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Chapter 6: Q12E (page 331)

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

12. \(A = \left[ {\begin{array}{{}{}}{\bf{1}}&{\bf{1}}&{\bf{0}}\\{\bf{1}}&{\bf{0}}&{ - {\bf{1}}}\\{\bf{0}}&{\bf{1}}&{\bf{1}}\\{ - {\bf{1}}}&{\bf{1}}&{ - {\bf{1}}}\end{array}} \right]\), \({\bf{b}} = \left( {\begin{array}{{}{}}{\bf{2}}\\{\bf{5}}\\{\bf{6}}\\{\bf{6}}\end{array}} \right)\)

Short Answer

Expert verified

(a) The orthogonal projection of b onto Col A is \(\left[ {\begin{array}{{}{}}5\\2\\3\\6\end{array}} \right]\).

(b) The least-square solution is \(\left[ {\begin{array}{{}{}}{\frac{1}{3}}\\{\frac{{14}}{3}}\\{ - \frac{5}{3}}\end{array}} \right]\).

Step by step solution

01

Find the orthogonal projection of \({\bf{\hat b}}\)

The orthogonal projection of b onto \({\rm{Col}}A\) is:

\(\begin{aligned}{}{\bf{\hat b}} &= {\rm{pro}}{{\rm{j}}_{{\rm{col}}A}}{\bf{b}}\\ &= \frac{{{\bf{b}} \cdot {{\bf{v}}_1}}}{{{{\bf{v}}_1} \cdot {{\bf{v}}_1}}}{{\bf{v}}_1} + \frac{{{\bf{b}} \cdot {{\bf{v}}_2}}}{{{{\bf{v}}_2} \cdot {{\bf{v}}_2}}}{{\bf{v}}_2} + \frac{{{\bf{b}} \cdot {{\bf{v}}_3}}}{{{{\bf{v}}_3} \cdot {{\bf{v}}_3}}}{{\bf{v}}_3}\\ &= \frac{{\left( {\begin{aligned}{{}{}}2&5&6&6\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1\\1\\0\\{ - 1}\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}1&1&0&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1\\1\\0\\{ - 1}\end{aligned}} \right)}}{{\bf{v}}_1} + \frac{{\left( {\begin{aligned}{{}{}}2&5&6&6\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1\\0\\1\\1\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}1&0&1&1\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1\\0\\1\\1\end{aligned}} \right)}}{{\bf{v}}_2} + \frac{{\left( {\begin{aligned}{{}{}}2&5&6&6\end{aligned}} \right)\left( {\begin{aligned}{{}{}}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right)}}{{\left( {\begin{aligned}{{}{}}0&{ - 1}&1&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right)}}{{\bf{v}}_3}\end{aligned}\)

Solve further,

\(\begin{aligned}{}{\bf{\hat b}} &= \frac{1}{3}\left( {\begin{aligned}{{}{}}1\\1\\0\\{ - 1}\end{aligned}} \right) + \frac{{14}}{3}\left( {\begin{aligned}{{}{}}1\\0\\1\\1\end{aligned}} \right) + \frac{{\left( { - 5} \right)}}{3}\left( {\begin{aligned}{{}{}}0\\{ - 1}\\1\\{ - 1}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}5\\2\\3\\6\end{aligned}} \right)\end{aligned}\)

The orthogonal projection of b onto ColA is \(\left( {\begin{aligned}{{}{}}5\\2\\3\\6\end{aligned}} \right)\).

02

Find the normal equation 

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

\(\begin{aligned}{}{A^T}A &= \left( {\begin{aligned}{{}{}}1&1&0&{ - 1}\\1&0&1&1\\0&{ - 1}&1&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}1&1&0\\1&0&{ - 1}\\0&1&1\\{ - 1}&1&{ - 1}\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}3&0&0\\0&3&0\\0&0&3\end{aligned}} \right)\end{aligned}\)

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

\(\begin{aligned}{}{A^T}{\bf{b}} &= \left( {\begin{aligned}{{}{}}1&1&0&{ - 1}\\1&0&1&1\\0&{ - 1}&1&{ - 1}\end{aligned}} \right)\left( {\begin{aligned}{{}{}}2\\5\\6\\6\end{aligned}} \right)\\ &= \left( {\begin{aligned}{{}{}}1\\{14}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

The normal equation can be written as:

\(\begin{aligned}{}\left( {{A^T}A} \right){\bf{x}} &= {A^T}{\bf{b}}\\\left( {\begin{aligned}{{}{}}3&0&0\\0&3&0\\0&0&3\end{aligned}} \right)\left( {\begin{aligned}{{}{}}{{x_1}}\\{{x_2}}\\{{x_3}}\end{aligned}} \right) &= \left( {\begin{aligned}{{}{}}1\\{14}\\{ - 5}\end{aligned}} \right)\end{aligned}\)

03

Find the least square solution

The least-square solution can be calculated as follows:

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

Thus, the least square solution is \(\left( {\begin{aligned}{{}{}}{\frac{1}{3}}\\{\frac{{14}}{3}}\\{ - \frac{5}{3}}\end{aligned}} \right)\).

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Most popular questions from this chapter

Find the distance between \({\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{10}\\{ - 3}\end{aligned}} \right)\) and \({\mathop{\rm y}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\{ - 5}\end{aligned}} \right)\).

Show that if an \(n \times n\) matrix satisfies \(\left( {U{\bf{x}}} \right) \cdot \left( {U{\bf{y}}} \right) = {\bf{x}} \cdot {\bf{y}}\) for all x and y in \({\mathbb{R}^n}\), then \(U\) is an orthogonal matrix.

In Exercises 17 and 18, all vectors and subspaces are in \({\mathbb{R}^n}\). Mark each statement True or False. Justify each answer.

17. a.If \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},{{\bf{v}}_3}} \right\}\) is an orthogonal basis for\(W\), then multiplying

\({v_3}\)by a scalar \(c\) gives a new orthogonal basis \(\left\{ {{{\bf{v}}_1},{{\bf{v}}_2},c{{\bf{v}}_3}} \right\}\).

b. The Gram鈥揝chmidt process produces from a linearly independent

set \(\left\{ {{{\bf{x}}_1}, \ldots ,{{\bf{x}}_p}} \right\}\)an orthogonal set \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) with the property that for each \(k\), the vectors \({{\bf{v}}_1}, \ldots ,{{\bf{v}}_k}\) span the same subspace as that spanned by \({{\bf{x}}_1}, \ldots ,{{\bf{x}}_k}\).

c. If \(A = QR\), where \(Q\) has orthonormal columns, then \(R = {Q^T}A\).

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

\(\left[ {\begin{align} 2\\{-5}\\{-3}\end{align}} \right]\), \(\left[ {\begin{align}0\\0\\0\end{align}} \right]\), \(\left[ {\begin{align} 4\\{ - 2}\\6\end{align}} \right]\)

For a matrix program, the Gram鈥揝chmidt 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.
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