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May 23rd, 2018, 07:31 AM  #1 
Newbie Joined: May 2018 From: israel Posts: 3 Thanks: 0  help with Linear spaces:)
Hi everyone, I need linear help. Given (R ^ 3): https://ibb.co/eLAgO8 Find a finite finite set for W cutting U. Thanks. Last edited by skipjack; May 24th, 2018 at 06:52 AM. 
May 24th, 2018, 06:57 AM  #2 
Global Moderator Joined: Dec 2006 Posts: 20,098 Thanks: 1905 
More readably, $\text{U = Span{(1,1,2),(2,2,1)}}$ $\text{.W = Sp{(1,3,4),(2,5,1)}}$ 
May 27th, 2018, 10:10 AM  #3 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,261 Thanks: 895 
So any vector in U is of the form (a+ 2b, a+ 2b, 2a+ b) and any vector in W is of the form (x+ 2y, 3x+ 5y, 4x+ y). In order that a vector be in both (I assume that "W cutting U" means "vectors that are in both U and W") we must have a+ 2b= x+ 2y, a+ 2b= 3x+ 5y, and 2a+ b= 4x+ y. From the first two equations, x+ 2y= 3x+ 5y. 2x= 3y so y= (2/3)x. Then a+ 2b= x+ 2y= x (4/3)x= (1/3)x and 2a+ b= 4x+ y= 4x (2/3)x= (10/3)x. Subtract 2 times 2a+ b= (10/3)x from a+ 2b= (1/3)x to get 3a= (1/3)x (20/3)x= (21/3)x, a= (7/3)x. And then b= (10/3)x 2a= (10/3)x+ (14/3)x= 8x. We can write (a+ 2b, a+ 2b, 2a+ b)= ((7/3)x+ (16/3)x, (7/3)x+ (16/3)x, (14/3)x+ (10/3)x)= (3x, 3x, 8x)= (3, 3, 8)x. Every vector in the intersection of U and W is a multiple of the vector (3, 3, 8).
Last edited by skipjack; May 27th, 2018 at 12:31 PM. 

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