# Thread: linear algebra

1. ## linear algebra

find the basis of annihilator of subspace W in R5, which is spanned by (2,-2,3,4,1), (0,0,-1,-2,3), (-1,1,2,5,2), (1,-1,2,3,0)?

please explain

i have reduced to the echo-matrix then i don't have the idea.

2. ## Re: linear algebra

Use the Gauss–Jordan elimination

1) reverse rows 1 and 3;
2) add row 1 to row 3, multiply row 1 by –2 and add row 4;
3) multiply row 2: by 4 and add to row 3, by –1 and add row 3 to row 1;
4) devide row 3 by 14 and row 4 by –2;
5) subtract row 3 from row 4;
6) multiply row 3 by –3 and add to row 2;
7) multiply row 2 by 2 and add to row 1

So, $\displaystyle (1,1,0,0,0)$ and $\displaystyle (1,0,-2,1,0)$.

3. ## Re: linear algebra

by inspecting the rank of the matrix DeMath has produced, we see that dim(W) = 3, hence dim(ann(W)) = 2. the vectors DeMath has produced aren't elements of (R5)*, unless you interpret them as the coordinates of the basis elements of ann(W) in the dual basis to the standard basis of R5.

that is to say, there is a natural isomorphism between ann(W) and W:

fx given by fx(y) = <x,y> is in ann(W) iff x is in W.

the chief difference is: ann(W) is a subspace of the dual space, whereas W is a subspace of the original vector space.

4. ## Re: linear algebra

by the equations
x1=x2+x4
x3=-2x4
x5=0
how you are concluding that (1,1,0,0,0) & (1,0,-2,1,0) are in W(per)

5. ## Re: linear algebra

Originally Posted by Deveno

that is to say, there is a natural isomorphism between ann(W) and W:

fx given by fx(y) = <x,y> is in ann(W) iff x is in W.
you are saying that from (1,1,0,0,0) and (1,0,-2,1,0), we are able to find fx and fy

is it right.