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Homework Help: Linear Algebra Matrix Proof

  1. Oct 2, 2013 #1
    1. The problem statement, all variables and given/known data

    Prove or disprove the following statements. I and 0 denote respectively the identity and zero matrix of the same size as A. If A is a square matrix such that A^2 - 3A +2I = 0 then A-cI is invertible whenever c is not equal to 1 and c is not equal to 2.

    2. Relevant equations

    3. The attempt at a solution

    I have factored the function to: (A-2I)(A-I)=0.

    However, we can't assume that A=2I and A=I because we are dealing with matrices (i.e. two non zero matrices can produce the zero matrix when multiplied together). I have a feeling that Eigenvectors might be related to this question, but I don't know how to apply that concept in this scenario.

    Thanks for your help.
  2. jcsd
  3. Oct 2, 2013 #2


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    Do you have the Cayley-Hamilton theorem available?
  4. Oct 2, 2013 #3


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    Yes, think about eigenvalues indirectly. Suppose A-cI is not invertible. That means there is a nonzero vector v such that (A-cI)v=0. So Av=cv. Can you use that to show (A^2-3A+2I)v=0 can only be true if c=2 or c=1?
    Last edited: Oct 2, 2013
  5. Oct 2, 2013 #4
    LCKurtz, I do not have the Cayley-Hamilton theorem available. It has not been presented in class yet.

    Dick, I have understood the following:

    So Av=cv
    I imagine that we need det(A-cI)=0 in order to prove that A-cI is invertible.

    I am stuck here. I don't know how to relate A^2 -3A +2I to the previous equations.
  6. Oct 2, 2013 #5


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    Go with Dick's suggestion. If Av = cv, try to evaluate (A2 - 3A + 2I)v, which you know has to be zero by the statement of the problem.
  7. Oct 2, 2013 #6
    Ok. Here is what I have:

    (A^2 -3A +2I) = 0
    A^2v -3Av +2Iv = 0
    AAv -3Av + 2v = 0
    (c^2)v - 3cv + 2v = 0
    (c^2 - 3c + 2)v = 0

    Therefore (A-cI) is not invertible when c=1 c=2.

    I think I understand now. Does everything seem to be in order in my procedure?

    Thanks for your help.
  8. Oct 2, 2013 #7


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    Yes, that's it. The important point is that (c-2)(c-1)v=0 is ONLY possible if c=1 or c=2, since v is nonzero.
  9. Oct 2, 2013 #8
    Perfect. Thanks for your help!
  10. Oct 3, 2013 #9


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    Your last statement is wrong- you are asserting that "(A-cI) is not invertible when c=1 c=2" when what you want to prove is "A- cI is invertible as long as long as [itex]c\ne 1[/itex] and [itex]c\ne 2[/itex]", the converse. Did you mean to say "(A-cI) is not invertible only when c=1 c=2"?

  11. Oct 3, 2013 #10
    OP, you should definitely not take this as excessively pedantic. What you have written, though somewhat correct in the vernacular language, is incorrect in terms of mathematical formalism.
  12. Oct 3, 2013 #11
    So then how *can* you show that (A-cI) **is** invertible when c /= 1, 2?
  13. Oct 3, 2013 #12
    It is shown. What he has said is different from that.
  14. Oct 3, 2013 #13
    Sorry "/= 1,2" is meant as "not equal to 1, 2" which, as HallsofIvy pointed out, is not proven.
  15. Oct 3, 2013 #14


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    It is if you the organize the thoughts correctly. We assumed A-cI is NOT invertible and concluded that then there is a nonzero vector such that (c-1)(c-2)v=0. That's a contradiction unless c=1 or c=2. Hence if c is not equal to 1 or 2 then A-cI IS invertible. Treat it as a proof by contradiction.
  16. Oct 8, 2013 #15
    Thanks Dick - don't know why I missed that; guess I got disoriented in the semantics. Thanks for clarifying it explicitly :D.
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