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Homework Help: Finding basis for null(A) and null(AP)

  1. Jan 26, 2010 #1
    1. The problem statement, all variables and given/known data
    A is a mxn matrix, and P is an invertible nxn matrix.

    So I want to prove that the bases of null(A) and null(AP) have the same number of elements.


    2. Relevant equations



    3. The attempt at a solution
    I was going to start off by assuming that {X1, X2, ... Xm} is a basis of null(A). This is my first issue. Can I really just assume that?

    So AXi = 0, following the definition of null space. Then V-1Xi is in null(AV), since AV(V-1Xi) = 0. Somehow I need to get from that to the fact that {V-1X1, V-1X2, ... V-1Xm} is independent. But how?

    I can prove that {V-1X1, V-1X2, ... V-1Xm} spans null(A), I just need to know that it's independent for it to be a basis. And if it is, then it has m elements and so does null(A)'s basis.
     
  2. jcsd
  3. Jan 26, 2010 #2

    HallsofIvy

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    Yes, null(A) is a vector space and every vector space has a basis.

    Where did "V" come from? There was no mention of V before. Did you mean P?

    I can prove that {V-1X1, V-1X2, ... V-1Xm} spans null(A), I just need to know that it's independent for it to be a basis. And if it is, then it has m elements and so does null(A)'s basis.[/QUOTE]
    Proof by contradiction: Suppose [itex]\{P^{-1}X1, P^{-1}X2, ..., P^{-1}Xm}[/itex] is NOT independent. Then there exist numbers [itex]a_1, a_2, ..., a_m[/itex], not all 0, such that [itex]a_1P^{-1}X1+ a_2P^{-1}X2+ ...+ a_mP^{-1}Xm= 0[/itex]. Since P, and so [itex]P^{-1}[/itex], is a linear tranformation, that is the same as [itex]P^{-1}a_1X1+ a_2P^{-1}X2+ ...a_mP^{-1}Xm= P^{-1}(a_1X1+ a_2X2+ ...+ a_mXm)= 0[/itex]. Now take P of both sides.
     
  4. Jan 26, 2010 #3
    I'm confused about where the contradiction is.

    Taking P of both sides, you'd end up with PP-1{a1X1 + a2X2 + ... amXm} = P0

    Then {a1X1 + a2X2 + ... + amXm) = 0

    Doesn't that support the initial assumption? How do you know that a1, a2, ... am above aren't all zero?

    EDIT: I think I just figured it out. Because a1X1 + a2X2... was my first basis, it has to be independent. Correct?
     
    Last edited: Jan 26, 2010
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