Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: 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?
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook