Question

10. Consider the probability transition matrix $T = \begin{bmatrix} 3/4 & 1/2 \ 1/4 & 1/2 \end{bmatrix}$ for a two-state Markov chain. It is a fact that $det(T - \lambda I) = \lambda^2 - \frac{5}{4}\lambda + \frac{1}{8} = \frac{1}{8}(\lambda - 1)(\lambda - \frac{1}{4})$ so that the eigenvalues of T are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{4}$. Find corresponding eigenvectors for the eigenvalues above. For the eigenvalue 1, scale your eigenvector so that the two entries sum to 1. Given this information, show that for any initial population distribution vector $\vec{s} = \begin{bmatrix} s_0 \ s_1 \end{bmatrix}$, $\lim_{n \to \infty} T^n \vec{s} = \begin{bmatrix} 2/3 \ 1/3 \end{bmatrix}$.

          10. Consider the probability transition matrix $T = \begin{bmatrix} 3/4 & 1/2 \ 1/4 & 1/2 \end{bmatrix}$ for a two-state Markov chain.
It is a fact that $det(T - \lambda I) = \lambda^2 - \frac{5}{4}\lambda + \frac{1}{8} = \frac{1}{8}(\lambda - 1)(\lambda - \frac{1}{4})$ so that the eigenvalues
of T are $\lambda_1 = 1$ and $\lambda_2 = \frac{1}{4}$. Find corresponding eigenvectors for the eigenvalues above.
For the eigenvalue 1, scale your eigenvector so that the two entries sum to 1. Given this
information, show that for any initial population distribution vector $\vec{s} = \begin{bmatrix} s_0 \ s_1 \end{bmatrix}$,
$\lim_{n \to \infty} T^n \vec{s} = \begin{bmatrix} 2/3 \ 1/3 \end{bmatrix}$.

10. Consider the probability transition matrix T =
< b m a t r i x > for a two-state Markov chain.
It is a fact that det(T - λ I) = λ^2 - (5)/(4)λ + (1)/(8) = (1)/(8)(λ - 1)(λ - (1)/(4)) so that the eigenvalues
of T are λ1 = 1 and λ2 = (1)/(4). Find corresponding eigenvectors for the eigenvalues above.
For the eigenvalue 1, scale your eigenvector so that the two entries sum to 1. Given this
information, show that for any initial population distribution vector s⃗ =
< b m a t r i x >,
limn →∞ T^n s⃗ =
< b m a t r i x >.

Added by Kathryn S.

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