Let u and v be linearly independent vectors in \({\mathbb{R}^n}\) that are not orthogonal. Describe how to find the best approximation to z in \({\mathbb{R}^n}\) by vectors of the form \({{\bf{x}}_1}{\mathop{\rm u}\nolimits} + {{\bf{x}}_2}{\mathop{\rm u}\nolimits} \) without first constructing an orthogonal basis for \({\mathop{\rm Span}\nolimits} \left\{ {{\bf{u}},{\bf{v}}} \right\}\).

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 \({\mathbb{R}^n}\).

Consider that \(W = {\mathop{\rm Span}\nolimits} \left\{ {{\bf{u}},{\bf{v}}} \right\}\).

It is given that, \({\bf{z}}\) is in \({\mathbb{R}^n}\), so assumes that \(\widehat {\bf{z}} = {{\mathop{\rm proj}\nolimits} _W}{\bf{z}}\).

Then, \(\widehat {\bf{z}}\) is in \({\mathop{\rm Col}\nolimits} A\), with \(A = \left[ {\begin{array}{*{20}{c}}{\bf{u}}&{\bf{v}}\end{array}} \right]\). Therefore, the vector \(\widehat {\bf{x}}\) in \({\mathbb{R}^2}\), with \(A\widehat {\bf{x}} = \widehat {\bf{z}}\). Hence, the least-square solution of \(A{\bf{x}} = {\bf{z}}\) is \(\widehat {\bf{x}}\) in \({\mathbb{R}^2}\).

The normal equations can be solved to determine \(\widehat {\bf{x}}\) and then \(A\widehat {\bf{x}}\) can be computed to determine \(\widehat {\bf{z}}\).