In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

11.\[y = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\],\[{{\bf{v}}_1} = \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{v}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

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

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

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

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 point in \[w\] 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\[{\bf{y}} \cdot {{\bf{v}}_1}\],\[{\bf{y}} \cdot {{\bf{v}}_2}\],\[{{\bf{v}}_1} \cdot {{\bf{v}}_1}\], and\[{{\bf{v}}_2} \cdot {{\bf{v}}_2}\]first, where\[{\bf{y}} = \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\\ &= 3\left( 3 \right) + 1\left( 1 \right) + 5\left( { - 1} \right) + 1\left( 1 \right)\\ &= 6\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}3\\1\\5\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= 3\left( 1 \right) + 1\left( { - 1} \right) + 5\left( 1 \right) + 1\left( { - 1} \right)\\ &= 6\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}3\\1\\{ - 1}\\1\end{aligned}} \right]\\ &= 3\left( 3 \right) + \left( 1 \right)\left( 1 \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right)\\ &= 12\end{aligned}\]

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

Now, find the closest to \[{\bf{y}}\] by substituting the obtained values into \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat y}} &= \frac{6}{{12}}{{\bf{v}}_1} + \frac{6}{4}{{\bf{v}}_2}\\ &= \frac{1}{2}\left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right] + \frac{3}{2}\left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}3\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\end{aligned}\]

Hence,the closest point is\[{\bf{\hat y}} = \left[ {\begin{aligned}3\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\].