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What is the general way of finding the basis for intersection of two vector spaces in $\mathbb{R}^n$?

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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    @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.2012-11-05

6 Answers 6

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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.2012-01-07
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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 set $ \{ \textbf{w}_1,\ \textbf{w}_2,...,\ \textbf{w}_r \} $ constitute the basis for the intersection space $ span(\textbf{w}_1,\ \textbf{w}_2,...,\ \textbf{w}_r) $.

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    @helios321 This is so because the basis vectors $\textbf{u}_i$ are linearly independent as those are part of basis themselves. You can check this one directly form the definition, just do $c_i \textbf{w}_i=\textbf{0}$ and show it is only possible when all $c_i$'s are zero.2018-05-16
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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...2011-03-06