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Idempotent Matrix

  1. Apr 18, 2010 #1
    An nxn matrix A is said to be idempotent if [tex]A^2=A[/tex]. Show that if [tex]\lambda[/tex] is an eigenvalue of an idempotent matrix, then [tex]\lambda[/tex] must be 0 or 1.

    The only reason I can think of is that it must 0 or 1 because if you square the values 0 and 1 don't change.
  2. jcsd
  3. Apr 18, 2010 #2
    Alright so let x be an eigenvalue to A. That means that Av=xv for some vector v. Then xv=Av=A^2v=A(Av)=A(xv)=x(Av)=x^2v which implies x^2=x so x(x-1)=0 which implies that the eigenvalues to A are either 1 or 0
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