# Transforms that Preserve The Dominant Eigenvector?

1. Jun 30, 2008

### csguy

Hi,

I'm working with stochastic matrices (square matrices where each entry is a probability of moving to a different state in a Markov chain) and I am looking for transforms that would preserve the dominant eigenvector (the "stationary distribution" of the chain). What I want to do is to cause the antidiagonal of the matrix to be zero.

I remember studying a host of methods that would preserve the spectrum (e.g. QR method, Jacobi rotation, Householder matrices, etc.), but which methods preserve the dominant eigenvector?

Any suggestions?

2. Jul 2, 2008

### maze

Supposing $A \textbf{v} = \lambda \textbf{v}$, where $\textbf{v}, \lambda$ is the dominant eigenvector/eigenvalue pair with components $v_1, v_2, ..., v_n$. Then you could do something like
$$B = \lambda \left[\begin{matrix} 1 & 0 & 0 & ... & 0 \\ \frac{v_2}{v_1} & 0 & 0 & ... & 0 \\ \frac{v_3}{v_1} & 0 & 0 & ... & 0 \\ \vdots & \vdots & \vdots & \ddots & 0 \\ \frac{v_{n-1}}{v_1} & 0 & 0 & ... & 0 \\ 0 & \frac{v_n}{v_2} & 0 & ... & 0 \end{matrix}\right]$$

I think $B \textbf{v} = \lambda \textbf{v}$ if you work out the multiplication.