## The Deflation Method for Finding Eigenvalues and Eigenvectors

The method of deflation proceeds by finding the largest eigenvalue by iteration, then reducing the
$n \times n$
matrix to an
$(n-1) \times (n-1)$
matrix, finding the largest eigenvalue of this matrix, reducing the matrix to an
$(n-2) \times (n-2)$
matrix, and so on.
Let
$A$
be an
$n \times n$
matrix with largest eigenvalue
$\lambda_1$
and associated eigenvector
$\mathbf{v}_1$
. If
$\mathbf{v}_1$
does not have 1 as the component of largest modulus, multiply
$\mathbf{v}_1$
by a permutation matrix
$P$
which interchanges the largest element and the first element. Suppose
$P \mathbf{v}_1 = \mathbf{w_1}$
. We must find an elementary matrix
$R$

$R \mathbf{w}_1= \mathbf{e}_1$
, the elementary vector with first component 1 and all other components 0.
Let
$B=RPAP^{-1}R^{-1}=RPAPR^{-1}$

Then

\begin{aligned} B \mathbf{e}_1 &= RPAPR^{-1} \mathbf{e}_1 \\ &=RPAP \mathbf{w}_1 \\ &=RPA \mathbf{v}_1 \\ &=\lambda_1 RP \mathbf{v}_1 \\ &=\lambda_1 \mathbf{e}_1 \end{aligned}

Thus
$\mathbf{e}_1$
is an eigenvector of
$B$
with eigenvalue
$\lambda_1$
and
$B$
must be upper triangular with
$\lambda_1$
as the first element on the leading diagonal.
$B=\left( \begin{array}{ccc} \lambda_1 & \dots & x^1 \\ \vdots & \ddots & \vdots \\ 0 & \dots & x^n \end{array} \right)$

Delete the first column and row to give an
$(n-1) \times (n-1)$
matrix
$B_1$
.
$A, \: B$
are similar so have the same eigenvalues
$\lambda_1, \: \lambda_2, \: \lambda_3, ..., \: \lambda_n$
. The eigenvalues
$\lambda_2, \: \lambda_3, ..., \: \lambda_n$
are eigenvalues of the
$(n-1) \times (n-1)$
matrix
$B_1$
.
We can find
$RPA$
by elementary row operations on
$A$
.
$B=RPAP^{-1}R^{-1}$
can then be found by applying
$P^{-1}$
and
$R^{-1}$
to the columns of
$RPA$

Let
$A=\left( \begin{array}{ccc} 0 & 5 & -6 \\ -4 & 12 & -12 \\ -2 & -2 & 10 \end{array} \right)$

The largest eigenvalue of
$B$
is
$\lambda_1=16$
with corresponding eigenvector
$\mathbf{v}_1 = \begin{pmatrix}1\\2\\-1\end{pmatrix}$

Interchange rows 1 and 2 of
$A$
, which will make 1.0 the first component of
$\mathbf{v}_1$

$\left( \begin{array}{ccc} -4 & 12 & -12 \\ 0 & 5 & -6 \\ -2 & -2 & 10 \end{array} \right)$

Subtract half of row 1 from row 2 and add half of row 1 to row 3, transforming
$\mathbf{v}_1$
into
$\mathbf{e}_1$

$\left( \begin{array}{ccc} -4 & 12 & -12 \\ 2 & -1 & 0 \\ -4 & 4 & 4 \end{array} \right)$

Interchange columns 1 and 2.
$\left( \begin{array}{ccc} 12 & -4 & -12 \\ -1 & 2 & 0 \\ 4 & -4 & 4 \end{array} \right)$

The deflated matrix is
$\left( \begin{array}{cc} 2 & 0 \\ -4 & 4 \end{array} \right)$

Using the iteration method for
$B$
returns the eigenvalue
$\lambda_2=4$
and the eigenvector of
$A$
is
$\mathbf{v}_2= \begin{pmatrix}1\\2\\1\end{pmatrix}$

Deflating the matrix
$B_1$
gives
$\left( \begin{array}{cc} 4 & -4 \\ 0 & 2 \end{array} \right)$
then
$B_2=\left( \begin{array}{c} 2 \end{array} \right)$

The eigenvalue= is
$\lambda_3=2$
then and the eigenvector, using
$A$
is
$\begin{pmatrix}2\\2\\1\end{pmatrix}$
.