Spectral representation to find $P^n$

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

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

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

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
$$\begin{bmatrix} x & .5y \\ \end{bmatrix} \begin{bmatrix} 1 \\ 1 \\ \end{bmatrix} = .8$$
$$x + .5y = .8$$ $$x + y = 1$$
and when solving $$x = .6$$ and $$y = .4$$.