91影视

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,\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\]and\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\].

Find\[{{\bf{u}}_1} \cdot {{\bf{u}}_2}\],\[{{\bf{u}}_2} \cdot {{\bf{u}}_3}\]and\[{{\bf{u}}_1} \cdot {{\bf{u}}_3}\].

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

\[\begin{aligned}{{\bf{u}}_2} \cdot {{\bf{u}}_3} &= \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= 1\left( 0 \right) + 0\left( { - 1} \right) + 1\left( 1 \right) + 1\left( { - 1} \right)\\ &= 0\end{aligned}\]

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_3} &= \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= 1\left( 0 \right) + 1\left( { - 1} \right) + 0\left( 1 \right) + \left( { - 1} \right)\left( { - 1} \right)\\ &= 0\end{aligned}\]

As \[{{\bf{u}}_1} \cdot {{\bf{u}}_2} = 0\], \[{{\bf{u}}_2} \cdot {{\bf{u}}_3} = 0\] and \[{{\bf{u}}_1} \cdot {{\bf{u}}_3} = 0\], so the set of vectors \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},{{\bf{u}}_3}} \right\}\] 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 every \[{\bf{y}}\] can be written as \[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\] when \[{\bf{z}} \in {w^ \bot }\].

Find\[{\bf{y}} \cdot {{\bf{u}}_1}\],\[{\bf{y}} \cdot {{\bf{u}}_2}\],\[{\bf{y}} \cdot {{\bf{u}}_3}\],\[{{\bf{u}}_1} \cdot {{\bf{u}}_1}\],\[{{\bf{u}}_2} \cdot {{\bf{u}}_2}\], and\[{{\bf{u}}_3} \cdot {{\bf{u}}_3}\]first, where\[{\bf{y}} = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\\ = 3\left( 1 \right) + 4\left( 1 \right) + 5\left( 0 \right) + 6\left( { - 1} \right)\\ = 1\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\\ = 3\left( 1 \right) + 4\left( 0 \right) + 5\left( 1 \right) + 6\left( 1 \right)\\ &= 14\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_3} &= \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= 3\left( 0 \right) + 4\left( { - 1} \right) + 5\left( 1 \right) + 6\left( { - 1} \right)\\ &= - 5\end{aligned}\]

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

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

\[\begin{aligned}{{\bf{u}}_3} \cdot {{\bf{u}}_3} &= \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= 0\left( 0 \right) + \left( { - 1} \right)\left( { - 1} \right) + 1\left( 1 \right) + \left( { - 1} \right)\left( { - 1} \right)\\ &= 3\end{aligned}\]

Now, find the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2},{{\bf{u}}_3}} \right\}\] by substituting the obtained values into \[{\bf{\hat y}} = \frac{{{\bf{y}} \cdot {{\bf{u}}_1}}}{{{{\bf{u}}_1} \cdot {{\bf{u}}_1}}}{{\bf{u}}_1} + \cdots + \frac{{{\bf{y}} \cdot {{\bf{u}}_p}}}{{{{\bf{u}}_p} \cdot {{\bf{u}}_p}}}{{\bf{u}}_p}\].

\[\begin{aligned}{\bf{\hat y}} = \frac{1}{3}{{\bf{u}}_1} + \frac{{14}}{3}{{\bf{u}}_2} - \frac{5}{3}{{\bf{u}}_3}\\ &= \frac{1}{3}\left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right] + \frac{{14}}{3}\left[ {\begin{aligned}1\\{ - 1}\\1\\{ - 1}\end{aligned}} \right] - \frac{5}{3}\left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\\ &= \left[ {\begin{aligned}5\\2\\3\\6\end{aligned}} \right]\end{aligned}\]

Hence, \[{\bf{\hat y}} = \left[ {\begin{aligned}5\\2\\3\\6\end{aligned}} \right]\].

Find\[{\bf{z}}\], which is orthogonal to\[W\]by using\[{\bf{z}} = {\bf{y}} - {\bf{\hat y}}\].

\[\begin{aligned}{\bf{z}} &= {\bf{y}} - {\bf{\hat y}}\\ &= \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right] - \left[ {\begin{aligned}5\\2\\3\\6\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{ - 2}\\2\\2\\0\end{aligned}} \right]\end{aligned}\]

So, \[{\bf{z}} = \left[ {\begin{aligned}{ - 2}\\2\\2\\0\end{aligned}} \right]\].

Write\[{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\], where\[{\bf{\hat y}} \in w\]and\[{\bf{z}} \in {w^ \bot }\].

\[\begin{aligned}{c}{\bf{y}} = {\bf{\hat y}} + {\bf{z}}\\ = \left[ {\begin{aligned}5\\2\\3\\6\end{aligned}} \right] + \left[ {\begin{aligned}{ - 2}\\2\\2\\0\end{aligned}} \right]\end{aligned}\]

Hence, \[{\bf{y}}\] as the sum of a vector in \[W\] is \[{\bf{\hat y}} + {\bf{z}} = \left[ {\begin{aligned}5\\2\\3\\6\end{aligned}} \right] + \left[ {\begin{aligned}{ - 2}\\2\\2\\0\end{aligned}} \right]\].