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I diagonalised $A=$ $\left[\begin{array}[c]{rr} 0.6 & 0.9\\ 0.4 & 0.1\end{array}\right]$

and got

$SAS^-$$^1$$= V = (-1/13)$ $\left[\begin{array}[c]{rr} -1 & -1\\ -4 & 9\end{array}\right]$ $\left[\begin{array}[c]{rr} 1 & 0\\ 0 & -0.3\end{array}\right]$ $\left[\begin{array}[c]{rr} 9 & 1\\ 4 & -1\end{array}\right]$

now I am supposed to somehow deduce what matrix $V^k$ approaches as $k$ approaches $∞$. How am I supposed to figure that out?

Also, I am supposed to find the limit for $SV^kS^-$$^1$ when $k$ approaches $∞$ and specify what the columns of this matrix portray.

What came to mind is that when all $|λ|<1$ then $A^k$ approaches $0$, and $λ_1=1$ and $λ_2 = -0.3$. I'm not really sure if that's of any use though. I'm somewhat lost and any help is much appreciated!

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    $\begingroup$ First, the diagonalization does not look correct. For the second part, see: math.stackexchange.com/questions/731808/… . Once you have the diagonalization, the rest is straightforward. $\endgroup$ – Amzoti Apr 3 '14 at 22:12
  • $\begingroup$ The diagonalization is correct. Yours thoughts on out to proceed are also correct. What is the problem? $\endgroup$ – Git Gud Apr 3 '14 at 22:17
  • $\begingroup$ I fixed the diagonalisation, I realised it was wrong after Amzoti pointed it out. Well from what I understand for $λ_1$ the matrix approaches infinity and for $λ_2$ it approaches 0. That is for matrix A though I assume, so I don't know how that helps with the matrices I need to find, nor am I sure how to put it into the form of a matrix once I do find a limit. $\endgroup$ – mangopancake Apr 3 '14 at 22:30
  • $\begingroup$ @mangopancake $\lim \limits_{k\to\infty}\left(SV^kS^{-1}\right)=S\lim \limits_{k\to\infty}\left(V^k\right)S^{-1}$. Does this help? $\endgroup$ – Git Gud Apr 3 '14 at 22:34
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Your solution looks close.

Write the matrix as fractions:

$$A=\left[\begin{array}[c]{rr} \dfrac{6}{10} & \dfrac{9}{10}\\ \dfrac{4}{10} & \dfrac{1}{10}\end{array}\right]$$

I have my eigenvalues and eigenvectors swapped from the order you show yours in.

Diagonalization yields:

$$A = PJP^{-1} = \left(\begin{array}{cc} -1 & \frac{9}{4} \\ 1 & 1 \\\end{array} \right)\left(\begin{array}{cc} -\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right) \left(\begin{array}{cc} -\frac{4}{13} & \frac{9}{13} \\ \frac{4}{13} & \frac{4}{13} \\ \end{array} \right)$$

This yields:

$$A^k = PJ^kP^{-1} = \left( \begin{array}{cc} \frac{1}{13} 2^{2-k} \left(-\frac{3}{5}\right)^k+\frac{9}{13} & ~\frac{9}{13}-\frac{1}{13} \left(-\frac{1}{10}\right)^k 3^{k+2} \\ \frac{4}{13}-\frac{1}{13} \left(-\frac{3}{5}\right)^k 2^{2-k} & ~\frac{1}{13} 3^{k+2} \left(-\frac{1}{10}\right)^k+\frac{4}{13} \\ \end{array} \right)$$

Note: Once we have diagonalized the matrix, we have:

$$J^k = \left(\begin{array}{cc} -\frac{3}{10} & 0 \\ 0 & 1 \\\end{array}\right)^k = \left(\begin{array}{cc} \left(-\frac{3}{10}\right)^k & 0 \\ 0 & (1)^k \\\end{array}\right)$$

Now, as far as the limit goes, do you see what happens to the $k$ terms as $k$ approaches infinity? They approach zero, so we are left with:

$$\displaystyle \lim_{k \to \infty} A^k = \left( \begin{array}{cc} \frac{9}{13} & ~\frac{9}{13} \\ \frac{4}{13} & ~\frac{4}{13} \\ \end{array} \right) $$

An easier approach would have been to take the limit of the diagonalized matrix first, and then multiply out, which yields:

$$\displaystyle \lim_{k \to \infty} J^k = \left( \begin{array}{cc} 0 & 0 \\ 0 & 1 \\ \end{array} \right) $$

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