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Suppose that firms in a particular industry fall into one of three size categories: large, medium and small. If a firm is large one year, the probabilities that it will remain large, fall into the medium size category, or become small in the next year are, respectively $0.7$, $0.2$, and $0.1$. For a firm of medium size, the corresponding probabilities are, respectively, $0.1$, $0.8$, and $0.1$. For a small firm, the probabilities are, respectively, $0$, $0.1$, and $0.9$.

I figured out the transition matrix stated below.

a) Express transition matrix $T = \begin{bmatrix} 0.7 & 0.1 & 0 \\ 0.2 & 0.8 & 0.1 \\ 0.1 & 0.1 & 0.9 \end{bmatrix}$ in the form $QDQ^{-1}$, where $D$ is a diagonal matrix.

b) Suppose that the total number of firms in the industry remains fixed at $4000$. By considering what happens to $D^n$ as $n \to \infty$, determine how many firms fall into each category in the long term.

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  • $\begingroup$ What have you tried? Do you know how to diagonalize a matrix? To begin, can you find the eigenvalues of $T$? $\endgroup$ – Sammy Black Feb 21 '14 at 2:40
  • $\begingroup$ @SammyBlack I don't know how to diagonalise a 3x3 matrix. The examples I see online show diagonalisation of matrices that have 2 zeros in a row so it is less complicated. I am not sure if this would be more complex. I can't find the eigenvalues of T without that. $\endgroup$ – user115636 Feb 21 '14 at 2:45
  • $\begingroup$ In what context have you encountered this exercise? Does this come from a linear algebra course? I would be shocked if you didn't learn about eigenvalues, eigenvectors, and diagonalization in the sections preceding the one in which you found this exercise. $\endgroup$ – Sammy Black Feb 21 '14 at 2:47
  • $\begingroup$ You have to solve the polynomial equation $\det (\lambda I - T) = 0$ for $\lambda$ to find the eigenvalues... $\endgroup$ – Sammy Black Feb 21 '14 at 2:49
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(a) $D$ is the diagonal matrix of eigenvalues, and $Q$ is normalized eigenvectors. Perform the decomposition. If you do it right, $D$ will have $1, 0.8, 0.6$ on its diagonal.

(b) $D^n$ is the same diagonal matrix but the entrees will be $1^n = 1, 0.8^n \to 0, 0.6^n \to 0$ so the last remaining result will converge to the eigenvector corresponding to the eigenvalue 1.

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