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.