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Homework Help: Proofs about Matrix

  1. Sep 18, 2011 #1
    1. The problem statement, all variables and given/known data
    See question 5

    2. Relevant equations

    3. The attempt at a solution

    For part a, it is very easy.
    Multiply the inverse of A 2 times on both side, we can see the B=inverse of A.
    i.e. The required B is inverse of A, then the proof is finished.

    But how about part b?
    It seems it is the same part a.

    Is part b also correct?

    Attached Files:

  2. jcsd
  3. Sep 18, 2011 #2


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    Part b is quite different from part a- and the difference is important to learn. Mathematics must be very very precise in its wording- unlike science we don't have observations and experiments to fall back on. In other words, we can't just look at the real world- words are everything!

    In part a it ask if, given a non-singular matrix A, there exist a matrix B such that [itex]AB^2= A[/itex]. You are right- just multiply, on the left, on both sides by [itex]A^{-1}[/itex], which exists because A is non-singular, and the equation becomes AB= I. Yes, B exists and is the inverse of A.

    In part B, it asks if there exists a matrix B such that, for any non-singular matrix, A, [itex]A^2B= A[/itex]. "Any" is the crucial word there. Is there a single matrix B that is the inverse of all invertible matrices?
  4. Sep 18, 2011 #3
    Well, after listening to your explaination, I know part b is obvious wrong.
    However, I wonder how to write it out.
  5. Sep 18, 2011 #4


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    "No, there does not exist a single matrix, B, such that [itex]A^2B= A[/itex] for all non-singular matrices, A."
  6. Sep 18, 2011 #5
    Oh, this is the prove?
  7. Sep 18, 2011 #6
    Let me try for part d.

    Since [itex]A[/itex] is non-singular, [itex]A^{-1}[/itex] exists.
    So [itex]\vec{x}[/itex]=[itex]A^{-1}[/itex][itex]\vec{y}[/itex] exists.

    So, there exists [itex]\vec{x}[/itex] s.t. [itex]A[/itex][itex]\vec{x}[/itex]=[itex]\vec{y}[/itex]

    Again, how to disprove part c?
    By simply saying NO, there doesn't exist?
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