Linear Algebra, Vector Space: how to find intersection of two subspaces ?

The answer is Sp(1, 1, -10).

Call the vectors a = (1,1,2); b = (2,2,1); c = (1, 3, 4); d = (2, 5, 1) then any vector in $U \bigcap W$ must be a linear combination of a and b, and at the same time a linear combination of c and d.

This gives you the following simultaneous equations:

$\alpha$(1,1,2) + $\beta$(2,2,1) = $\gamma$(1,3,4) + $\delta$(2,5,1). Solve them by the method of your choice to get a parametric equation for a line.

Let $U$ and $V$ be two sub spaces(in matrix form: columns as basis vectors). Let $z$ be a vector that lies in intersection of these two sub spaces. Then $\exists$ two coeff vectors $x,y$ such that \begin{align*} z = Ux &= Vy \\ Ux &= Vy \\ U^T Ux &= U^T Vy \\ x &= (U^T U)^{-1} U^TVy \quad \text{ and similarly } \quad y = (V^T V)^{-1} V^TUx \\ \text{Thus} \quad x &= (U^T U)^{-1} U^TV(V^T V)^{-1} V^TUx \\ x &= \hat{M}x, \end{align*} where, $$\hat{M} =(U^T U)^{-1} U^TV(V^T V)^{-1} V^TU $$ We can see that $x$ is the Eigen vector of $\hat{M}$ corresponding to Eigen value $1$. Thus required basis is the set of independent vectors such that

$$\{Ux:\hat{M}x = x\}$$

In another way, let $\hat{M_1} = \Big(U(U^T U)^{-1} U^T \Big) \Big(V(V^T V)^{-1} V^T \Big) = P_UP_V$. The required basis is the set of independent vectors such that

$$\{s:\hat{M_1}s = s\}$$

Geometrically $P_U=U(U^T U)^{-1} U^T$ and $P_V =V(V^T V)^{-1} V^T $ are projection matrices onto the sub spaces $U$ and $V$ respectively. So we can see that the basis elements are those independent vectors, which remain unchanged after two projections, corresponding to the given two sub spaces.

Even though this is years after the question, posting following for completeness. This is very similar to solution by Pavan Karjol.

Let $P_u$ and $P_v$ be two projection matrices onto the two sub-spaces spanned by the columns of $U$ and $V$. These are calculated as $P_u=U(U^T U)^{-1} U^T$ and $P_v =V(V^T V)^{-1} V^T.$

Let $z$ be a vector that lies in intersection of these two sub-spaces. Since $z$ is on both sub-spaces, projecting it onto each subspace doesn't change the vector.

Therefore, $z = P_uz$ and $z = P_vz$.

Substituting one in the other gives $z = P_uP_vz$ therefore, $0 = (P_uP_v-I)z$ .

Meaning, $z$ is in the null space of $M=(P_uP_v - I)$. Perform the eigenvalue decomposition on $M$ and select the eigenvectors corresponding to zero or nearly zero eigenvalues and that forms an orthogonal basis for the intersection of given sub-spaces.