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Assuming we have 2 subspaces, $\mathbb W$ and $\mathbb U$ of $\mathbb V$.

how to get thier intersection?

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What do you mean by "get their intersection"? – Orat Feb 28 '13 at 9:46
Like this. $\mathbb{W} \cap \mathbb{U}$ – AJY Jun 10 '15 at 23:56
up vote 9 down vote accepted

Suppose that $\mathbb{W}=\operatorname{span}(w_1,w_2,\ldots,w_m)$ and $\mathbb{U}=\operatorname{span}(u_1,u_2,\ldots,u_n)$ where each of these is a minimum basis for the respective sets. If a vector $a$ is in the intersection of $\mathbb{W}$ and $\mathbb{U}$, it must be able to be expressed as a linear combination of the spanning set of $\mathbb{W}$ and $\mathbb{U}$. i.e. $$ a=c_1w_1+c_2w_2+\ldots+c_kw_m \text{ and } a=d_1u_1+d_2u_2+\ldots+d_ku_n $$ for some $c_i$ and $d_i$ to be determined. Since $a$ is arbitrary, you must have $$ c_1w_1+c_2w_2+\ldots+c_kw_m=d_1u_1+d_2u_2+\ldots+d_ku_n $$ which is equivalent to $$ c_1w_1+c_2w_2+\ldots+c_kw_m+e_1u_1+e_2u_2+\ldots+e_ku_n=0 $$ where $e_k=-d_k$. Since $a$ was an arbitrary element of $\mathbb{W}$ and $\mathbb{U}$, this system of linear equations must have a solution. Thus, the intersection of $\mathbb{W}$ and $\mathbb{U}$ is the nullspace of the matrix $$ [w_1,w_2,\ldots,w_m,u_1,\ldots,u_n]. $$

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If your subspaces are defined by two systems of homogeneous linear equation, just combine them in a single system, and the set of solutions will be the intersection.

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Andreas Cartani's solution only works when the subspace is given as the solution of a linear equation system.

Daryl's solution is incomplete. Let $\operatorname{span}(x_1,\dots,x_l)$ be the nullspace of the last matrix, let $y_1, \dots, y_l$ be the first $m$ components of each vector $x_1, \dots, x_l$, i. e. $x_i \in \mathbb R^{m+n}$ but $y_i \in \mathbb R^m$. Let $z_i = U y_i$ where $U = (u_1, \dots, u_m)$ is the matrix with columns $u_1$ to $u_m$, then $\operatorname{span}(z_1, \dots, z_l) = \mathbb U \cap \mathbb W$. (The $x_i$ are only the $c_j$ and $e_j$ but the quantities of interests are the $a$'s which correstpond to the $z_i$.)

Alternatively, let $\mathbb U = \operatorname{span}(u_1, \dots, u_n)$ and $\mathbb W = \operatorname{span}(w_1, \dots, w_m)$. Let $U = (u_1, \dots, u_n)$ be the matrix with columns $u_1$ to $u_n$ and $W = (w_1, \dots, w_m)$ be the matrix with columns $w_1$ to $w_m$. Solve the equation system $U^T x = 0$ for $x$ and let $\mathbb X = \operatorname{span}(x_1, \dots, x_k)$ be its solution. Solve analogously $W^T y = 0$ for $y$ and let $\mathbb Y = \operatorname{span}(y_1, \dots, y_l)$ its solution. Let $Z = (x_1, \dots, x_k, y_1, \dots, y_l)$ be the matrix consisting of the columns $x_1$ to $x_k$ and $y_1$ to $y_l$. The solution of the equation system $Z^T z = 0$ in $z$ is the subspace $\mathbb Z$ in question for which $\mathbb Z = \mathbb U \cap \mathbb W$.

The reason for the last solution to work ist that it holds in general $\ker(A^T) = \operatorname{ran}(A)^\perp$ and $(\mathbb A \cap \mathbb B)^\perp = \mathbb A^\perp + \mathbb B^\perp$ for matrices $A$ and subspaces $\mathbb A$ and $\mathbb B$ (as well as ${(\mathbb A^\perp)}^\perp = \mathbb A$). Here we have $\mathbb U = \operatorname{ran}(U)$, $\mathbb W = \operatorname{ran}(W)$, $\mathbb X = \ker(U^T)$, $\mathbb Y = \ker(W^T)$ and $\mathbb Z = \ker(Z^T)$. $\operatorname{ran}(Z) = \mathbb X + \mathbb Y$ completes the puzzle.

Daryl's solution is to prefer when the two subspaces have low dimension (compared to their orthogonal complement). The other solution is to prefer for high dimensional subspaces (compared to their orthogonal complement).

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