## maggieL 2 years ago Find a such that th subspace-SPAN{(a, 1, 1), (0, 1, 2), (1, -1, 0)} of R^3 has dimension 2.

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1. alexandercpark

i believe you just use some way to solve a system of equations (i would use RREF) and find a value of a such that one of those will zero out

2. maggieL

I tried but couldn't get the zeros:(

3. RaphaelFilgueiras

|dw:1372827825197:dw|

4. RaphaelFilgueiras

if the column space has dim( 2) the rank of matrice must be 2 and the null space must have dim 1 then a=-1/2

5. maggieL

um I'm trying to digest this... I'm actually a bit confused with the "null space". so does dimension2 mean 2 leading 1s with 1 free variable or am i just mixing things up?

6. RaphaelFilgueiras

Dim(R(A)=C(A'))=Dim(C(A))=rank A is mxn Dim(C(A))+dim(N(A'))=m Dim(R(A))+dim(N(A))=n dim(N(A))=n-r (number of free variables)

7. RaphaelFilgueiras

dimension is the number of vectors in a basis

8. RaphaelFilgueiras

the third column is the linear combination of the first two ones

9. RaphaelFilgueiras

if a =-1/2, then the column space is spaned by(1,-1,0) and (0,1,2)

10. maggieL

does that mean: if it has dimension 3, there can't be any zero rows?

11. RaphaelFilgueiras

in this case yes

12. maggieL

so a would be -1?

13. RaphaelFilgueiras

if a =-1 then the dim(C(A))=3 and the nullspace contains only zero vector

14. RaphaelFilgueiras

C(A)=R³

15. maggieL

so if it has dimension 3, I would have to write a=-1 with the column space spanned by (1, -1, 0) (0, 1, 2) (-1, 1, 1)? sry I'm so smart with this:(

16. maggieL

I'm not*

17. RaphaelFilgueiras

yes

18. maggieL

thanks so much! ur so patient:)

19. RaphaelFilgueiras

you are welcome