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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 $3w_1-2w_2$. Tags algebra, linear, spaces, vector Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post alphaknight61 Linear Algebra 1 November 26th, 2012 04:54 AM rayman Abstract Algebra 0 January 27th, 2011 08:35 AM remeday86 Linear Algebra 1 July 10th, 2010 04:20 AM conradtsmith Linear Algebra 4 May 26th, 2009 12:48 AM supercali Abstract Algebra 0 November 27th, 2007 12:40 PM

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