Question

Question: Let\[y\],\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\]be as in Exercise 12. Find the distance from\[{\bf{y}}\]to the subspace of\[{\mathbb{R}^4}\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

Short answer

The distance is 8.

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

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

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

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

The Best Approximation Theorem

Here,\[{\bf{\hat y}}\]is said to be the closest pointin\[w\]to\[{\bf{y}}\]for which\[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\| < \left\| {{\bf{y}} - {\bf{v}}} \right\|\]for all\[{\bf{v}}\]in\[w\]distinct from\[{\bf{\hat y}}\], 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 the distance

The distance from\[{\bf{y}}\]to the closest point is also the distance between the point\[{\bf{y}}\]in\[{\mathbb{R}^3}\]to subspace\[w\], so the distance can be found by\[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\|\].

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\[{\bf{y}} = \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right]\].

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\\ &= 3\left( 1 \right) + \left( { - 1} \right)\left( { - 2} \right) + 1\left( { - 1} \right) + 13\left( 2 \right)\\ &= 30\end{aligned}\]

\[\begin{aligned}{\bf{y}} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= 3\left( { - 4} \right) + \left( { - 1} \right)\left( 1 \right) + 1\left( 0 \right) + 13\left( 3 \right)\\ &= 26\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_1} \cdot {{\bf{v}}_1} &= \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right] \cdot \left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right]\\ &= 1\left( 1 \right) + \left( { - 2} \right)\left( { - 2} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 2 \right)\left( 2 \right)\\ &= 10\end{aligned}\]

\[\begin{aligned}{{\bf{v}}_2} \cdot {{\bf{v}}_2} &= \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right] \cdot \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= - 4\left( { - 4} \right) + \left( 1 \right)\left( 1 \right) + 0\left( 0 \right) + \left( 3 \right)\left( 3 \right)\\ &= 26\end{aligned}\]

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

\[\begin{aligned}{\bf{\hat y}} &= \frac{{30}}{{10}}{{\bf{v}}_1} + \frac{{26}}{{24}}{{\bf{v}}_2}\\ &= 3\left[ {\begin{aligned}1\\{ - 2}\\{ - 1}\\2\end{aligned}} \right] + \left[ {\begin{aligned}{ - 4}\\1\\0\\3\end{aligned}} \right]\\ &= \left[ {\begin{aligned}{ - 1}\\{ - 5}\\{ - 3}\\9\end{aligned}} \right]\end{aligned}\]

Hence,the closest point is\[{\bf{\hat y}} = \left[ {\begin{aligned}{ - 1}\\{ - 5}\\{ - 3}\\9\end{aligned}} \right]\].

Now, find the distance by using\[\left\| {{\bf{y}} - {\bf{\hat y}}} \right\|\].

\[\begin{aligned}\left\| {{\bf{y}} - {\bf{\hat y}}} \right\| &= \left\| {\left[ {\begin{aligned}3\\{ - 1}\\1\\{13}\end{aligned}} \right] - \left[ {\begin{aligned}{ - 1}\\{ - 5}\\{ - 3}\\9\end{aligned}} \right]} \right\|\\ &= \left\| {\left[ {\begin{aligned}4\\4\\4\\4\end{aligned}} \right]} \right\|\\ &= \sqrt {{4^2} + {4^2} + {4^2} + {4^2}} \\ = \sqrt {64} \\ &= 8\end{aligned}\]

So, the distance is 8.