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Is it worth fighting this? (null and row spaces)

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    If A is an m x n matrix, show that null(A) = [row(A)]_|_ (meaning rowA perp).

    I handed in an assignment with this question on it, and got zero points. I think what I did is mostly right, but I want someone to make sure I'm not out to lunch before I go to my prof.

    2. Relevant equations

    3. The attempt at a solution
    The dim[null(A)] = n -r
    dim [row(A)] = r

    So combining the above equations, we get dim[null(A)] + dim[row(A)] = n

    Both null(A) and row(A) are subspaces of Rn. So I can use the theorem that dim(U) + dim(U perp) = n.

    Then I wrote that by the above condition, null(A) = (rowA) perp.

    And I got zero points out of eight. I think I should have included this line:
    U = null(A)
    U perp = row(A).

    But (rowA) perp = U perp perp = U = null(A), which proves it. In the previous question in the assignment I proved that U perp perp = U, so I don't need to show it again.

    This is worth 20 % of my final grade - do you think I should approach my prof about this? My TA was the one who marked it, but she told me what I was doing was wrong. So I'm not sure... thanks!
  2. jcsd
  3. Apr 6, 2010 #2


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    Your mistake is in assuming that because the dimensions of null(A) and row(A) perp are the same, the two subspaces are the same. That's like saying because both the xz and xy planes are two-dimensional, they are the same subspace of R3.

    Unfortunately, your mistake likely caused you to completely miss the meat of the real argument, which is why the TA didn't give you any credit.
  4. Apr 6, 2010 #3
    I'm confused... when did I assume that the dimensions of null(A) and [row(A)]perp were the same? All I did was apply the theorem "Let U be a fixed subspace of R^n. Then dim(U) + dim(U perp) = n.

    It's definitely true that dim(nullA) + dim(rowA) = n, right? And nullA, rowA are both supspaces of R^n. So why doesn't the above theorem apply?

    Your reasoning with the xz and xy planes makes total sense, I just can't see how to apply it here.
  5. Apr 6, 2010 #4


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    The theorem you used gives you:

    If U=null(A), then dim(null(A))+dim(null(A) perp)=n.
    If U=row(A), then dim(row(A))+dim(row(A) perp)=n.

    It doesn't tell you anything about the subspaces except their dimension. At best, you can conclude that dim(null(A))=dim(row(A) perp). I assumed you had reached this conclusion about the dimensions and then made the leap to the subspaces being equal. If you didn't, it's completely unclear how you went from applying the theorem to saying the two subspaces are the same.
    Last edited: Apr 6, 2010
  6. Apr 6, 2010 #5
    I didn't reach that conclusion... I think I must have applied the theorem incorrectly. I'm not getting what you're saying :(

    My reasoning was that if you have dim(A) + dim(B) = n, and A, B are both supspaces of R^n, then B = A perp.

    Is it because you can't go backwards? The theorem says that dim(U) + dim(U perp) = n, but maybe you can only use it if you already know that the thing in the second dim( ) is perp. Maybe you can't just know that it's equal to n...
  7. Apr 6, 2010 #6
    Yeah I think I got it. You can't say that just because dim(nullA) + dim(rowA) = n that row(A) is necessarily nullA perp.

    The best you can say is that dim(nullA) + dim(rowA) = dim(nullA) + dim(nullA perp).

    Then you would get dim(rowA) = dim(nullA perp), but the spaces aren't necessarily equal just because their dimensions are equal.

    What's a better way to do this problem, then?
  8. Apr 6, 2010 #7


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    Yeah, you got it. I don't actually know what a correct proof is offhand. It's easier to see where an argument is wrong than to come up with the correct one. Perhaps one of the math experts will chime in with a suggestion.
  9. Apr 6, 2010 #8
    That's okay! Thanks for your help and patience :)
  10. Apr 7, 2010 #9


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    how about this as a start framework...

    say nxn matrix [itex] A [/itex] has non-zero row vectors [itex] a_i \in \mathbb{R}^n[/itex].

    multiplying [itex] A.a_i [/itex] shows cleary the ith component of the product will be [itex] |a_i|^2 [/itex], so none of the row vectors are in the null space.

    now consider the subspace [itex] R \subset \mathbb{R}^n [/itex] spanned by the row vectors

    assuming it exists, take a non-zero vector [itex] p \in R^{\perp} [/itex], and have a look at the product [tex] A.p [/tex]
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