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Follow up to B=I-2A-A^2 where mu=1-2lambda+(lambda)^2

  1. Apr 18, 2010 #1
    Show that if [tex]\lambda=1[/tex] is an eigenvalue of A, then the matrix B will be singular.

    [tex]\mu(1)=1-2+1=0[/tex]

    [tex]B\mathbf{x}=0\mathbf{x}[/tex]

    or does this need to be done via determinant?

    [tex]det(B-\mu I)=0[/tex] since [tex]\lambda=1, \mu=0[/tex].

    [tex]det(B-0I)=det(B)=0[/tex] Hence B is singular.
     
  2. jcsd
  3. Apr 18, 2010 #2

    Mark44

    Staff: Mentor

    This is true, but how did you arrive at it? Also, what's the significance of this equation in terms of what you're supposed to show?
    This is another way to show it.
     
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