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Lin. Alg. Invertility

  1. May 15, 2009 #1
    To find out if 4 in an eigenvalue of A, decide if A-4I is invertible...

    So, if A-4I is invertible, then its cols are lin ind by IMT, and also there is only the trivial solution to A-4I=0, so thus 4 is not an eigenvalue of A

  2. jcsd
  3. May 15, 2009 #2


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    ???? The definition of "eigenvalue" is that [itex]\lambda[/itex] is an eigenvalue if and only if [itex]Av= \lambda v[/itex] has non-trivial solutions. That is the same as saying that [itex]Av- \lambda v= (A- \lambda I)v= 0[/itex] has non-trivial solutions. Since v=0 is obviously a solution, saying it has non-trivial solutions means it does NOT have a "unique" solution. If the matrix M has an inverse, then the equation Mv= u has the unique solution [math]v= M^{-1}u[/math].
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