# Determine the steady state from a discrete dynamic system with only the eigenvalue of the diagonalized transition matrix.

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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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 $B$.
$B = P \pmatrix{0 & 0 & 0\cr 0 & 0 & 0\cr 0 & 0 & 1\cr} P^{-1}$ – Robert Israel Mar 5 '12 at 16:05
$A^k = P D^k P^{-1}$. Consider each nonzero element of $D^k$... – Robert Israel Jun 25 '15 at 19:07