\[n \times n\]
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
\[\begin{equation} \begin{aligned} B \mathbf{e}_1 &= RPAPR^{-1} \mathbf{e}_1 \\ &=RPAP \mathbf{w}_1 \\ &=RPA \mathbf{v}_1 \\ &=\lambda_1 RP \mathbf{v}_1 \\ &=\lambda_1 \mathbf{e}_1 \end{aligned} \end{equation}\]
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 \[\begin{pmatrix}2\\2\\1\end{pmatrix}\]
.