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I need help solving this problem Assume $\lambda =-1$ is an eigenvalue of a 3x3 matrix A and $x= [2\ 3\ 4]^{T}$ is an eigenvector corresponding to this $\lambda$. Find $A^{101}x$. I have no idea how to use the eigenvalue and vector to work my way up to $A$. I do know that $A^{101}=PD^{101}P^{-1}$ where $P$ is a matrix made from the eigenvectors as its columns and $D$ is a diagonal matrix

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Hint:

For eigenvalues $\lambda$, you would have $Ax = \lambda x$, so you would have $A^2x=A\lambda x=\lambda^2x$. Then you can go on from this logic.

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  • $\begingroup$ is there any way of solving it using $PD^{101}P^{-1}$? $\endgroup$ – pkid58 Nov 27 '16 at 23:28
  • $\begingroup$ Probably, but you don't need to. You've been given everything you need already. $\endgroup$ – ConMan Nov 27 '16 at 23:34
  • $\begingroup$ You can not do it in $PD^{101}P^{-1}$ since you only know one eigenvalue, if you know other twos, then it is very easy to calculate in your way. $\endgroup$ – duanduan Nov 27 '16 at 23:52
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Hint: If $Ax = \lambda x$, then $$A^2x = A\lambda x = \lambda Ax = \lambda^2 x$$ Can you take it from here?

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  • $\begingroup$ is there any way of solving it using $PD^{101}P^{-1}$? $\endgroup$ – pkid58 Nov 27 '16 at 23:28
  • $\begingroup$ @pkid58 Well, if your matrix can not be diagonalized, then you can't assume that such a decomposition exists. If it is, then yes you can use it but this approach is more complicated. $\endgroup$ – Hello Nov 27 '16 at 23:46

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