91Ó°ÊÓ

Exercises 11 and 12 state that when \(A{\bf{x}} = \lambda {\bf{x}}\) then \({{\bf{x}}^T}A{\bf{x}} = {{\bf{x}}^T}\left( {\lambda {\bf{x}}} \right) = \lambda \left( {{{\bf{x}}^T}{\bf{x}}} \right)\) and the Rayleigh quotient \(R\left( {\bf{x}} \right) = \frac{{{{\bf{x}}^T}A{\bf{x}}}}{{{{\bf{x}}^T}{\bf{x}}}}\) equals \(\lambda \).

When \(A\) is an \(m \times n\) matrix and \({\bf{b}}\) in \({\mathbb{R}^m}\), then \(\widehat {\bf{x}}\) in \({\mathbb{R}^n}\) is aleast-squares solutionof \(A{\bf{x}} = {\bf{b}}\) such that;

\(\left\| {{\bf{b}} - A\widehat {\bf{x}}} \right\| \le \left\| {{\bf{b}} - A{\bf{x}}} \right\|\)for every \({\bf{x}}\) in \({\mathbb{R}^n}\)

The equation (1) can be expressed as \(V\lambda = {\bf{b}}\), with \(V\) is the single nonzero column vector \({\bf{v}}\) and \({\bf{b}} = A{\bf{v}}\).

The precise solution of the normal equations \({V^T}V\lambda = {V^T}{\bf{b}}\) is the least-square solution \(\widehat \lambda \) of \(V\lambda = {\bf{b}}\). The equation is written in the original notation as \({{\bf{v}}^T}{\bf{v}}\lambda = {{\bf{v}}^T}A{\bf{v}}\). The least-square solution \(\widehat \lambda \) of \(V\lambda = {\bf{b}}\) is \(\frac{{{{\bf{v}}^T}A{\bf{v}}}}{{{{\bf{v}}^T}{\bf{v}}}}\) because \({{\bf{v}}^T}{\bf{v}}\) is nonzero. The obtained expression is known as the Rayleigh quotient that was described in section 5.8.

Thus, the least-square solution \(\widehat \lambda \) of \(V\lambda = {\bf{b}}\) is \(\frac{{{{\bf{v}}^T}A{\bf{v}}}}{{{{\bf{v}}^T}{\bf{v}}}}\) because \({{\bf{v}}^T}{\bf{v}}\) is nonzero.