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Solve A^2+i=0

  1. Feb 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A be an n x n matrix. Is it possible for A2+I=0 in the case when n is odd?

    So A is a 2p+1 x 2p+1; however, I don't see this making a difference to the proof if n is odd or even.

    The only way I view A2+I=0 is if A has zero has every elements except when i=j where all a11 to a(2p+1)(2p+1) elements are equal to i=[tex]\sqrt{-1}[/tex].

    Other then this observation I have made I am lost on this problem.
    Last edited: Feb 12, 2010
  2. jcsd
  3. Feb 12, 2010 #2
    Re: A^2+i=0

    Are we talking about real matrices or complex matrices? If complex matrices are allowed then your construction is perfectly valid. If not note that det(-I) = -1 when n is odd (this is where the oddness is required), and therefore we have,
    [tex]-1=det(-I) = det(A^2) = det(A)^2[/tex]
    Is this possible for a real matrix A?
  4. Feb 12, 2010 #3
    Re: A^2+i=0

    The book doesn't state whether it is speaking of real or complex. But when -I2, we will obtain I+I not 0.
  5. Feb 12, 2010 #4
    Re: A^2+i=0

    Due to the fact that the condition of being odd is necessary for the real case, but not the complex I would assume the real case is meant. I'm sorry but I don't understand your last sentence. If it's about my solution, then maybe I should expand a bit. I simply subtracted I to get [itex]A^2 = -I[/itex] and then took the determinant of both sides.
  6. Feb 12, 2010 #5
    Re: A^2+i=0

    I saw it a different way but given your method and how I thought of it ((-I2)+I=0). When is a real number squared to obtain -1? So I am guessing A2+I=0 when n is odd is never unless we consider complex solutions; however, isn't this the same when n is even?
  7. Feb 12, 2010 #6
    Re: A^2+i=0

    No because if I is the n x n identity matrix, then -I is the nxn diagonal matrix with -1 as its only diagonal element. Thus the determinant is,
    [tex]det(-I) = (-1)^n[/tex]
    In the odd case this gives us -1 which as you rightly observed is impossible for real matrices. However in the even case we get 1 and then my equation would simply say
    [tex]1 = det(A)^2[/tex]
    which is easy to find matrices satisfying. Consider for instance the matrix,
    [tex]A = \left[\begin{array}{cc} 1 & 1 \\ -2 & -1 \end{array} \right][/tex]
    which actually satisfy A^2 = -I (if my quick calculations are correct).
  8. Feb 12, 2010 #7
    Re: A^2+i=0

    This only works then when n is even?
  9. Feb 12, 2010 #8
    Re: A^2+i=0

    What does "this" refer to? I only made two claims regarding the case where n is even:
    1) If n =2, then your statement definitely isn't true (as I provided a counterexample).
    2) The proof I gave does not work when n is even because det(I)=1 IF n is even.
    As I have mentioned previously det(-I) is the product of the diagonal elements and therefore it's (-1)^n which is -1 if n is odd. Thus if n is odd and A^2 = -I we get:
    [tex]-1 = det(A)^2[/tex]
    As a real matrix has a real determinant and no real number squared is -1 we get a contradiction.
  10. Feb 12, 2010 #9
    Re: A^2+i=0

    You example matrix A when squared is -I and is a 2 x 2
  11. Feb 12, 2010 #10
    Re: A^2+i=0

    That just shows that it's possible for A^2 + I =0 when n is even (namely when n is 2) so a proof could never possibly prove it in the even case too.
  12. Feb 12, 2010 #11
    Re: A^2+i=0

    That doesn't make sense to me. "That just shows that it's possible...so a proof could never possibly prove it..."

    How can you show it and then say it isn't possible?
  13. Feb 12, 2010 #12
    Re: A^2+i=0

    I haven't proved it. I presented a proof that says the following:
    - (1) If n is odd, then there does not exist a real nxn matrix A such that A^2 + I = 0.
    I then gave an example that shows,
    - (2) If n is 2, then there does exist a real nxn matrix A such that A^2 + I = 0.
    What I simply tried to emphasize is that my proof doesn't work when n is even, and you can't possibly construct a proof that would work when n is even because then it would contradict (2).
  14. Feb 12, 2010 #13
    Re: A^2+i=0

    Based on det(A2)=(-1)n couldn't there exist matrices when n is odd?
  15. Feb 12, 2010 #14
    Re: A^2+i=0

    No. Because imagine that such a matrix A existed. When n is odd [itex](-1)^n=-1[/itex] so,
    [tex]det(A^2) = (-1)^n = -1[/tex]
    Also since det is multiplicative we have [itex]det(A^2) = det(AA) = det(A)det(A) = det(A)^2[/itex] so we would have,
    [tex]det(A)^2 = -1[/tex]
    However det(A) is real so in that case we would have a real number whose square is negative, but this is a contradiction so such a matrix can't exist.
  16. Feb 12, 2010 #15
    Re: A^2+i=0

    So it isn't true in either case but there are some in the even case? However, if the elements can be complex, it is true for both cases then?
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