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Chapter 6: Q4E (page 331)

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 \({\bf{\hat x}}\).

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

Short Answer

Expert verified

(a) \(\left( {\begin{aligned}{{}{}}3&3\\3&{11}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}6\\{14}\end{aligned}} \right)\)

(b)\(\left( {\begin{aligned}{{}{}}1\\1\end{aligned}} \right)\)

Step by step solution

01

(a) Step 1: Find the products \({A^T}A\) and \({A^T}{\bf{b}}\)

Find the product \({A^T}A\).

\(\begin{aligned}{}{A^T}A &= \left( {\begin{aligned}{{}{}}1&1&1\\3&{ - 1}&1\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1&3\\1&{ - 1}\\1&1\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}3&3\\3&{11}\end{aligned}} \right)\end{aligned}\)

Find the product \({A^T}{\bf{b}}\).

\(\begin{aligned}{}{A^T}{\bf{b}} &= \left( {\begin{aligned}{{}{}}1&1&1\\3&{ - 1}&1\end{aligned}} \right)\left( {\begin{aligned}{{}{}}5\\1\\0\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}6\\{14}\end{aligned}} \right)\end{aligned}\)

02

Find the solution by constructing the normal equations

The normal equations can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} = {A^T}{\bf{b}}\\\left( {\begin{aligned}{{}{}}3&3\\3&{11}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\end{aligned}} \right) = \left( {\begin{aligned}{{}{}}6\\{14}\end{aligned}} \right)\end{aligned}\)

03

(b) Step 3: Find the component \({\bf{\hat x}}\)

The component \({\bf{\hat x}}\) can be calculated as:

\(\begin{aligned}{}{\bf{\hat x}} &= {\left( {{A^T}A} \right)^{ - 1}}\left( {{A^T}{\bf{b}}} \right)\\ &= {\left( {\begin{aligned}{{}{}}3&3\\3&{11}\end{aligned}} \right)^{ - 1}}\left( {\begin{aligned}{{}{}}6\\{14}\end{aligned}} \right)\\ &= \frac{1}{{24}}\left( {\begin{aligned}{{}{}}{24}\\{24}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}1\\1\end{aligned}} \right)\end{aligned}\)

The \({\bf{\hat x}}\) component is \(\left( {\begin{aligned}{{}{}}1\\1\end{aligned}} \right)\).

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Most popular questions from this chapter

Question: In exercises 1-6, determine which sets of vectors are orthogonal.

\(\left[ {\begin{align}1\\{ - 2}\\1\end{align}} \right]\), \(\left[ {\begin{align}0\\1\\2\end{align}} \right]\), \(\left[ {\begin{align}{ - 5}\\{ - 2}\\1\end{align}} \right]\)

For a matrix program, the Gram–Schmidt process worksbetter with orthonormal vectors. Starting with \({x_1},......,{x_p}\) asin Theorem 11, let \(A = \left\{ {{x_1},......,{x_p}} \right\}\) . Suppose \(Q\) is an\(n \times k\)matrix whose columns form an orthonormal basis for

the subspace \({W_k}\) spanned by the first \(k\) columns of A. Thenfor \(x\) in \({\mathbb{R}^n}\), \(Q{Q^T}x\) is the orthogonal projection of x onto \({W_k}\) (Theorem 10 in Section 6.3). If \({x_{k + 1}}\) is the next column of \(A\),then equation (2) in the proof of Theorem 11 becomes

\({v_{k + 1}} = {x_{k + 1}} - Q\left( {{Q^T}T {x_{k + 1}}} \right)\)

(The parentheses above reduce the number of arithmeticoperations.) Let \({u_{k + 1}} = \frac{{{v_{k + 1}}}}{{\left\| {{v_{k + 1}}} \right\|}}\). The new \(Q\) for thenext step is \(\left( {\begin{aligned}{{}{}}Q&{{u_{k + 1}}}\end{aligned}} \right)\). Use this procedure to compute the\(QR\)factorization of the matrix in Exercise 24. Write thekeystrokes or commands you use.

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

6.\(A = \left( {\begin{aligned}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\\{\bf{1}}&{\bf{0}}&{\bf{1}}\end{aligned}} \right)\),\({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{7}}\\{\bf{2}}\\{\bf{3}}\\{\bf{6}}\\{\bf{5}}\\{\bf{4}}\end{aligned}} \right)\)

a. Rewrite the data in Example 1 with new \(x\)-coordinates in mean deviation form. Let \(X\) be the associated design matrix. Why are the columns of \(X\) orthogonal?

b. Write the normal equations for the data in part (a), and solve them to find the least-squares line, \(y = {\beta _0} + {\beta _1}x*\), where \(x* = x - 5.5\).

Show that if \(U\) is an orthogonal matrix, then any real eigenvalue of \(U\) must be \( \pm 1\).
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