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 April 24th, 2014, 12:16 PM #1 Newbie   Joined: Apr 2014 From: Israel Posts: 3 Thanks: 0 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?
 April 24th, 2014, 12:23 PM #2 Newbie   Joined: Apr 2014 From: Israel Posts: 3 Thanks: 0 Solved the question about complexes. any Ideas about the fist question regarding Vector spaces?
April 24th, 2014, 12:36 PM   #3
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Quote:
 Originally Posted by AnaRhisT94 W = Sp{(1,3,4),(2,5,1)} , U = Span({1,1,2),(2,2,1)} Find a group that spreads U^W.
Do you need to find a basis of $U\cap W$?

Quote:
 Originally Posted by AnaRhisT94 (The^ is like an Upside down "U", ^ means Union in other words)
$\cap$ means intersection, while $\cup$ means union. Which one is used in this problem?

 April 24th, 2014, 12:43 PM #4 Newbie   Joined: Apr 2014 From: Israel Posts: 3 Thanks: 0 yes, intersection.
 April 24th, 2014, 01:11 PM #5 Senior Member   Joined: Dec 2013 From: Russia Posts: 327 Thanks: 108 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 $x_1u_1+x_2y_2=-z_1w_1-z_2w_2\in U\cap W$ 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 $-z_1w_1-z_2w_2=3w_1-2w_2$ 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 $3w_1-2w_2$.

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