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Math Help - Eigenvalue Proof

  1. #1
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    Eigenvalue Proof

    I am suppose to prove that if R is a matrix, R^2 = I and lambda is an eigenvalue, then lamda is 1 or - 1

    Do I somehow need to show that an eigenvector, say it is v that Rv and R^2 v ??? compared but I am lost!

    Is it just that since R is squared then either 1 or -1 are values??? Frostking
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    Quote Originally Posted by Frostking View Post
    I am suppose to prove that if R is a matrix, R^2 = I and lambda is an eigenvalue, then lamda is 1 or - 1

    Do I somehow need to show that an eigenvector, say it is v that Rv and R^2 v ??? compared but I am lost!

    Is it just that since R is squared then either 1 or -1 are values??? Frostking
    R^2=I\Longrightarrow (R-I)(R+I)=0\Longrightarrow \mbox{R is a zero of the polynomial }(x-1)(x+1)\Longrightarrow \mbox{the minimal pol. of R divides }(x-1)(x+1)\Longrightarrow\,...

    Tonio
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    A slightly different proof, without using the minimal polynomial (directly):

    If \lambda is an eigenvalue of R then there exist non-zero v such that Rv= \lambda v. Apply R to both sides of that equation- R^2v= R(\lambda v)= \lambda Rv= \lambda(\lambda v)= \lambda^2 v. But it is given that R^2= I so we are saying that v= \lambda^2 v so that (1- \lambda^2)v= 0. Since v is not 0, we must have 1- \lambda^2= 0 and, from that, \lambda= 1 or \lambda= -1.
    Last edited by HallsofIvy; November 4th 2009 at 02:06 AM.
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    Quote Originally Posted by HallsofIvy View Post
    A slightly different proof, without using the minimal polynomial (directly):

    If \lambda is an eigenvalue of R then there exist non-zero v such that Rv= \lambda v. Apply R to both sides of that equation- R^2v= R(\lambda v)= \lambda Rv= \lambda(\lambda v)= \lambda^2 v. But it is given that R^2= I so we are saying that v= \lambda^2 v so that (1- \lambda^2)v= 0. Since v\neq 0, we must have 1- \lambda^2= 0 and, from that, \lambda= 1 or \lambda= -1.
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