# How to find a basis for the intersection of two vector spaces in $\mathbb{R}^n$?

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

I will use the same ideas as this other answer, but will add some more detail on some of the steps.

Let $$\mathcal U$$ and $$\mathcal V$$ be two finite-dimensional vector spaces. I want to find a basis for the intersection $$\mathcal U\cap\mathcal V$$.

Let $$U$$ and $$V$$ be matrices whose columns are the basis vectors of $$\mathcal U$$ and $$\mathcal V$$, respectively. The problem is then equivalent to that of characterising $$\operatorname{Range}(U)\cap \operatorname{Range}(V)$$. In other words, the problem is that of finding the non-zero solutions for $$x,y$$ to the matrix equation $$Ux=Vy.\tag A$$ Indeed, $$z\in\operatorname{Range}(U)\cap \operatorname{Range}(V)$$ iff there are $$x,y$$ such that $$z=Ux=Vy$$.

Now, to solve (A) we can define $$A\equiv(U|-V)$$ (this is the matrix with columns the full set of the vectors in both the bases of $$\mathcal U$$ and $$\mathcal V$$), and find its nullspace. Indeed, $$AX=0$$ where $$X\equiv\begin{pmatrix}x\\y\end{pmatrix}$$ implies $$Ux=Vy$$.

Once we have a full basis set for the nullspace of $$A$$, in the form of an orthonormal set of vectors $$\{X_i\}$$ (with each $$X_i$$ corresponding to a pair $$x_i,y_i$$), we can compute the corresponding set of vectors in the intersection that we are looking for, by simply computing $$w_i\equiv Ux_i=Vy_i$$ for each $$i$$.

Now to prove that $$\{w_i\}_i$$ is linearly independent. Suppose $$\sum_i c_i w_i=0$$. Then $$U(\sum_i c_i x_i)=0$$ and $$V(\sum_i c_i y_i)=0$$. But because $$\operatorname{Ker}(U)=\operatorname{Ker}(V)=\{0\}$$, this means that $$\sum_i c_i x_i=\sum_i c_i y_i=0$$. But this is, in turn, equivalent to $$\sum_i c_i X_i=0$$, and because $$\{X_i\}$$ is a linearly independent set, this implies $$c_i=0$$.

We conclude that $$\{w_i\}$$, where $$w_i\equiv U x_i=V y_i$$ and $$\{(x_i, y_i)\}_i$$ is a basis for $$\operatorname{Ker}[(U|-V)]$$, is a basis for $$\mathcal U \cap\mathcal V$$.