W is the subspace of R^{4} with basis { w1 , w2, w3}

where w1 = (1 ,1 ,3,2) w2=(1,1,-2,-3), w3=(2,1,-3,-1)

Find the orthogonal complement of W is R^{4}

Am not sure how to find it, do i multiply w1*w2*w2 = 0 ?

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- May 8th 2013, 09:23 AMTweetyorthogonal complement
W is the subspace of R^{4} with basis { w1 , w2, w3}

where w1 = (1 ,1 ,3,2) w2=(1,1,-2,-3), w3=(2,1,-3,-1)

Find the orthogonal complement of W is R^{4}

Am not sure how to find it, do i multiply w1*w2*w2 = 0 ? - May 8th 2013, 09:32 AMHallsofIvyRe: orthogonal complement
I would hope that you at least know that you

**can't**multiply three vectors like that!

The orthogonal complement of a three dimensional subspace is a**one**dimesional subspace so you want to find a single vector (a, b, c, d) such that (a, b, c d).(1, 1, 3, 2)= a+ b+ 3c+ d= 0, (a, b, c, d).(1, 1, -2, -3)= a+ b- 2c- 3d= 0, and (a, b, c, d)(2, 1, -3, -1)= 2a+ b- 3c- d= 0. The orthogonal complement is the suspace of all multiples of that vector. - May 8th 2013, 09:48 AMTweetyRe: orthogonal complement
and how do i find that vector? Do I put those vectors in a matrix and row reduce it?

- May 8th 2013, 02:14 PMHallsofIvyRe: orthogonal complement
You have, as I said, three equations in four variables.

a+ b+ 3c+ d= 0, a+ b- 2c- 3d= 0, and 2a+ b- 3c- d= 0

Find all a, b, c, and d that satisfy those equations.

Personally, I prefer to solve systems of equations as I learned way back in high school rather than using matrices. For example, if you subtract the second equation from the first, you eliminate both a and b: 5c+ 4d= 0. If you subtract the second equation from the third you eliminate b: a- c+2d= 0. From 5c+ 4d= 0, c= -(4/5)d. Then a- c+ 2d= a+(4/5)d+ 2d= a+ 14/5d= 0 so a= -(14/5)d. The first equation becomes a+ b+ 3c+ d= -(14/5)d+ b- (4/5)+ d= b- (13/5)d= 0 so b= (13/5)d.

(a, b, c, d)= (-(14/5)d, (13/5)d, -(4/5)d, d) where d can be any number. If you don't like fractions, take d= 5 so your vector is (-14, 13, -4, 5) and the "orthogonal complement" is spanned by that vector.