# by finding the eigenvalues and eigenvectors the evaluate the following.

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$=\begin{bmatrix}0\\0\end{bmatrix}$

• There are directions in the FAQ section to write properly mathematics in this site with LaTex. Please follow them. – DonAntonio Apr 24 '13 at 12:52
• Eigenvectors are non-zero vectors. The eigenvector with eigenvalue $1$ should be $\begin{bmatrix}1\\0\end{bmatrix}$. – robjohn Apr 24 '13 at 15:48
• I am sorry for the bad representation of my question. thanks @robjohn for fixing that !! :D – nicolas Apr 24 '13 at 19:13
• 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. – Kosta Apr 7 '18 at 0:42

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.

• Thanks for your answer !! I completely got it now thanks. – nicolas Apr 24 '13 at 19:12
• @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 – Amzoti Apr 24 '13 at 19:57
• Good job, Amzoti! Nice feedback, too! +1 – Namaste Apr 25 '13 at 0:20
• Hmmm...slow...a few questions I've answered...but need something to did my teeth into! – Namaste Apr 25 '13 at 3:18
• Yes, I agree. And the OPs whom I've answered seem to have gone "awol" this evening! – Namaste Apr 25 '13 at 3:21