Finding Dominant Eigenvalue By Considering Powers of a Matrix

Suppose we have a matrix  
\[A\]
. The limiting form of high powers of  
\[A\]
  depends on the nature of the eigenvalues. Where there is a single real dominant eigenvalue  
\[\lambda\]
  the ratio of elements of powers of  
\[A\]
  tends to that eigenvalue:  
\[lim_{m \rightarrow \infty} \frac{a^{m+1}_{ij}}{a^m_{ij}}= \lambda\]
  (1)
This means that the roots of the equation  
\[a^m_{ij} \lambda^2-2 a^{m+1}_{ij} \lambda +a^{m+2}_{ij}=0\]
  get closer the the eigenvalue  
\[\lambda\]
  of  
\[A\]
.
Example: Let  
\[A=\left( \begin{array}{ccc} 4 & 1 & 2 \\ 2 & 4 & -3 \\ 3 & 1 & 3 \end{array} \right)\]
.
\[A^8=\left( \begin{array}{ccc} 952149 & 625000 & 63476 \\ -463868 & -234375 & -161132 \\ 952148 & 625000 & 63477 \end{array} \right)\]
.
\[A^9=\left( \begin{array}{ccc} 5249024 & 3515625 & 219726 \\ -2807618 & -1562500 & -708007 \\ 5249023 & 3515625 & 219727 \end{array} \right)\]
.
\[A^{10}=\left( \begin{array}{ccc} 28686524 & 19531250 & 610351 \\ -16479493 & -9765625 & -3051757 \\ 28586523 & 19531250 & 610352 \end{array} \right)\]
.
Let  
\[m=8\]
  in (1) and use as  
\[a_{ij}\]
  the element  
\[a_{11}\]
.
We have  
\[952149 \lambda^2-2 \times 5249024 \lambda +28686524=0\]
.
The roots are  
\[\lambda+5.000008, \: 6.025628\]
.
Using  
\[a_{33}\]
  instead gives the equation  
\[62477 \lambda^2 - 2 \times 219729 \lambda +610352\]
  gives  
\[\lambda=4.999956, \: 1.923078\]
.
Hence the dominant eigenvalue is  
\[\lambda=5\]
.