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Introductory Linear Alebra proof question

  1. Jul 5, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A be an n x n matrix and let x and y be vectors in R^n. Show that if Ax = Ay and x [tex]\neq[/tex] y, then the matrix A must be singular.

    2. Relevant equations
    So far we have learned the definition of a matrix that has an inverse to be one where: if there exists a matrix B and AB = BA = I. The matrix B is said to be the multiplicative inverse of A.

    3. The attempt at a solution

    I have done earlier problems that involved proving things have an inverse with the above definition, however I can not think of how to apply it to this problem.

    So what I did was use my knowledge from earlier classes (also this concept was touched upon a few problems earlier, but has not yet been defined), that if the determinant = 0 then the matrix does not have an inverse. But I have not found anything that helps me.

    A = |a11 a12| x = x1 y = y1
    |a21 a22| x2 y2

    Ax = Ay

    (this is supposed to read as Ax = Ay expanded, sorry.
    |a11*x1 + a12*x2| |a11*x1 + a12*x2|
    | | = | |
    |a21*x1 + a22*x2| |a21*x1 + a22*x2|

    I've been getting it into equations like: a11(x1-y1) + a12(x2-y2) = 0, and a few similar things, but I have not yet found something that will fit into the formula for the determinant.

    So am I on the right track? If i am please give me some hints on how to proceed, if not then let me know what to do please. Thanks a lot, and sorry if anything is unclear.
  2. jcsd
  3. Jul 5, 2010 #2


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    If Av=0 for some nonzero vector v, then A is not invertible. Can you see why that's true from your definition of invertible? Can you see why that might be true from what you are given?
  4. Jul 6, 2010 #3


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    If A has a zero eigenvalue then A is singular (definition) This has shown to be the case in previous posts.
  5. Jul 6, 2010 #4
    If you're still stuck, try a proof by contradiction. Assume A is nonsingular, i.e. A-1 exists. Since this is a proof by contradiction, you should expect to either contradict one of your premises (e.g. x [tex]\neq[/tex] y) or to arrive at something that clearly doesn't make sense (e.g. 1 = 0).
    You can easily arrive at the former contradiction if you recall that A-1A = AA-1 = I.

    We often prefer straightforward proofs over proofs by contradiction, but either method seems fine here since no real insight in the theory is lost in this case by the latter approach. I hope this helped a bit.
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