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Matrix manipulations/rank of a matrix

  1. Feb 5, 2008 #1
    My question has 5 short parts and for each, i'm supposed to give a counterexample if the statement is false or give an argument to prove that a statement is true.

    1.Q: Linear system with m equations and n variables. If m> or = to n, then the system can have at most 1 solution.
    A: I think this one is false because if the # of parameters= n-rank, since rank < or = n, if we take it to be less than n, then # of parameters = n - (some # less than n), which gives a # > 0, then this means there is at least 1 parameter and infinate solutions possible.

    2.Q: A and B are matrices, and the product AB is defined. Then rank (AB) = rank A
    A: I would say that this is false because if A and B are different sizes, say A is 5x4 and B is 4x6, the rank of both A and B would have to be < or = 4, but AB produces a 5x6 matrix, meaning that the rank must be < or = 5. So couldn't A and B each have a rank of 4, while AB has a rank of 5? But if this is true, how can it be proved?

    3.Q: A and B are nxn matricies and AB=0. Then at least one of A,B must have a determinant 0.
    4.Q: An nxn matrix A satisfies AB=BA for every nxn matrix B, then A must be the identity matrix.

    I'm not sure about 3 or 4, any hints/points in the right direection would be appreciated.

    5.Q: If the system Ax=b has no solution, then the system Ax=0 has only the trivial solution.
    A: I think this is false because the statements: The system Ax=0 has the only trivial solution x=0 and Ax=b has a unique solution for any vector b, according to the invertible matrix theorem. So since the bolded parts contradict each other, I assume its false but i'm not sure how to go about proving it. Any ideas?
     
  2. jcsd
  3. Feb 5, 2008 #2

    HallsofIvy

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    If all the equations were independent (i.e. rank m) (which is not given [and could only happen if m=n]) then there would bo no solutions so this is correct.

    Even if A and B were of the same size, it might happen that AB= 0 which has rank 0!

    3. Do you know that det(AB)= det(A)det(B)? That makes 4 trivial. Even if you don't know that, you should know that a matrix has an inverse if and only if its determinant is not 0. If A does not have determinant 0, multiply both sides of AB= 0 by A-1.

    4. Look at AB and BA for A any diagonal matrix.

    In fact, it is exactly the other way around. If Ax= b has no solution, then A is not invertible. Ax= 0 has an infinite number of solutions
     
  4. Feb 6, 2008 #3
    I have tried quite a few different things and it seems that If both A and B are diagonals this works, but since diagonal matrices can be row reduced to the Identity matrix, does this make number 3 true? When only one of A or B are diagonals, it doesn't seen to work out unless one is the identity matrix. Does this mean that it must be true, or are there special cases that i am missing here?
     
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