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maggieL

  • one year 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
    • one year ago
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    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
    • one year ago
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    I tried but couldn't get the zeros:(

  3. RaphaelFilgueiras
    • one year ago
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    |dw:1372827825197:dw|

  4. RaphaelFilgueiras
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    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
    • one year ago
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    dimension is the number of vectors in a basis

  8. RaphaelFilgueiras
    • one year ago
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    the third column is the linear combination of the first two ones

  9. RaphaelFilgueiras
    • one year ago
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    if a =-1/2, then the column space is spaned by(1,-1,0) and (0,1,2)

  10. maggieL
    • one year ago
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    does that mean: if it has dimension 3, there can't be any zero rows?

  11. RaphaelFilgueiras
    • one year ago
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    in this case yes

  12. maggieL
    • one year ago
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    so a would be -1?

  13. RaphaelFilgueiras
    • one year ago
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    if a =-1 then the dim(C(A))=3 and the nullspace contains only zero vector

  14. RaphaelFilgueiras
    • one year ago
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    C(A)=R³

  15. maggieL
    • one year ago
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    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
    • one year ago
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    I'm not*

  17. RaphaelFilgueiras
    • one year ago
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    yes

  18. maggieL
    • one year ago
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    thanks so much! ur so patient:)

  19. RaphaelFilgueiras
    • one year ago
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    you are welcome

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