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The eigen values a matrix A are 0,1,3. The corresponding eigen vectors are $$\begin{bmatrix} 1 \\ 1 \\ 1 \\ \end{bmatrix}, \begin{bmatrix} 1 \\ 0 \\ -1 \\ \end{bmatrix}, \begin{bmatrix} 1 \\ -2 \\ 1 \\ \end{bmatrix} $$

Need to find $A^4$.

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  • $\begingroup$ eigenvalues are $0^4, 1^4$ and $3^4$ and eigenvectors are the same $\endgroup$ – SSepehr Feb 2 '17 at 11:38
  • $\begingroup$ Can you share a proof of the above result? $\endgroup$ – Pruthvi Bharadwaj Feb 2 '17 at 11:40
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    $\begingroup$ $A^2v = A(Av)=A(\lambda v) = \lambda (Av) = \lambda^2 v$ $\endgroup$ – SSepehr Feb 2 '17 at 11:40
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Writing the matrix defined by the eigenvalues is trivial and is given by $$D =\left( \begin{array}{ccc} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 3 \end{array} \right).$$ Now the matrix whose columns are the eigenvectors is given by $$P =\left( \begin{array}{ccc} 1 & 1 & 1 \\ 1 & 0 & -2 \\ 1 & -1 & 1 \end{array} \right).$$ The required matrix can be obtained by changing to the standard coordinate bases by computing $A = PDP^{-1} $. Now we thus get $$\boxed {A^4 =PD^4P^{-1}} $$

Can you take it from here? Hope it helps.

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