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Matrix invertibility question

  1. Dec 6, 2011 #1
    I just finished a final in a linear algebra course and was unsure about one of the questions. The question was:

    If A^2 - A + I = 0 , show that A is invertible.

    My approach was that det(A^2 + I) = det(A)

    det(A^2 + I) will never be zero, so det(A) is non-zero and therefore A is invertible.

    Is this the right way of doing this problem?

  2. jcsd
  3. Dec 6, 2011 #2
    Wh will [itex]det(A^2+I)[/itex] never be 0??

    Perhaps the easiest way to approach this is to calculate the eigenvalues...
  4. Dec 6, 2011 #3
    I cannot calculate the eigenvalues as A is an arbitrary matrix. The information in my original post is all that was given.
  5. Dec 6, 2011 #4
    Have you heard about minimal/characteristic polynomials??
  6. Dec 6, 2011 #5
    Oh I see now.

    A2 - A + I = 0

    A2 - A + det(A)*I = 0 <-- from Cayley-Hamilton

    So det(A)=1 and therefore A is invertible

    Is this where you were leading me?

    I can't recall now whether we were told A is 2x2. Does that matter?
  7. Dec 6, 2011 #6
    Yes, but it needs some tweaking.

    For example, the characteristic polynomial doesn't need to be [itex]x^2-x+1[/itex]. But can we say something about the minimal polynomial??
  8. Dec 6, 2011 #7
    Sorry to butt in, but I can't see why you need to use the minimal polynomial or Cayley-Hamilton Theorem here..
    You have:
    A^2 - A + I = 0

    I = A - A^2
    I = A(I-A)

    So inv(A) = (I - A) and A must be invertible..?
    Sorry if I've made a mistake but this is sound as far as I can tell..

    Also: this may not be the most obvious way of doing it- you can use minimal polynomials but Jonny I have no idea how you're arguing that the Cayley-Hamilton Theorem implies A2 - A + det(A)*I = 0 ..
    The question tells you that A^2 - A + I = 0, what does this imply about the minimal polynomial? And from there what can you deduce about the characteristic polynomial, and hence about the eigenvalues?
    Last edited: Dec 6, 2011
  9. Dec 6, 2011 #8


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    I'll buy that.

    I don't see where the characteristic polynomial would lead to. You know a factor of it, but if the order of A is > 2, you don't know anything about the other factors.

    ... unless Micromass has seen something that I haven't
  10. Dec 6, 2011 #9
    Of course the method Zoe posted is superior. But the method with the characteristic polynomial works too. The point is that the minimal polynomail is the least P(X) such that P(A)=0. So if a polynomial is 0 in A then the minimal polynomial divides it.

    In our case, we know that the minimal polynomial will divide [itex]A^2-A+I[/itex].

    The crucial part now is that the eigenvalues of our matrix are exactly the roots of the minimal polynomial. So seeing [itex]A^2-A+I=0[/itex] immediately gives us information about the eigenvalues.
  11. Dec 6, 2011 #10
    I believe you've been foiled micromass. As I said, I didn't think the problem mentioned that A was 2x2, which if I'm wrong, and I may well be, you really can't imply anything about the eigenvalues from the given equation.

    Mind you, this is a 600 level linear algebra course, and also my first course in linear algebra, so I'm obviously naive on the topic. It appears however, that Zoe-b's solution is THE solution in this case.

    Feel free to give your whole spiel now micromass, as I'm satisfied on the topic with the given solution.
  12. Dec 6, 2011 #11


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    OK. After thinking about the special case where A is diagonal, I get it.

    Before, I was thinking "yeah, but the other N-2 eigenvalues might be anything...."
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