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For a discrete dynamical system I know that the transition matrix A is diagonalizable with the eigenvalues of 0.1, 0.2, and 0.3. The question asks what I can say about the long term behaviour of the system.

I know that $$ X_k = A^{k}X_0 $$ and $$ A = PDP^{-1} $$ $$ A^k = PD^kP^{-1} $$

Thus I choose to approach the question with by combining the two equations to: $$ X_k = PD^kP^{-1}X_0 $$ as I know that D is \begin{pmatrix}0.1&0&&0\\ 0&0.2&&0\\ 0&0&&0.3\end{pmatrix}

Though as 0.1, 0.2, and 0.3 to a large number (k) all equal 0. Thus my equation becomes: $$ X_k = P \begin{pmatrix}0&0&&0\\\ 0&0&&0\\\ 0&0&&0\end{pmatrix} P^{-1}X_0 $$ so I get $$X_k = \begin{pmatrix}0&0&&0\\\ 0&0&&0\\\ 0&0&&0\end{pmatrix} $$

My question is that this seems too simple and the steady state should not be a zero matrix. I feel like I am missing something but the eigenvalues of the diagonalizable transition matrix is the only information we are given and I have no idea what else the answer can be. Is this right or am I missing something?

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up vote 1 down vote accepted

It is true that the limit is $0$, but you can say more about the long-term behaviour. The leading term will come from the largest eigenvalue. Specifically, you should have $A^k \sim (.3)^k B$ as $k \to \infty$ for a certain matrix $B$.

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What exactly is B? Is it Xo? Otherwise that makes sense. Thank you – Kyra Mar 5 '12 at 15:38
$B = P \pmatrix{0 & 0 & 0\cr 0 & 0 & 0\cr 0 & 0 & 1\cr} P^{-1}$ – Robert Israel Mar 5 '12 at 16:05
@RobertIsrael I realize that this is an old question but could you elaborate why "The leading term will come from the largest eigenvalue."? I have seen this coming up in different scenarios but I don't know why it's happening. – Yasmin Jun 25 '15 at 17:21
$A^k = P D^k P^{-1}$. Consider each nonzero element of $D^k$... – Robert Israel Jun 25 '15 at 19:07

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