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Write the subspace spanned by vectors as a kernel of a matrix.

  1. Oct 8, 2013 #1

    Lets say I have a vectorspace in Rn, that is called V.
    V = span{v1,v2,..... vk}

    Is it then possible to create an m*n matrix A, whose kernel is V.
    That is Ax = 0, x is a sollution if and only if x is an element of V.

    Also if this is possible, I imagine that k may not b equal to m?
  2. jcsd
  3. Oct 8, 2013 #2
    Also, if it makes the question easier, we can assume that v1,...,vk is linerly independent.
  4. Oct 8, 2013 #3


    Staff: Mentor

    Homework problem?
  5. Oct 8, 2013 #4
    Not really. In my book they talk about a column space for R3. It is obvious that the column space is spanned by two particular vectors([3,3,4] and [-1,-1,5]). But then they make the statment that the columnspace is those vectors who have the property that x1-x3 = 0, this is the same as saying the column space is the vectors who satisfy Av where A = [1 0 -1], so I am wondering how we can do this generally. That is, change the description of a vector space, from a spanning-set, to a Kernel. And if there is an algorithm for doing this.

    Another way of asking is to compare with solving the equaion Bx =0. Then we can find vectors that span Null B. I want to go the opposite way, we have the vectors that span Null B, but I want to find the matrix B(I suspect this matrix may not be unique.).
    Last edited: Oct 8, 2013
  6. Oct 8, 2013 #5


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    The first thing I think that should be pointed out is that you aren't changing the description of a vector space in general, just giving a different way of describing vector subspaces of Rn. Obviously you can't define a vector space in general as the kernel of a linear transformation because a linear transformation doesn't make sense if you don't know what a vector space is already.

    Anyway ,back to the task at hand, if my matrix B has rows b1,....,bm, then Bv = 0 means that v is perpendicular to each row of B. So a matrix for which B vj = 0 for each j has rows which are perpendicular to every vj, and then to make sure that the kernel does not have a larger subspace than you want you need to make sure B has enough rows in it - I'll let you think a bit on what procedure you can do to generate B algorithmically (it only requires a standard linear algebra result).
  7. Oct 8, 2013 #6

    Is this proof correct?, I have found a matrix, whose Kernel contains the spanning set, but I do not yet see how to prove that it does not contain any other vectors.

    I assume that the vectors v1, vk are linerly independent.

    If b is a row vector in B, then we must have

    b*vk=0, where * is matrix multiplication(not dot-product)

    We can transpose this to.


    And we can put it all together in a Matrix.

    |...........| * b' = 0(vector)

    When we solve this we get n-k linerly independent b', because the rank of the matrix is k, and the rank-theorem gives dim Null(Matrix)=n-k.

    So we can put these b row vectors together to form B, which is then (n-k)*n. And we have B*x =0(vector), if x is in span{v1,...,vk}.

    But we also have to show that if vector is not in span{v1,..,vk} than Bx is not 0(vector).
    To show this i can use contrapositive proof. That is if Bx=0(vector), then x is in span{v1,..,vk}.
    This last part is where I get stuck, do you see how to proceed?
  8. Oct 8, 2013 #7


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    For this part I recommend considering the span of [itex] \left{ v_1,...,v_k, b_1,...,b_{n-k} \right} [/itex].
  9. Oct 10, 2013 #8
    First extend the vectors ## \{ v_{1} v_{2} ... v_{k} \} ## to ## \{ v_{1} v_{2} ... v_{k} v_{k+1} ... v_{n} \} ##, a basis for ##ℝ^{n}##. Then define the following linear map ## T : ℝ^{n} → ℝ^{n} ## by the following operations on that basis:
    ## T(v_{1}) = 0 ##
    ## T(v_{2}) = 0 ##
    ## T(v_{k}) = 0 ##
    ## T(v_{k+1}) = v_{k+1} ##
    ## T(v_{n}) = v_{n} ##.

    Then ##[T]_{β}## where ##β## is the standard ordered basis for ##ℝ^{n}## will have a kernel consisting of V. Try proving this result yourself.

    Last edited: Oct 10, 2013
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