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 Post subject: Basis vector representationPosted: Thu, 12 Feb 2009 06:54:44 UTC
 S.O.S. Oldtimer

Joined: Sun, 16 Sep 2007 19:14:24 UTC
Posts: 265
If you have the following basis vectors for a polynomial vector space
c_1*[-1,1,0]
c_2*[-1,0,1]

then how do you get to the form

(x-1)
(x^2 - 1)

as representing the same basis?

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 Post subject: Posted: Thu, 12 Feb 2009 10:12:59 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Tue, 20 Nov 2007 04:36:12 UTC
Posts: 826
Location: Las Cruces
Sometimes in 3d, people write the x,y,z unit vectors as i,j,k. So the vector (1,2,3) is 1 i + 2 j + 3 k.

In the vector space of polynomials of degree <= 2, the "natural" basis vectors are the polynomials { 1, x, x^2} and the vector (1,2,3) is 1 (1) + 2 (x) + 3(x^2) .

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 Post subject: Posted: Thu, 12 Feb 2009 17:06:55 UTC
 S.O.S. Oldtimer

Joined: Sun, 16 Sep 2007 19:14:24 UTC
Posts: 265
I don't see the connection between my post and your post...

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 Post subject: Posted: Thu, 12 Feb 2009 19:19:23 UTC
 Member of the 'S.O.S. Math' Hall of Fame

Joined: Tue, 20 Nov 2007 04:36:12 UTC
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Location: Las Cruces
(1,2,3) = 1(1) + 2(x) + 3(x^2) = x^2 + 2x + 1
(-1,1,0) = (-1)(1) + 1(x) + 0 (x^2) = x -1
(-1,0,1) = (-1)(1) = 0 (x) + 1 (x^2) = -1 + x^2 = x^2 - 1

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