Question

Question 20: Let \({{\mathop{\rm u}\nolimits} _1}\) and \({{\mathop{\rm u}\nolimits} _2}\) be as in Exercise 19, and let \({{\mathop{\rm u}\nolimits} _4} = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\). It can be shown that \({{\mathop{\rm u}\nolimits} _4}\) is not in the subspace \(W\) spanned by \({{\mathop{\rm u}\nolimits} _1}\), and \({{\mathop{\rm u}\nolimits} _2}\). Use this fact to construct a nonzero vector v in \({\mathbb{R}^3}\) that is orthogonal to \({{\mathop{\rm u}\nolimits} _1}\), and \({{\mathop{\rm u}\nolimits} _2}\).

Short answer

The vector \({\bf{v}}\) is \({\bf{v}} = \left( {\begin{array}{*{20}{c}}0\\{\frac{4}{5}}\\{\frac{2}{5}}\end{array}} \right)\).

Step-by-step solution

The Orthogonal Decomposition Theorem

Suppose that \(W\) is a subspace of \({\mathbb{R}^n}\). Then each \({\bf{y}}\) in \({\mathbb{R}^n}\) can be expressed uniquely in the form

\({\bf{y}} = \widehat {\bf{y}} + {\bf{z}}\) … (1)

With \(\widehat {\bf{y}}\) is in \(W\) and \({\bf{z}}\) is in \({W^ \bot }\). In particular, when \(\left\{ {{{\mathop{\rm u}\nolimits} _1}, \ldots ,{{\mathop{\rm u}\nolimits} _p}} \right\}\) is anorthogonal basis of \(W\), then;

\(\widehat {\bf{y}} = \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{{{\mathop{\rm u}\nolimits} _1} \cdot {{\mathop{\rm u}\nolimits} _1}}}{{\mathop{\rm u}\nolimits} _1} + \ldots + \frac{{{\mathop{\rm y}\nolimits} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{{{\mathop{\rm u}\nolimits} _p} \cdot {{\mathop{\rm u}\nolimits} _p}}}{{\mathop{\rm u}\nolimits} _p}\) … (2)

And, \({\bf{z}} = {\bf{y}} - \widehat {\bf{y}}\).

Construct a nonzero vector v in \({\mathbb{R}^3}\)

Consider that \({{\bf{u}}_1} = \left( {\begin{array}{*{20}{c}}1\\1\\{ - 2}\end{array}} \right),{\rm{ }}{{\bf{u}}_2} = \left( {\begin{array}{*{20}{c}}5\\{ - 1}\\2\end{array}} \right)\), and \({{\bf{u}}_3} = \left( {\begin{array}{*{20}{c}}0\\0\\1\end{array}} \right)\). Let, \({{\bf{u}}_4} = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right)\).

According to the Orthogonal Decomposition Theorem, the sum of two vectors in \(W = {\mathop{\rm Span}\nolimits} \left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\) is \({{\bf{u}}_4}\) and a vectorv orthogonal to \(W\).

Construct a nonzero vector \({\mathop{\rm v}\nolimits} \) in \({\mathbb{R}^3}\) as shown below:

\(\begin{array}{c}{\bf{v}} = {{\bf{u}}_4} - {{\mathop{\rm proj}\nolimits} _W}{{\bf{u}}_4}\\ = {{\bf{u}}_4} - \left( {\frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \frac{{{\bf{y}} \cdot {{\bf{u}}_2}}}{{{{\bf{u}}_2} \cdot {{\bf{u}}_2}}}{{\bf{u}}_2}} \right)\\ = {{\bf{u}}_4} - \left( {\frac{1}{6}{{\bf{u}}_1} - \frac{1}{{30}}{{\bf{u}}_2}} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right) - \left( {\frac{1}{6}\left( {\begin{array}{*{20}{c}}1\\1\\{ - 2}\end{array}} \right) - \frac{1}{{30}}\left( {\begin{array}{*{20}{c}}5\\{ - 1}\\2\end{array}} \right)} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right) - \left( {\left( {\begin{array}{*{20}{c}}{\frac{1}{6}}\\{\frac{1}{6}}\\{ - \frac{2}{6}}\end{array}} \right) - \left( {\begin{array}{*{20}{c}}{\frac{5}{{30}}}\\{ - \frac{1}{{30}}}\\{\frac{2}{{30}}}\end{array}} \right)} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\1\\0\end{array}} \right) - \left( {\begin{array}{*{20}{c}}0\\{\frac{1}{5}}\\{ - \frac{2}{5}}\end{array}} \right)\\ = \left( {\begin{array}{*{20}{c}}0\\{\frac{4}{5}}\\{\frac{2}{5}}\end{array}} \right)\end{array}\)

Therefore, each multiple of the vector \({\bf{v}}\) is also in \({W^ \bot }\).

Thus, the nonzero vector \({\bf{v}}\) is \({\bf{v}} = \left( {\begin{array}{*{20}{c}}0\\{\frac{4}{5}}\\{\frac{2}{5}}\end{array}} \right)\).