Diagonalization without calculating eigenvectors

Question

In many examples in stochastic processes that I came across to , make matrix diagonalization (in order to find probabilities) without calculating eigenvectors , or without even calculating $Q\Lambda Q^{-1}$, $Q$ eigenvectors matrix , $\Lambda$ eigenvalues matrix.So I give the following example and if someone could explain the steps it would be very helpful.

Suppose that we have the following stochastic matrix

$$P= \left[ \begin {array}{ccc} 0&1&0 \\ 0&2/3&1/3\\ 1/16&15/16&0\end {array} \right] $$

with eigenvalues $\lambda_{1}=1,\lambda_{2}=-5/24,\lambda_{3}=-1/8$ and $u_{i},i=1,2,3$ the corresponding eigenvectors .

So we have that

$P^{n}=Q\left[ \begin {array}{ccc} 1^{n}&0&0 \\ 0&(-5/24)^{n}&0\\ 0&0&(-1/8)^{n}\end {array} \right]Q^{-1}$

and we are asked to calculate $p_{11}^{n}$

But in order to avoid the calculation of $Q\Lambda^{n}Q^{-1}$ , we say that $p_{11}^{n}$ can be written as

$p_{11}^{n}=a+b(-5/24)^{n}+c(-1/8)^{n}$, where $a,b,c\in \mathbb{R}$.

Until now everything is fine , my problem starts when it says that vectors $u_{i}$ form a base for $R^{3}$ and if $v_{i}$ are the rows of matrix $Q^{-1}$ then the matricies $\Pi_{i}=u_{i}v_{i}^{T}$ are projections with the direction of $u_{i}$ and furthermor we have that

$I=\sum_{i=1}^{3}u_{i}v_{i}^{T}$

Also we know that $P^{n}u_{i}=\lambda^{n}u_{i}$

so combining the last two equations together we have that

$P^{n}=\sum_{i=1}^{3}\lambda^{n}u_{i}v_{i}^{T}$

After that it calculates $p_{11}^{n}$ (where I assume that $p_{11}^{n}$ is going to be the (1,1) element of matrix $P^{n}$) as

$p_{11}^{n}=\sum_{i=1}^{3}\lambda^{n}(u_{i}v_{i}^{T})_{11}$

and beucase $p^{0}_{11}=1$ (identity matrix) ,$p^{1}_{11}=0$ (from our initial matrix) and $p^{2}_{11}=(0,1,0)*(0,0,p)^{T}$ manages to calculate a,b,c.

So to summarize my questions are:

1)Why $u_{i}$ form a base ?? ( I mean they are eigenvectors of $P$)

2)if someone can explain the calculation of matricies $\Pi_{i}$ and why identity matrix is equal to there sum.

3)The calculation of $p_{11}^{2}$

Answer 1

1. $u_i$ are a basis for $\mathbb{R}^3$ because $P$ is assumed diagonalizable. "Most" matrices are diagonalizable, so you can pretty much just assume this until it fails. In this particular case it's free because the eigenvalues are distinct.

2. This is basically the relation $I=Q (Q^T Q)^{-1} Q^T$. This says that you can write the identity as the sum of the orthogonal projections onto each column of the invertible matrix $Q$. The reason the "weights" are given by multiplying by $v_k^T$ instead of $u_k^T$ (which is what you might be more familiar) is because the columns of $Q$ aren't orthogonal.

3. $p^2_{11}$ is just being calculated by direct matrix multiplication.