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May 23rd, 2018, 07:31 AM   #1
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help with Linear spaces:)

Hi everyone,
I need linear help.
Given (R ^ 3):

Find a finite finite set for W cutting U.

Last edited by skipjack; May 24th, 2018 at 06:52 AM.
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May 24th, 2018, 06:57 AM   #2
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More readably,
$\text{U = Span{(1,1,2),(2,2,1)}}$
$\text{.W = Sp{(1,3,4),(2,5,1)}}$
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May 27th, 2018, 10:10 AM   #3
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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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