Find a basis for the intersection of 2 vector spaces

So I'm given 2 bases for the vector spaces $U$ and $V$. Suppose dim$(U \cap V) \geq2$, then how do I find a basis for it? Thanks in advance.

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First find $U\cap V$ – Belgi Nov 19 '12 at 16:02
Thanks but how? Isn't it like let $x \in U \cap V$ then $x \in U$ and $x \in V$? – drawar Nov 19 '12 at 16:18
That depands on whats $U,V$...you need to work with the definition. thats what you wrote in the comment. – Belgi Nov 19 '12 at 16:28

Presumably $U,V$ are contained in a finite dimensional vector space $Z$. Let $W = U \cap V \subset Z$.

Let $u_1,...,u_{n_U}$ and $v_1,...,v_{n_V}$ be bases of $U,V$ respectively. Define $A: \mathbb{R}^{n_U} \to Z$ by $A x = \sum_k x_k u_k$ and similarly $B: \mathbb{R}^{n_V} \to Z$ as $B y = \sum_k y_k v_k$. Note that ${\cal R} A = U$, and ${\cal R} B = V$ and so $W = {\cal R} A \cap {\cal R} B$.

Define $\Gamma: \mathbb{R}^{n_U} \times \mathbb{R}^{n_V}\to Z$ by $\Gamma (x,y) = Ax-By$, and let $N = \ker \Gamma$. Let $\Pi_x : \mathbb{R}^{n_U} \times \mathbb{R}^{n_V}\to \mathbb{R}^{n_U}$ be the projection $\Pi_x (x,y) = x$.

Note that $\Gamma$ is given by the matrix $\Gamma = \begin{bmatrix} u_1 & \cdots & u_{n_U} & v_1 & \cdots v_{n_V} \end{bmatrix}$.

Then $p \in W$ $\iff$ there exists $x,y$ such that $Ax=By$ and $p=Ax$ $\iff$ there exists $x,y$ such that $\Gamma (x,y) = 0$ and $p=Ax$ $\iff$ $n \in N$ and $p=A\Pi_x n$ $\iff$ $p \in A \Pi_x N$.

So a procedure would be to find a basis $\nu_1,...,\nu_{n_N}$ for $N = \ker \Gamma$, and select a maximal linearly independent subset of $A \Pi_x \nu_1,...,A \Pi_x \nu_{n_N}$.

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Thank you. Anyway in the book 'Schaum's outline of theory and problems of linear algebra', the author presented another method to find a basis for the intersection of 2 vector spaces as below: 1. Find a homogeneous system whose nullspace is generated by the basis vectors of U. 2. Find another homogeneous system whose nullspace is generated by the basis vectors of W. 3. Combine the two systems into a new system and solve for the basis of its nullspace, this basis is a basis for the intersection of U and W. Can you help me discuss the validity of this method? – drawar Nov 20 '12 at 2:36
Yes, it involves finding a basis for $V^\bot$, say $v_1,...,v_k$. Then $x \in V$ $\iff$ $\begin{bmatrix} v_1^T \\ \cdots \\ v_k^T \end{bmatrix} x =0$. Repeat for $U$ and stack one matrix on top of the other. Then any element in the stacked matrix null space must be in the null space of the original ones and vice versa. – copper.hat Nov 20 '12 at 3:03