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Linear Algebra - Ranks and dimension

  1. Jan 6, 2012 #1
    1. The problem statement, all variables and given/known data
    given two matrices A, B of 5x5 order:
    [itex]\rho(A)<\rho(B)[/itex]

    the nullspace for A is [itex]Sp{(0, 3, -1, 2, 1),(4, -2, 1, 4, 0)}[/itex]
    prove that
    [itex]AB\neq 0[/itex]


    2. Relevant equations
    [itex]\rho(A)=n- \rho(P_a)[/itex]


    3. The attempt at a solution
    so I now that the nullity is 2 and so [itex]\rho(A)=3[/itex]
    and because of that [itex]4\leq \rho(B)\leq 5[/itex]

    but I don't know how to prove that [itex]AB\neq 0[/itex]
     
  2. jcsd
  3. Jan 6, 2012 #2

    micromass

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    Can you find an x such that [itex]ABx\neq 0[/itex]?? That is: can you find an x such that Bx is not contained in the nullspace of A?
     
  4. Jan 7, 2012 #3
    Thanks for the reply
    the null space of A is [itex]Sp[(1,0,0,0,0),(0,1,0,0,0)][/itex]
    the nullity of B is 1 but it can be either Sp(1,0,0,0,0) or Sp(0,0,0,0,1) etc. or am I wrong?
    I guess I am because this way it can be contained in the null space of A or not, and there's no way to determine.


    let's say that the null space of B is Sp(0,0,1,0,0) so it's not contained in the null space of A,

    and let's call it [itex]Bx=(0,0,s,0,0) |s \in \mathbb{F}[/itex]
    so I just say that

    [itex]ABx = (0,0,a_{33}s,a_{43}s,a_{53}s)[/itex]

    and if AB was 0 for every x it would have been (0,0,0,0,0)?
    if so how can I determine if Bx is contained in Ax or not?
    Thanks again!
     
  5. Jan 7, 2012 #4

    micromass

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    No, it's not.

    No, it's not.

    What made you think that?
     
  6. Jan 7, 2012 #5
    I guess I'm wrong, let me tell you what my train of thought was:

    I took the original null space given, put it in a matrix and got to that, that's the basis of the
    null space.
    so the nullity is 2, and then rank(A) is 3 (because it's a 5x5 matrix)
    because rank(B)>rank(A) , rank(B) is either 4 or 5. and it's nullity is 1 or 0.

    I thought about a different way to prove it, but I need to prove that nullspace A, nullspace B,
    are subspaces of nullspace AB.

    if so: [itex]nullity(AB)=nullity(A)+nullity(B)-nullity(A\cap B)[/itex]

    which means:
    [itex]nullity(AB)\leq nullity(A)+nullity(B)[/itex]

    because we know the nullity of both A and B (assuming I'm right)
    [itex]nullity(AB)\leq 3[/itex]
    using the rank nullity theorem:

    [itex]\rho(AB)=n-nullity(AB)[/itex]

    [itex]\rho(AB) \geq 2[/itex]

    what do you think?
     
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