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Chapter 6: Q8E (page 331)

Exercise 3-8 refer to \({{\bf{P}}_{\bf{2}}}\) with the inner product given by evaluation at \( - {\bf{1}}\), 0, and 1. (See Example 2).

8. Compute the orthogonal projection of q onto the subspace spanned by p, for p and q in Exercise 4.

Short Answer

Expert verified

The orthogonal projection is \( - \frac{3}{2}t + \frac{1}{2}{t^2}\).

Step by step solution

01

Write the results from Exercise 4

\(\begin{align*}p\left( { - 1} \right) &= 3\left( { - 1} \right) - {\left( { - 1} \right)^2}\\ &= - 3 - 1\\ &= - 4\end{align*}\)

\(\begin{align*}p\left( 0 \right) &= 3\left( 0 \right) - {\left( 0 \right)^2}\\ &= 0\end{align*}\)

\(\begin{align*}p\left( 1 \right) &= 3\left( 1 \right) - {\left( 1 \right)^2}\\ &= 3 - 1\\ &= 2\end{align*}\)

And,

\(\begin{align*}q\left( { - 1} \right) &= 3 + 2{\left( { - 1} \right)^2}\\ &= 5\end{align*}\)

\(\begin{align*}q\left( 0 \right) &= 3 + 2{\left( 0 \right)^2}\\ &= 3\end{align*}\)

\(\begin{align*}q\left( 1 \right) &= 3 + 2{\left( 1 \right)^2}\\ &= 5\end{align*}\)

02

Find the inner product of q and p

The inner product \(\left\langle {q,p} \right\rangle \) can be calculated as follows:

\(\begin{align*}\left\langle {q,p} \right\rangle &= \left\langle {p,q} \right\rangle \\ &= p\left( { - 1} \right)q\left( { - 1} \right) + p\left( 0 \right)q\left( 0 \right) + p\left( 1 \right)q\left( 1 \right)\\ &= \left( { - 4} \right)\left( 5 \right) + \left( 0 \right)\left( 3 \right) + \left( 2 \right)\left( 5 \right)\\ &= - 10\end{align*}\)

03

Find the inner product of p and p

The inner product \(\left\langle {p,p} \right\rangle \) can be calcaulted as follows:

\(\begin{align*}\left\langle {p,p} \right\rangle &= p\left( { - 1} \right)p\left( { - 1} \right) + p\left( 0 \right)p\left( 0 \right) + p\left( 1 \right)p\left( 1 \right)\\ &= \left( { - 4} \right)\left( { - 4} \right) + \left( 0 \right)\left( 0 \right) + \left( 2 \right)\left( 2 \right)\\ &= 16 + 0 + 4\\ &= 20\end{align*}\)

04

Find the orthogonal projection of q onto subspace spanned by p

The orthogonal projection can be calculated as follows:

\(\begin{align*}\hat q &= \frac{{\left\langle {q,p} \right\rangle }}{{\left\langle {p,p} \right\rangle }}p\\ &= - \frac{{10}}{{20}}\left( {3t - {t^2}} \right)\\ &= - \frac{3}{2}t + \frac{1}{2}{t^2}\end{align*}\)

Thus, the orthogonal projection is \( - \frac{3}{2}t + \frac{1}{2}{t^2}\).

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Most popular questions from this chapter

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

8. \(\left\| {\mathop{\rm x}\nolimits} \right\|\)

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\({\bf{Span}}\left\{ {{{\bf{u}}_2},{{\bf{u}}_3},{{\bf{u}}_{\bf{4}}}} \right\}\).

Given \(A = QR\) as in Theorem 12, describe how to find an orthogonal\(m \times m\)(square) matrix \({Q_1}\) and an invertible \(n \times n\) upper triangular matrix \(R\) such that

\(A = {Q_1}\left[ {\begin{aligned}{{}{}}R\\0\end{aligned}} \right]\)

The MATLAB qr command supplies this “full” QR factorization

when rank \(A = n\).

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

12. \(\left( {\begin{aligned}{{}{}}1&3&5\\{ - 1}&{ - 3}&1\\0&2&3\\1&5&2\\1&5&8\end{aligned}} \right)\)

In Exercises 1-4, find a least-sqaures solution of \(A{\bf{x}} = {\bf{b}}\) by (a) constructing a normal equations for \({\bf{\hat x}}\) and (b) solving for \({\bf{\hat x}}\).

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)
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