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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 $1$.

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

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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 $(1,2,2,-2)$

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.

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.

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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