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Chapter 6: Q9E (page 331)

Let \({{\bf{P}}_{\bf{3}}}\) have the inner product given by evaluation at \( - {\bf{3}}\), \( - {\bf{1}}\),1, and 3. Let \({p_{\bf{0}}}\left( t \right) = {\bf{1}}\), \({p_{\bf{1}}}\left( t \right) = t\), and \({p_{\bf{2}}}\left( t \right) = {t^{\bf{2}}}\).

  1. Compute the orthogonal projection of \({p_{\bf{2}}}\) onto the sub-spaced spanned by , and \({p_{\bf{1}}}\).
  2. Find the polynomial q that is orthogonal to \({p_{\bf{0}}}\) and , such that \(\left\{ {{p_{\bf{0}}},{p_{\bf{1}}},q} \right\}\) is an orthogonal basis for . Scale the polynomial q so that its vector of values at \(\left( { - {\bf{3}}, - {\bf{1}},{\bf{1}},{\bf{3}}} \right)\) is \(\left( {{\bf{1}}, - {\bf{1}}, - {\bf{1}},{\bf{1}}} \right)\).

Short Answer

Expert verified

a. 5

b. \(\frac{1}{4}\left( {{t^2} - 5} \right)\)

Step by step solution

01

Find the values of polynomials

The value of \({p_0}\left( t \right)\) is 1 for all values of \(t\).

The values of \({p_1}\left( t \right) = t\) are:

\(\begin{aligned}{p_1}\left( { - 3} \right) = - 3\\{p_1}\left( { - 1} \right) = - 1\\{p_1}\left( 1 \right) = 1\\{p_1}\left( 3 \right) = 3\end{aligned}\)

The values of \({p_2}\left( t \right) = {t^2}\) are:

\(\begin{aligned}{p_2}\left( { - 3} \right) = 9\\{p_2}\left( { - 1} \right) = 1\\{p_2}\left( 1 \right) = 1\\{p_2}\left( 3 \right) = 9\end{aligned}\)

02

Find the inner products

Find the inner product \(\left\langle {{p_2},{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_0}} \right\rangle &= {p_2}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_2}\left( 1 \right){p_0}\left( 1 \right) + {p_2}\left( 3 \right){p_0}\left( 3 \right)\\ &= \left( 9 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 1 \right)\\ &= 20\end{aligned}\)

Find the inner product \(\left\langle {{p_2},{p_1}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_2},{p_1}} \right\rangle &= {p_2}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_2}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_2}\left( 1 \right){p_1}\left( 1 \right) + {p_2}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( 9 \right)\left( { - 3} \right) + \left( 1 \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 9 \right)\left( 3 \right)\\ &= 0\end{aligned}\)

Find the inner product \(\left\langle {{p_0},{p_0}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_0},{p_0}} \right\rangle &= {p_0}\left( { - 3} \right){p_0}\left( { - 3} \right) + {p_0}\left( { - 1} \right){p_0}\left( { - 1} \right) + {p_0}\left( 1 \right){p_0}\left( 1 \right) + {p_0}\left( 3 \right){p_1}\left( 3 \right)\\ &= 1 + 1 + 1 + 1\\ &= 4\end{aligned}\)

Find the inner product \(\left\langle {{p_1},{p_1}} \right\rangle \).

\(\begin{aligned}\left\langle {{p_1},{p_1}} \right\rangle &= {p_1}\left( { - 3} \right){p_1}\left( { - 3} \right) + {p_1}\left( { - 1} \right){p_1}\left( { - 1} \right) + {p_1}\left( 1 \right){p_1}\left( 1 \right) + {p_1}\left( 3 \right){p_1}\left( 3 \right)\\ &= \left( { - 3} \right)\left( { - 3} \right) + \left( { - 1} \right)\left( { - 1} \right) + \left( 1 \right)\left( 1 \right) + \left( 3 \right)\left( 3 \right)\\ &= 20\end{aligned}\)

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Most popular questions from this chapter

In Exercises 7–10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]’s, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

10.\[y = \left[ {\begin{aligned}3\\4\\5\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\1\\0\\{ - 1}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\0\\1\\1\end{aligned}} \right]\],\[{{\bf{u}}_3} = \left[ {\begin{aligned}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right]\]

Compute the least-squares error associated with the least square solution found in Exercise 3.

Find an orthogonal basis for the column space of each matrix in Exercises 9-12.

9. \(\left[ {\begin{aligned}{{}{}}3&{ - 5}&1\\1&1&1\\{ - 1}&5&{ - 2}\\3&{ - 7}&8\end{aligned}} \right]\)

Question: In Exercises 1 and 2, you may assume that\(\left\{ {{{\bf{u}}_{\bf{1}}},...,{{\bf{u}}_{\bf{4}}}} \right\}\)is an orthogonal basis for\({\mathbb{R}^{\bf{4}}}\).

2.\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{2}}\\{\bf{1}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{2}}}\\{\bf{1}}\\{ - {\bf{1}}}\\{\bf{1}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{3}}} = \left[ {\begin{aligned}{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\\{ - {\bf{1}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{4}}} = \left[ {\begin{aligned}{ - {\bf{1}}}\\{\bf{1}}\\{\bf{1}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({\bf{x}} = \left[ {\begin{aligned}{\bf{4}}\\{\bf{5}}\\{ - {\bf{3}}}\\{\bf{3}}\end{aligned}} \right]\)

Write v as the sum of two vectors, one in\({\bf{Span}}\left\{ {{{\bf{u}}_1}} \right\}\)and the other in\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Question: In Exercises 3-6, verify that\(\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\)is an orthogonal set, and then find the orthogonal projection of y onto\({\bf{Span}}\left\{ {{{\bf{u}}_{\bf{1}}},{{\bf{u}}_{\bf{2}}}} \right\}\).

4.\(y = \left[ {\begin{aligned}{\bf{6}}\\{\bf{3}}\\{ - {\bf{2}}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{1}}} = \left[ {\begin{aligned}{\bf{3}}\\{\bf{4}}\\{\bf{0}}\end{aligned}} \right]\),\({{\bf{u}}_{\bf{2}}} = \left[ {\begin{aligned}{ - {\bf{4}}}\\{\bf{3}}\\{\bf{0}}\end{aligned}} \right]\)
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