The Rayleigh Quotient

The Rayleigh quotient arises from attempting to optimise quadratic forms subject to constraints.
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  
\[\mathbf{v}_1= \begin{pmatrix}1\\1.2\\0.8\end{pmatrix}\]
.

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