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Prove that Ax=Ix has only the trivial solution

  1. Oct 6, 2012 #1
    1. The problem statement, all variables and given/known data
    Let A be a square matrix of order n such that [itex]2{A^2} + A = 4I[/itex]. Prove that the only [itex]x \in ℝ^n[/itex] that satisfies [itex]Ax = Ix[/itex] is x=0.




    2. Relevant equations
    [itex]Ax = 0[/itex] has only the trivial solution iff A is invertible.


    3. The attempt at a solution
    The problem would be pretty trivial if the given equation was Ax=0, but how am I gonna tackle it when the RHS is Ix? TIA!
     
  2. jcsd
  3. Oct 6, 2012 #2

    tiny-tim

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    hi drawar! :smile:

    hint: if Ax = x, what is A2x ? :wink:
     
  4. Oct 6, 2012 #3
    Thank you for your reply :)
    It seems your hint works out quite nicely, please check my working:
    Observe that [itex]Ax = Ix = x \Rightarrow {A^2}x = Ax = x[/itex]
    Hence [itex]2{A^2} + A = 4I \Leftrightarrow 2{A^2}x + Ax = 4Ix \Leftrightarrow 2x + x = 4x \Leftrightarrow x = 0[/itex]
    This completes the proof.
     
  5. Oct 6, 2012 #4

    tiny-tim

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    fine! :smile:
     
  6. Oct 6, 2012 #5
    Yeah, thanks!
     
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