Question

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set in \({\mathbb{R}^n}\). Verify the following inequality, called Bessel’s inequality, which is true for each x in \({\mathbb{R}^n}\):

\({\left\| {\bf{x}} \right\|^2} \ge {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + {\left| {{\bf{x}} \cdot {{\bf{v}}_2}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\)

Short answer

It is verified that Bessel’s equality \({\left\| {\widehat {\bf{x}}} \right\|^2} = {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\) is true for all \({\bf{x}}\) in \({\mathbb{R}^n}\).

Step-by-step solution

Statement of Theorem 10

When the orthonormal basisfor a subspace \(W\) of \({\mathbb{R}^n}\) is \(\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}\),then;

\({{\mathop{\rm proj}\nolimits} _W}{\bf{y}} = \left( {{\bf{y}} \cdot {{\bf{u}}_1}} \right){{\bf{u}}_1} + \left( {{\bf{y}} \cdot {{\bf{u}}_2}} \right){{\bf{u}}_2} + \cdots + \left( {{\bf{y}} \cdot {{\bf{u}}_p}} \right){{\bf{u}}_p}\) … (4)

If \(U = \left[ {\begin{array}{*{20}{c}}{{{\bf{u}}_1}}&{{{\bf{u}}_2}}& \cdots &{{{\bf{u}}_p}}\end{array}} \right]\), then \({{\mathop{\rm proj}\nolimits} _W}{\bf{y}} = U{U^T}{\bf{y}}\) for every \({\bf{y}}\) in \({\mathbb{R}^n}\).

Verify Bessel’s inequality

Consider \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) as an orthonormal set in \({\mathbb{R}^n}\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \cdots + {c_p}{{\bf{v}}_p}\), then \({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\).

It is given that \({\bf{x}}\) and \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) is an orthonormal set in \({\mathbb{R}^n}\). Consider \(\widehat {\bf{x}}\) as the orthogonal projection of \({\bf{x}}\) onto the subspace spanned by the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}\).

According to Theorem 10, \(\widehat {\bf{x}} = \left( {{\bf{x}} \cdot {{\bf{v}}_1}} \right){{\bf{v}}_1} + \ldots + \left( {{\bf{x}} \cdot {{\bf{v}}_p}} \right){{\bf{v}}_p}\). According to Exercise 2, \({\left\| {\widehat {\bf{x}}} \right\|^2} = {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\). Bessel’s inequality arises from the fact that \({\left\| {\widehat {\bf{x}}} \right\|^2} \le {\left\| {\bf{x}} \right\|^2}\)that is stated before the proof of the Cauchy-Schwarz inequality.

Thus, it is verified that Bessel’s equality \({\left\| {\widehat {\bf{x}}} \right\|^2} = {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\) is true for all \({\bf{x}}\) in \({\mathbb{R}^n}\).