Linear Algebra, Vector spaces? W = Sp{(1,3,4),(2,5,1)} , U = Span({1,1,2),(2,2,1)} Find a group that spreads U^W.(The^ is like an Upside down "U", ^ means Union in other words) Also, another question about Complexes: If z1*z2 != 1, z1=z2=1, Does ( (z1+z2)/(1+z1*z2) ) Rational? 
Solved the question about complexes. any Ideas about the fist question regarding Vector spaces? 
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yes, intersection. 
Let the given vectors from $U$ be called $u_1,u_2$ and the vectors from $W$ be called $w_1,w_2$. A vector $a\in U\cap W$ iff there exist numbers $x_1,x_2$, $y_1,y_2$ such that \[ a=x_1u_1+x_2u_2=y_1w_1+y_2w_2 \] Let us consider the system of linear equations \[ x_1\begin{pmatrix} 1\\ 1\\ 2\end{pmatrix}+ x_2\begin{pmatrix} 2\\ 2\\ 1\end{pmatrix}+ z_1\begin{pmatrix} 1\\ 3\\ 4\end{pmatrix}+ z_2\begin{pmatrix} 2\\ 5\\ 1\end{pmatrix}= \begin{pmatrix} 0\\ 0\\ 0\end{pmatrix} \] Each solution $(x_1,x_2,z_1,z_2)$ gives a vector $x_1u_1+x_2y_2=z_1w_1z_2w_2\in U\cap W$ and conversely, every vector in the intersection determines a solution to the system. It is enough to find only the $z_1,z_2$ part of the solution. Convert the matrix of the system to row echelon form to find free variables. In my calculations, the last equation becomes $2z_1+3z_2=0$, so $z_2$ is a free variable and $z_1=(3/2)z_2$. One particular solution is $z_1=3$, $z_2=2$. So $z_1w_1z_2w_2=3w_12w_2$ is a vector in the intersection. All other solutions to the system are multiples of $(3,2)$, and so every vector in the intersection is a multiple of $3w_12w_2$. 
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