## 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
$\lambda=5$
.