91Ó°ÊÓ

Orthogonal set: If \[{{\bf{u}}_i} \cdot {{\bf{u}}_j} = 0\] for \[i \ne j\], then the set of vectors \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\} \in {\mathbb{R}^n}\] is said to be orthogonal.

Given vectors are:\[{{\bf{u}}_1} = \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\], and\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\].

Find the product\[{{\bf{u}}_1} \cdot {{\bf{u}}_2}\].

\[\begin{aligned}{c}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\\ &= 3\left( 1 \right) + \left( { - 1} \right)\left( { - 1} \right) + 2\left( { - 2} \right)\\ &= 0\end{aligned}\]

As \[{{\bf{u}}_1} \cdot {{\bf{u}}_2} = 0\], so the set of vectors \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is orthogonal.

If \[\left\{ {{{\bf{u}}_1}, \ldots ,{{\bf{u}}_p}} \right\}\] is any orthogonal basis of \[w\], then \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\], where \[w\] is a subspace of \[{\mathbb{R}^n}\] and \[{\bf{\hat y}} \in w\], for which each \[{\bf{y}}\] can be written as \[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\] when \[{\bf{z}} \in {w^ \bot }\].

Find\[{\bf{y}} \cdot {{\bf{u}}_1}\],\[{\bf{y}} \cdot {{\bf{u}}_2}\],\[{{\bf{u}}_1} \cdot {{\bf{u}}_1}\]and\[{{\bf{u}}_2} \cdot {{\bf{u}}_2}\]first, where\[{\bf{y}} = \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right] \cdot \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\\ &= \left( { - 1} \right)3 + 2\left( { - 1} \right) + 6\left( 2 \right)\\ &= 7\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\\ &= \left( { - 1} \right)1 + 2\left( { - 1} \right) + 6\left( { - 2} \right)\\ &= - 15\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\\ &= 3\left( 3 \right) + \left( { - 1} \right)\left( { - 1} \right) + 2\left( 2 \right)\\ &= 14\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\\ &= 1\left( 1 \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( { - 2} \right)\left( { - 2} \right)\\ &= 6\end{aligned}\]

Now, find the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] by substituting the obtained values into \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat y}} &= \frac{7}{{14}}{{\bf{u}}_1} - \frac{{15}}{6}{{\bf{u}}_2}\\ &= \frac{1}{2}\left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right] - \frac{{15}}{6}\left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\end{aligned}\]

Hence, the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is \[\left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\].