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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

A Householder matrix, or an elementary reflector, has the form \(Q = I - 2{\bf{u}}{{\bf{u}}^T}\) where u is a unit vector. (See Exercise 13 in the Supplementary Exercise for Chapter 2.) Show that Q is an orthogonal matrix. (Elementary reflectors are often used in computer programs to produce a QR factorization of a matrix A. If A has linearly independent columns, then left-multiplication by a sequence of elementary reflectors can produce an upper triangular matrix.)

In Exercises 5 and 6, describe all least squares solutions of the equation \(A{\bf{x}} = {\bf{b}}\).

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

Question: In Exercises 9-12, find (a) the orthogonal projection of b onto \({\bf{Col}}A\) and (b) a least-squares solution of \(A{\bf{x}} = {\bf{b}}\).

11. \(A = \left( {\begin{aligned}{{}{}}{\bf{4}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{ - {\bf{5}}}&{\bf{1}}\\{\bf{6}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{ - {\bf{1}}}&{ - {\bf{5}}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{9}}\\{\bf{0}}\\{\bf{0}}\\{\bf{0}}\end{aligned}} \right)\)

In Exercises 1-6, the given set is a basis for a subspace W. Use the Gram-Schmidt process to produce an orthogonal basis for W.

6. \(\left( {\begin{aligned}{{}}3\\{ - 1}\\2\\{ - 1}\end{aligned}} \right),\left( {\begin{aligned}{{}}{ - 5}\\9\\{ - 9}\\3\end{aligned}} \right)\)

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

5.\[y = \left[ {\begin{aligned}{ - 1}\\2\\6\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}3\\{ - 1}\\2\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}1\\{ - 1}\\{ - 2}\end{aligned}} \right]\]
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