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,\[{{\bf{v}}_1} = \left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right]\], and \[{{\bf{v}}_2} = \left[ {\begin{aligned}5\\{ - 2}\\4\\2\end{aligned}} \right]\].

Find \[{{\bf{v}}_1} \cdot {{\bf{v}}_2}\].

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right] \cdot \left[ {\begin{aligned}5\\{ - 2}\\4\\2\end{aligned}} \right]\\ &= 2\left( 5 \right) + \left( 0 \right)\left( { - 2} \right) + \left( { - 1} \right)\left( 4 \right) + \left( { - 3} \right)\left( 2 \right)\\ &= 0\end{aligned}\]

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

Here, \[{\bf{\hat y}}\] is said to be the closest pointin \[w\] to \[{\bf{y}}\], if \[{\bf{\hat y}}\] be the orthogonal projection of \[{\bf{y}}\] onto \[w\], where \[w\] is a subspace of \[{\mathbb{R}^n}\], and \[{\bf{y}} \in w\]. \[{\bf{\hat y}}\] can be found as:

\[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\]

Find \[{\bf{z}} \cdot {{\bf{v}}_1}\], \[{\bf{z}} \cdot {{\bf{v}}_2}\], \[{{\bf{v}}_1} \cdot {{\bf{v}}_1}\], and \[{{\bf{v}}_2} \cdot {{\bf{v}}_2}\] first, where \[{\bf{z}} = \left[ {\begin{aligned}2\\4\\0\\{ - 1}\end{aligned}} \right]\].

\[\begin{aligned}{\bf{z}} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}2\\4\\0\\{ - 1}\end{aligned}} \right] \cdot \left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right]\\ &= 2\left( 2 \right) + \left( 4 \right)\left( 0 \right) + 0\left( { - 1} \right) + \left( { - 1} \right)\left( { - 3} \right)\\ &= 7\end{aligned}\]

\[\begin{aligned}{\bf{z}} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}2\\4\\0\\{ - 1}\end{aligned}} \right] \cdot \left[ {\begin{aligned}5\\{ - 2}\\4\\2\end{aligned}} \right]\\ &= 2\left( 5 \right) + \left( 4 \right)\left( { - 2} \right) + 0\left( 4 \right) + \left( { - 1} \right)\left( 2 \right)\\ &= 0\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right] \cdot \left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right]\\ &= 2\left( 2 \right) + \left( 0 \right)\left( 0 \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( { - 3} \right)\left( { - 3} \right)\\ &= 14\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_2} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}5\\{ - 2}\\4\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}5\\{ - 2}\\4\\2\end{aligned}} \right]\\ &= 5\left( 5 \right) + \left( { - 2} \right)\left( { - 2} \right) + 4\left( 4 \right) + 2\left( 2 \right)\\ &= 49\end{aligned}\]

Now, find the best approximation to \[{\bf{z}}\], that is \[{\bf{\hat z}}\] by using the formula \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat z}} &= \frac{7}{{14}}{{\bf{v}}_1} + \frac{0}{{49}}{{\bf{v}}_2}\\ &= \frac{1}{2}{{\bf{v}}_1} - 0{{\bf{v}}_2}\\ &= \frac{1}{2}\left[ {\begin{aligned}2\\0\\{ - 1}\\{ - 3}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}1\\0\\{ - \frac{1}{2}}\\{ - \frac{3}{2}}\end{aligned}} \right]\end{aligned}\]

Hence,thebest approximation to \[{\bf{z}}\] is\[\frac{1}{2}{{\bf{v}}_1} + 0{{\bf{v}}_2} = \left[ {\begin{aligned}1\\0\\{ - \frac{1}{2}}\\{ - \frac{3}{2}}\end{aligned}} \right]\].