## Finding the Smallest Eigenvalue and Corresponding Eigenvector By Iteration

We can find an eigenvalue and corresponding eigenvalue of a matrix
$M$
by iteration, choosing any initial vector
$\mathbf{v}_1$
and iterating as shown here.
This method iterates only the the eigenvector with the largest eigenvalue.
We can adapt the method though to iterate to the eigenvector with the smallest eigenvalue. The eigenvalues
$\lambda$
and eigenvectors
$\mathbf{v}$
of
$M$
satisfy:
$M \mathbf{v} \lambda \mathbf{v}$
.
Multiply on the left by
$M^{-1}$
to give
$M^{-1}M \mathbf{v}=\mathbf{v}=M^{-1} \lambda \mathbf{v}=\lambda M^{-1} \mathbf{v}$

Hence
$\frac{1}{\lambda} \mathbf{v}=M^{-1} \mathbf{v}$

Now iterate by repeatedly operating on a vector
$\mathbf{v}$
with
$M^{-1}$
and this will iterate to the eigenvector with smallest eigenvalue.
Example: Let
$M= \left( \begin{array}{ccc} 4 & 3 & -1 \\ 2 & 3 & -1 \\ -2 & 1 & 5 \end{array} \right)$
then
$M^{-1}= \left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right)$
and let an initial estimate of the unknown eigenvector be
$\mathbf{v}_1 = \begin{pmatrix}1\\0\\0\end{pmatrix}$
.
\begin{aligned} M^{-1} \mathbf{v}_1 &= \left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right) \begin{pmatrix}\\0\\0\end{pmatrix} \\ &=\begin{pmatrix}0.333\\-0.167\\0.167\end{pmatrix} \\ &= 0.333 \begin{pmatrix}1\\-0.5\\0.5\end{pmatrix} \\ &=0.333 \mathbf{v}_2 \end{aligned}

\begin{aligned} M^{-1} \mathbf{v}_2 &= \left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right) \begin{pmatrix}1\\-0.5\\0.5\end{pmatrix} \\ &=\begin{pmatrix}0.417\\-0.334\\0.334\end{pmatrix} \\ &= 0.417 \begin{pmatrix}1\\-0.8\\8\end{pmatrix}\\ &=0.417 \mathbf{v}_3 \end{aligned}

\begin{aligned} M^{-1} \mathbf{v}_3 &= \left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right) \begin{pmatrix}1\\-0.8\\0.8\end{pmatrix} \\ &=\begin{pmatrix}0.467\\-0.433\\0.433\end{pmatrix} \\ &=0.467 \begin{pmatrix}1\\-0.927\\0.927\end{pmatrix}\\ &=0.467 \mathbf{v}_4 \end{aligned}

\begin{aligned} M^{-1} \mathbf{v}_4 &= \left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right) \begin{pmatrix}1\\-0.927\\0.927\end{pmatrix} \\ &=\begin{pmatrix}0.488\\-0.476\\0.476\end{pmatrix} \\ &= 0.488 \begin{pmatrix}1\\-0.979\\0.979\end{pmatrix} \end{aligned}

In this way the eigenvector converges to
$\begin{pmatrix}1\\-1\\1\end{pmatrix}$
with corresponding eigenvalue
$1/0.5=2$
. In fact:
$\left( \begin{array}{ccc} 0.333 & -0.125 & 0.042 \\ -0.167 & 0.375 & 0.042 \\ -0.167 & -0.125 & 0.208 \end{array} \right) \begin{pmatrix}1\\1\\-1\end{pmatrix}=\begin{pmatrix}0.5\\-0.5\\0.5\end{pmatrix} = 0.5 \begin{pmatrix}1\\-1\\1\end{pmatrix}$

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

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