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Matrix Question: A*B=A (B is non-identity)

  1. Nov 30, 2011 #1
    If two ℝeal valued, non-identity matrices A (dimension MxN) and B (dimension NxN)
    satisfy the condition A * B = A
    Is there a name for the relationship between B and A ?

    For example for: [itex]A =
    \left( \begin{array}{cc}
    \frac{1}{3} & \frac{2}{3} \end{array} \right), \quad B = \left( \begin{array}{cc}
    \frac{3}{5} & \frac{2}{5} \\
    \frac{1}{5} & \frac{4}{5} \end{array} \right) \quad A * B = \left( \begin{array}{cc}
    \frac{1}{3} & \frac{2}{3} \end{array} \right)
    [/itex]

    My Observations:
    • If such A exists for B, then for any real number k, k*A is also a solution for B
      since (k*A)*B = k*(A*B)=k*A
      implies, there will be infinite solutions for B.
      .
    • [itex]
      p\to+\infty{\left( \begin{array}{cc}
      \frac{3}{5} & \frac{2}{5} \\
      \frac{1}{5} & \frac{4}{5} \end{array} \right)}^p = \left( \begin{array}{cc}
      \frac{1}{3} & \frac{2}{3} \\
      \frac{1}{3} & \frac{2}{3} \end{array} \right)
      [/itex]

    My Question: Given a matrix B, how do we find A such that A*B=A ?
     
  2. jcsd
  3. Nov 30, 2011 #2

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    This looks a bit too much like homework for me to give a direct answer (now).

    I can give a hint in the form of a question: How is this related to eigenvalues and eigenvectors?
     
  4. Nov 30, 2011 #3
    Thank you very much for the hint. I swear this not my homework. I know I missed something very basic but didn't know which one.
     
  5. Nov 30, 2011 #4
    I am hitting a wall. May I request another hint or two?
     
  6. Dec 2, 2011 #5
    D_H gave a pretty good hint. I'll give you a little more.

    First consider the problem AB=B. Given A you can easily find all such B using eigenthings.

    You can transform AB=A into this form by taking the transpose of both sides.

    Very good problem if you came up with it yourself.
     
  7. Dec 2, 2011 #6

    D H

    User Avatar
    Staff Emeritus
    Science Advisor

    And one more hint: Other than the trivial solution A=0, there are no solutions if one is not an eigenvalue of B.
     
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