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Linear Algebra Kernel/Image/Rank

  1. Aug 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Suppose that A is an 8x11 matrix whose kernel is of dimension 5, and B is an 11x9 matrix whose image is of dimension 7. If the subspace kernel(A) + image(B) has dimension 10, what is the rank of AB?

    2. Relevant equations

    Rank Nullity Theorem: For an n x m matrix A, dim(ker(A)) + dim(img(A)) = m

    3. The attempt at a solution

    Ok, so we know the following about A and B.

    dim[ker(A)] = 5 (given)
    dim[img(A)] = 6 (by rank-nullity theorem)

    dim[img(B)] = 7 (given)
    dim[ker(B)] = 2 (by rank-nullity theorem)

    AB will be an 8x9 matrix, so the dimension of the image (which is the rank) can't be more than 8. I suppose the kernel of A is contained in AB, but I don't know what to make of that or how to use that information.

    For the subspace ker(A) + img(B), could it be that ker(A) and img(B) share two components because the dimension is 10, but dim[ker(A)] + dim[img(B)] = 12?

    This seems like a simple enough problem, but I have problems utilizing the meaning of kernel and image.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Aug 4, 2010 #2

    HallsofIvy

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    I'm not sure what you mean by "subspace kernel(A) + image(B)". What kind of "addition" are you doing here? More important is [itex]kernal(A)\cap image(B)[/itex], the intersection of the two spaces. u is in the kernel of AB if and only if Bu is in that intersection.
     
  4. Aug 4, 2010 #3
    If by "+" you mean "direct add" then I think we get this: There are two components shared between ker A and img B (because the dimension of the direct add of the two subspaces is two less than the sums of the dimensions).

    What we are looking for is the dimension of the image of A(Bv).

    Then we have: dim(img B) = 7. But since there are two components of the image of B which are in the kernel of A, the image of A when the source vector is Bv will be reduced by two (since those components will add the 0 vector to the result). Hence rank(AB) = rank(A) - 2 = 4.
     
  5. Aug 7, 2010 #4
    Ahh, I see. Thanks a lot for the help! I'm sorry my reply is so belated.

    I suppose I meant "direct add" when I said "kernel(A) + image(B)" but my weak command of mathematics at the moment prevented me from knowing any other type of "addition"- so I'm sorry for not specifying.

    I never remember to think of matrix multiplication in terms of a transformation of some sort- it clearly helps when trying to ascertain the image or kernel of some matrix.
     
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