Given a stochastic matrix, M, and the knowledge that a steady state exists in the process, can I find the initial state vector as a linear combination of the eigenvectors associated with that matrix? If so, which theorem or rule allows me to do that?
What happens if said matrix has a negative eigenvalue? Apparently, I can't use the eigenvector associated with an eigenvalue of -1 to find the initial state vector (why?), but is this a general rule for every negative eigenvalue or only -1?
I can't find this anywhere!
For example, given the matrix:
$ \begin{bmatrix}
\frac{1}{4} & \frac{1}{2} & \frac{1}{2} & \frac{1}{2} \\
\frac{1}{4} & \frac{1}{2} & 0 & 0 \\
\frac{1}{4} & 0 & \frac{1}{2} & 0 \\
\frac{1}{4} & 0 & 0 & \frac{1}{2}
\end{bmatrix} $
The eigenvalues (represented by $\lambda $) are:
$ \lambda_{1}=1, \lambda_{2}=\frac{1}{2}, \lambda_{3}=\frac{1}{2}; \lambda_{4}=-\frac{1}{4}. $
And the eigenvectors (represented by $v_{\lambda }$) for the positive eigenvalues:
$v_{1}=(2, 1,1,1); v_{\frac{1}{2}}= {(0,-1,0,1), (0,-1,1,0)}$
So, is it correct to write the initial state vector $X_{0}$ as a linear combination of the eigenvectors of the non-negative eigenvalues?:
$X_{0}=\alpha (2,1,1,1)+ \beta (0,−1,0,1) + \gamma (0,−1,1,0)$
Best Answer
Edit: The original question was about steady states. Now it turns out the OP really meant to ask about the initial state. Answering both questions:
A steady state is represented by a vector $v$ with $Mv=v$. That's the same as an eigenvector with eigenvalue $\lambda=1$. (So in that example the steady state vectors are exactly the scalar multiples of $v_1$.)
The question of how you "find" the initial state doesn't make much sense - the initial state can be anything. Yes, if M is diagonalizable then you can write the initial state as a linear combination of eigenvectors. This is more or less by definition: An $n\times n$ matrix is diagonalizatble if and only if the eigenvectors span $\Bbb R^n$. Whether the eigenvalues are all positive is irrelevant to this.
A second ago I said I thought $-1$ could not be an eigenvalue. That was wrong; $M=\begin{bmatrix}0&1\\1&0\end{bmatrix}$ is a counterexample.
If you have a process $v_k=M^kv_0$, if $M$ is diagonalizable and if $v_k\to v$ then yes, the initial vector must be a linear combination of eigenvectors for eigenvalues $\lambda\ne1$. This is clear, because if $$v_0=\sum_ja_jw_j$$where $w_j$ is an eigenvector with eigenvalue $\lambda_j$ then $$v_o^k=\sum_j a_j\lambda_j^kw_j,$$and $(-1)^k$ does not converge as $k\to\infty$. (Other negative eigenvales of modulus less than $1$ are no problem; if $-1<\lambda<0$ then $\lambda^k\to0$.)