{pmatrix}\textbf{u}_i \\ \textbf{v}_i\end{pmatrix}$ of its nullspace.

2) For each basis vector $\textbf{s}_i$ construct the vector $\textbf{w}_i=\mathrm{Base}(U)\textbf{u}_i=\mathrm{Base}(W)\textbf{v}_i$.

3) The $span(\textbf{w}_1,\ \textbf{w}_2,...,\ \textbf{w}_r)$ constitute the basis for the intersection space.

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