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so the question is : by finding the eigenvalues and eigenvectors of the matrix $$ P=\begin{bmatrix}1&6\\0&-2\end{bmatrix}\qquad\text{evaluate }P^{20}\begin{bmatrix}-2\\1\end{bmatrix} $$ I found the eigenvalues and eigenvectors but I can sort out how I will evaluate the second part.

eigenvalue$=-2$; eigenvector$=\begin{bmatrix}-2\\1\end{bmatrix}$

eigenvalue$=1$; eigenvector$=\begin{bmatrix}0\\0\end{bmatrix}$

Thank you in advance. Hugo

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  • 2
    $\begingroup$ There are directions in the FAQ section to write properly mathematics in this site with LaTex. Please follow them. $\endgroup$ – DonAntonio Apr 24 '13 at 12:52
  • $\begingroup$ Eigenvectors are non-zero vectors. The eigenvector with eigenvalue $1$ should be $\begin{bmatrix}1\\0\end{bmatrix}$. $\endgroup$ – robjohn Apr 24 '13 at 15:48
  • $\begingroup$ I am sorry for the bad representation of my question. thanks @robjohn for fixing that !! :D $\endgroup$ – nicolas Apr 24 '13 at 19:13
  • $\begingroup$ It would be helpful to see your attempt at the problem, so we can properly gauge where the problem is, rather than giving you a full solution. $\endgroup$ – Kosta Apr 7 '18 at 0:42
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I agree with the comments, that this was very hard to read and I hope I understood it properly.

You should get an Eigensystem as follows:

$$\lambda_1 = -2, v_1 = (-2, 1)$$

$$\lambda_2 = 1, v_2 = (1, 0)$$

Next, we can write the Jordan Normal Form as:

$$P = S J S^{-1} = \begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$$

What do you notice about $J$? What is it made from?

What do you notice about the columns of $P$? What is it made from?

Now that we have the Jordan Normal Form (diagonalized matrix), we can use to find the matrix power:

$$P^{20} = \left(S J S^{-1}\right)^{20} = \left(S J^{20} S^{-1}\right) = \begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} -2 & 0 \\ 0 & 1 \end{bmatrix}^{20} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$$

Of course, a diagonal matrix to a power is just each value of that diagonal entry to that power, so we get:

$$P^{20} = \begin{bmatrix} -2 & 1 \\ 1 & 0 \end{bmatrix} \cdot \begin{bmatrix} 1048576 & 0 \\ 0 & 1 \end{bmatrix} \cdot \begin{bmatrix} 0 & 1 \\ 1 & 2 \end{bmatrix}$$

You can now just multiply that out for the final result.

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  • $\begingroup$ Thanks for your answer !! I completely got it now thanks. $\endgroup$ – nicolas Apr 24 '13 at 19:12
  • $\begingroup$ @hugonicolasadario: You are very welcome. I should have also mentioned that there are other ways to do this, but I was trying to use your approach to provide guidance. Regards $\endgroup$ – Amzoti Apr 24 '13 at 19:57
  • $\begingroup$ Good job, Amzoti! Nice feedback, too! +1 $\endgroup$ – Namaste Apr 25 '13 at 0:20
  • $\begingroup$ Hmmm...slow...a few questions I've answered...but need something to did my teeth into! $\endgroup$ – Namaste Apr 25 '13 at 3:18
  • $\begingroup$ Yes, I agree. And the OPs whom I've answered seem to have gone "awol" this evening! $\endgroup$ – Namaste Apr 25 '13 at 3:21

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