Eigenvalues and Eigenvectors By Iteration

We can find the eigenvalues (and then the eigenvectors)
$\lambda$
of a square matrix
$M$
by solving the equation
$det(M- \lambda I)=0$
. This can be an ordeal when the entries and eigenvalues of the matrix are decimals, and the eigenvectors contain decimal components. We can however, find eigenvalues by iteration.
Example: Let
$\left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right)$
and let an initial estimate of the unknown eigenvector be
$\mathbf{v}_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}$
.
\begin{aligned} M \mathbf{v}_1 &= \left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right) \begin{pmatrix}1\\0\\0\end{pmatrix} \\ &=\begin{pmatrix}4\\2\\-2\end{pmatrix} \\ &= 4 \begin{pmatrix}1\\0.5\\-0.5\end{pmatrix} \\ &= 4 \mathbf{v}_2\end{aligned}

\begin{aligned} M \mathbf{v}_2 &= \left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right) \begin{pmatrix}1\\0.5\\-0.5\end{pmatrix} \\ &= \begin{pmatrix}5\\4\\-4\end{pmatrix} \\ &= 4 \begin{pmatrix}1\\0.8\\-8\end{pmatrix} \\ &=5 \mathbf{v}_3\end{aligned}

\begin{aligned} M \mathbf{v}_3 &= \left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right) \begin{pmatrix}1\\0.8\\-0.8\end{pmatrix} \\ &=\begin{pmatrix}5.6\\5.2\\-5.2\end{pmatrix} \\ &= 5.6 \begin{pmatrix}1\\0.93\\-0.93\end{pmatrix} \\ &=5.6 \mathbf{v}_4 \end{aligned}

\begin{aligned} M \mathbf{v}_4 &= \left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right) \begin{pmatrix}1\\0.93\\-0.93\end{pmatrix}\\ &=\begin{pmatrix}5.86\\5.72\\-5.72\end{pmatrix} \\ &= 5.86 \begin{pmatrix}1\\0.98\\-0.98\end{pmatrix}\end{aligned}

In this way the eigenvector converges to
$\begin{pmatrix}1\\1\\-1\end{pmatrix}$
with corresponding eigenvalue 6. In fact:
$\left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right) \begin{pmatrix}1\\1\\-1\end{pmatrix}=\begin{pmatrix}6\\6\\-6\end{pmatrix} = 6 \begin{pmatrix}1\\1\\-1\end{pmatrix}$

Iteration in this way always iterates to the eigenvector with the largest eigenvalue.