Question

In Exercises 3–6, verify that\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]is an orthogonal set, and then find the orthogonal projection of\[y\]onto Span\[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\].

6.\[{\rm{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\]

Short answer

The projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is \[\left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].

Step-by-step solution

Write the definition

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.

Check the set of given vectors is orthogonal or not

Given vectors are,\[{{\bf{u}}_1} = \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\]and\[{{\bf{u}}_2} = \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\].

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

\[\begin{aligned}{c}{{\bf{u}}_1} \cdot {{\bf{u}}_2} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= - 4\left( 0 \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 1 \right)\\ &= 0\end{aligned}\]

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

The Orthogonal Decomposition Theorem

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 the Orthogonal projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\]

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 \[{\bf{y}} = \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= \left( 6 \right)\left( { - 4} \right) + 4\left( { - 1} \right) + 1\left( 1 \right)\\ &= - 27\end{aligned}\]

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

\[\begin{aligned}{{\bf{u}}_1} \cdot {{\bf{u}}_1} &= \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right]\\ &= - 4\left( { - 4} \right) + \left( { - 1} \right)\left( { - 1} \right) + 1\left( 1 \right)\\ &= 18\end{aligned}\]

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

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

\[\begin{aligned}{\bf{\hat y}} &= - \frac{{27}}{{18}}{{\bf{u}}_1} + \frac{5}{2}{{\bf{u}}_2}\\ &= - \frac{3}{2}\left[ {\begin{aligned}{ - 4}\\{ - 1}\\1\end{aligned}} \right] + \frac{5}{2}\left[ {\begin{aligned}0\\1\\1\end{aligned}} \right]\\ &= \left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\end{aligned}\]

Hence, the projection of \[{\bf{y}}\] onto Span \[\left\{ {{{\bf{u}}_1},{{\bf{u}}_2}} \right\}\] is \[\left[ {\begin{aligned}6\\4\\1\end{aligned}} \right]\].