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Background:

A square matrix P, if diagonalizable, can be written $P = NDN^{-1}$. The columns of $N$ are $f_k$ and the rows of $N^{-1}$ are $\pi_k$. The diagonals of $D$ are the eigenvalues of $P$.

Define matrix $B_k$ as $B_k = f_k \pi_k$. (Although I'm not sure why this wouldn't simply be the identity matrix).

Then it follows that $$P = \lambda_1 B_1 + ... + \lambda_n B_n$$ where each $\lambda$ is an eigenvalue.

And, $$P^k = \lambda_1^kB_1 + ... + \lambda_n^kB_n$$

Then an example is provided, where I have my question:

Let $$P = \begin{bmatrix} .8 & .2 \\ .3 & .7 \\ \end{bmatrix} $$

The eigenvalues are $\lambda_1 = 1$ and $\lambda_2 = .5$. $f_1 = \mathbf 1$ is the eigenvector corresponding to $\lambda_1 = 1$. The row eigenvector corresponding to $\lambda_1 = 1$ and satisfying $\pi_1f_1 = 1$ is $\pi_1 = (.6, .4)$.

This is my question: why is $\pi_1 = (.6,.4)$? How is that calculated? Couldn't there be an infinite number of possibilities?

To finish the example, $$B_1 = f_1\pi_1 = \begin{bmatrix} .6 & .4 \\ .6 & .4 \\ \end{bmatrix} $$

$$P^0 = I = B_1 + B_2$$ thus

$$B_2 = \begin{bmatrix} .4 & -.4 \\ -.6 & .6 \\ \end{bmatrix} $$

$$P^k = \begin{bmatrix} .6 & .4 \\ .6 & .4 \\ \end{bmatrix} + .5^k \begin{bmatrix} .4 & -.4 \\ -.6 & .6 \\ \end{bmatrix} $$

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1 Answer 1

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I think I see how $\pi_1$ was determined. Using the spectral representation with $\pi_1$ and $f_1$,

$$\begin{bmatrix} x & y \\ \end{bmatrix} \begin{bmatrix} 1 & 0 \\ 0 & .5 \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} $$ which should be the first row, first column of $P$, which gives

$$\begin{bmatrix} x & .5y \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} = .8 $$

giving the equations

$$x + .5y = .8$$ $$x + y = 1$$

and when solving $x = .6$ and $y = .4$.

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