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Matrix ranks

  1. Apr 28, 2007 #1


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    1. The problem statement, all variables and given/known data
    A and B are nxn matricies. B =/= 0 and rank(A)=n-1
    Prove or disprove that:
    1) if AB=0 then rank(B)=1
    2) if BA=0 then rank(B)=1

    2. Relevant equations

    3. The attempt at a solution
    If n=2 and we define A as:
    [1 1]
    [1 1]
    and B as:
    [1 -1
    [-1 1]
    then rank(A) = 1 = 2-1 and B=/=0 but AB=BA=0 and rank(B)=2 whick disproves both (1) and (2). Is that right? It seems odd that they'd give to similar questions with identical answers.
  2. jcsd
  3. Apr 28, 2007 #2


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    What makes you think that r(B) = 2?
  4. Apr 29, 2007 #3


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    Oh ... rank(B) = 1 also... I guess I have to think about it somemore.
  5. Apr 29, 2007 #4


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    Ok, now I think that both of the statments are right:
    1) Since the rank of A is n-1 then the dimension to the space of all the solutions to Ax=0 is
    n - rank(A) = n-n+1=1. Since AB=0 then all the coloums of B are solution to Ax=0. But the dimension of that space is 1 so there cannot be more that 1 linearly independent vectors in the coloums of B so rank(B)<=1 but we know that rank(B)>0 so rank(B)=1

    2)If we look at the equation BA=0 then the coloums of A are solutions to Bx=0. So the dimension of the space of the solutions to Bx=0 (P) has to be bigger or equal than rank of A:
    dim(P) >= rank(A) --> n-rank(B) >= n-1 --> rank(B) <= 1
    But rank(B)>0 so rank(B)=1

    Are those answers right?
  6. Apr 29, 2007 #5
    Looks good to me!
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