linear algebramarkov chainsmatrices
Suppose to have the following 2 x 2 Markov Matrix.
$\begin{pmatrix}
.8 & .1\\
.2 & .9
\end{pmatrix}$
Considering the system
$\begin{pmatrix}
.8 \\
.2
\end{pmatrix} \cdot 150 + \begin{pmatrix}
.1\\
.9
\end{pmatrix} \cdot 150 = \begin{pmatrix}
a\\
b
\end{pmatrix}$
when $a= 100$ and $b = 200$ the entire system is in equilibrium.
Is there a theorem about reaching (always?) an equilibrium point for a Markov matrix?
The Markov chain converges to a unique equilibrium if there is only one recurrent class and it is aperiodic. In this case the directed graph corresponding to the Markov chain looks like this:
The Markov chain is irreducible (i.e. every state communicates with every other state), but it is periodic with period $2$, because from states $1$ and $2$ you can only go to $3$ and $4$ and vice versa. Therefore it does not converge to an equilibrium. In fact, the even powers of the matrix converge to $$ \frac{1}{27} \pmatrix{13 & 14 & 0 & 0\cr 13 & 14 & 0 & 0\cr 0 & 0 & 20 & 7\cr 0 & 0 & 20 & 7\cr}$$ and the odd powers converge to $$ \frac{1}{27} \pmatrix{0 & 0 & 20 & 7\cr 0 & 0 & 20 & 7\cr 13 & 14 & 0 & 0\cr 13 & 14 & 0 & 0\cr}$$
I dont know if you have to compute the conditioning $\kappa$ as the ratio of the eigenvalues $\max(\sigma)/\min(\sigma)$ under norm 2, which would not require to compute the actual inverse.
From the R manual: "The condition number takes on values between 1 and infinity, inclusive, and can be viewed as a factor by which errors in solving linear systems with this matrix as coefficient matrix could be magnified."
The condition number must be specified by the required resolution of your algorithm.
Double Precision Representation
If you are having computations under IEEE 754 64-bit double precision representation, i.e. 64 bits per value, such as MATLAB, you will have:
From here you can build some VERY rough estimates, which are only referencial about how deep the figures can go.
Significative Order
The smallest order is $10^{-308}$. If your condition number / eigenvalue ratio $\kappa$ is $10^{-6}$, you could only make 308/6=51 products of the power of A approx., before the smallest or the biggest eigenvalue got truncated.
For example, if $v=10^{6}$, $v^{51}=10^{306}$, and $v^{52}=$Inf.
From here, the allowable conditioning would be given as, with $n$ the number of expected iterations using the compromised values: $$\kappa=10^{-308/n}$$
Significative Digits
By other side, the representation allows 52 bits i.e. $10^{-15}$ significative digits on the fractions. Hence if you want to keep a resolution in significative digits of $10^{-6}$, you could only make 15-6=9 products approximatedly of the power of A before the smallest or the biggest eigenvalue got truncated.
For example:
From here, the allowable conditioning would be given as, with $n$ the number of expected iterations using the compromised values: $$\kappa=10^{15-n}$$
Hence, it must be observed from the problem factors which is the proper conditioning figure your calculation should retain, based on the usage of the data and the purpose of the algorithms.
https://stat.ethz.ch/R-manual/R-devel/library/Matrix/html/rcond.html https://www.mathworks.com/help/matlab/ref/cond.html https://en.wikipedia.org/wiki/Condition_number https://en.wikipedia.org/wiki/Double-precision_floating-point_format
Best Answer
This (stochastic) matrix $A$ has two eigen values $\lambda_1=1$ and $\lambda_2=0.7$ so it is diagonalizable, i.e. $A=P\begin{bmatrix}1&0\\0&0.7\end{bmatrix}P^{-1}=PDP^{-1}$, where $P=\begin{bmatrix}1&-1\\2&1\end{bmatrix}$
To get the equilibrium vector. We want to compute $v$ such that $\lim_{n \to \infty}A^nv=v$.
Now $v=c_1v_1+c_2v_2$, where $v_1,v_2$ are eigenvectors corresponding to the eigenvalues $1$ and $0.7$.
Thus $$A^nv=c_1A^nv_1+c_2A^nv_2 =c_1(1)^nv_1+c_2(0.7)^nv_2$$ $$\lim_{n \to \infty}A^nv=c_1v_1$$ So the multiple of first eigenvector (for eigenvalue $1$) is the equilibrium vector.