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Linear Algebra: Null Space and Dimension

  1. Apr 18, 2009 #1
    1. The problem statement, all variables and given/known data

    Prove that dim(nullA) = dim(null(AV))
    (A is a m x n matrix, V is a n x n matrix and is invertible

    2. Relevant equations

    AX=0 and AVX = 0
    Null(AV) = span{X1,..Xd}
    Null(A) = span{V-1X1,.., V-1Xd}


    3. The attempt at a solution

    so you need to prove that dim(null(AV) is a subset of nullA?

    Therefore d = dimA = dim(AV) and let nullA= span{X1,..Xd} and null(AV) = span{ X1,..Xd}.
    Null(AV) = span{X1,..Xd}
    Null(A) = span{V-1X1,.., V-1Xd}

    If a1(V-1X1 ) + …+a2(V-1Xd) = 0
    So 0 = VV-1(t1X1 +… + t1Xd)
    0 = t1X1 +… + t1Xd all of ti = 0 meaning it is linearly independent, and also span {V-1X1,.., V-1Xd} which is a basis of nullA

    AX = 0 so AVX=O then
    V0 = AVX which is in null(AV)
    This also means V0 = t1X1 +… + t1Xd
    which also means 0 = t1V-1X1 +…+ tdV-1Xd which spans nullA.

    Is this correct?

    Thank you, in advance :)
     
  2. jcsd
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