Suppose we have a quadratic form
\[\langle \mathbf{x}, A \mathbf{x} \rangle \]
subject to the constrain \[\langle \mathbf{x}, \mathbf{x} \rangle =1 \]
where \[A\]
is an \[n \times n\]
Hermitian matrix, so that \[A\]
is equal to its own complex conjugate transpose.The Rayleigh quotient is defined as
\[\rho = \rho( \mathbf{x})= \frac{\langle \mathbf{x},A \mathbf{x} \rangle}{\langle \mathbf{x}, \mathbf{x} \rangle}\]
(1)Furthermore, if
\[A\]
is Hermitian with eigenvalues \[\lambda_1 \leq \lambda_2 \leq ... \leq \lambda_n\]
and associated eigenvectors \[\mathbf{v}_1, \mathbf{v}_2,..., \mathbf{v}_n\]
then \[\lambda_1 \leq \rho( \mathbf{x}) \leq \lambda_n\]
and \[\lambda_1 = min(\rho(\mathbf{x}))= \rho(\mathbf{x}_1) \]
We can use (1) to find approximate eigenvalues. Let
\[\mathbf{x}_i\]
be an eigenvector associated with the eigenvalue \[\lambda_i\]
, which must be real since the matrix is Hermitian.Define
\[\mathbf{x}=\mathbf{x}_i + \epsilon \mathbf{z}, \: \| \epsilon \| \ll 1\]
.Then
\[\rho(\mathbf{x})= \lambda_i+[ \rho(\mathbf{z})- \lambda_i ] \frac{\langle \mathbf{z}, \mathbf{z} \rangle}{\langle \mathbf{x}, \mathbf{x} \rangle} \| \epsilon \|^2\]
Suppose
\[A= \left( \begin{array}{ccc} 1.7 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right) \]
.\[\begin{equation} \begin{aligned} \rho (\mathbf{x}) &= \frac{(x_1,x_2,x_3)\left( \begin{array}{ccc} 1.7 & -1 & 0 \\ -1 & 2 & -1 \\ 0 & -1 & 2 \end{array} \right) \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}}{\begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix} \cdot \begin{pmatrix}x_1\\x_2\\x_3\end{pmatrix}} \\ &= \frac{1.7x_1^2+2x_2^2+2x_3^2-2x_1x_2-2x_2x_3}{x_1^2+x_2^2+x_3^2} \end{aligned} \end{equation}\]
The eigenvector(s) associated with the smallest eigenvalue will have components of the same sign. Let
\[\mathbf{v}_1= \begin{pmatrix}1\\1\\1\end{pmatrix}\]
, then \[\rho(\mathbf{v}_1)=0.57\]
.If
\[\mathbf{v}_1= \begin{pmatrix}1\\2\\1\end{pmatrix}\]
, then \[\rho(\mathbf{v}_1)=0.62\]
.The estimates are upper bounds on
\[\lambda_1\]
so the best guess so far for \[\lambda_1\]
is 0.57.In fact
\[\lambda_1=0.5\]
with eigenvector \[\mathbf{v}_1= \begin{pmatrix}1\\1.2\\0.8\end{pmatrix}\]
.