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Chapter 6: Q-6-14SE (page 331)

Explain why an equation \(A{\bf{x}} = {\mathop{\rm b}\nolimits} \) has a solution if and only if b is orthogonal to all solutions of the equation \({A^T}{\mathop{\rm x}\nolimits} = 0\).

Short Answer

Expert verified

The equation \(A{\bf{x}} = {\bf{b}}\) has a solution if and only if \({\bf{b}}\) is orthogonal to all solutions of \({A^T}{\bf{x}} = 0\).

Step by step solution

01

Part (c) in Exercise 13

Exercise 13(c) states that let \(A\) be a \(m \times n\) matrix and \({\mathop{\rm Col}\nolimits} A = {\left( {{\mathop{\rm Nul}\nolimits} {A^T}} \right)^ \bot }\).

02

Explain why an equation \(A{\bf{x}} = {\mathop{\rm b}\nolimits} \) has a solution if and only if b is orthogonal to all solutions of the equation \({A^T}{\mathop{\rm x}\nolimits}  = 0\) 

The equation \(A{\bf{x}} = {\bf{b}}\) contains a solution such that if \({\bf{b}}\) is in \({\mathop{\rm Col}\nolimits} A\). According to part (c) of Exercise 13, the equation \(A{\bf{x}} = {\bf{b}}\) contains a solution such that if \({\bf{b}}\) is orthogonal to \({\mathop{\rm Nul}\nolimits} {A^T}\). This occurs only when \({\bf{b}}\) is orthogonal to every solution of the equation \({A^T}{\bf{x}} = 0\).

Thus, the equation \(A{\bf{x}} = {\bf{b}}\) has a solution if and only if \({\bf{b}}\) is orthogonal to all solutions of \({A^T}{\bf{x}} = 0\).

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Most popular questions from this chapter

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

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

Let \(\overline x = \frac{1}{n}\left( {{x_1} + \cdots + {x_n}} \right)\), and \(\overline y = \frac{1}{n}\left( {{y_1} + \cdots + {y_n}} \right)\). Show that the least-squares line for the data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\) must pass through \(\left( {\overline x ,\overline y } \right)\). That is, show that \(\overline x \) and \(\overline y \) satisfies the linear equation \(\overline y = {\hat \beta _0} + {\hat \beta _1}\overline x \). (Hint: Derive this equation from the vector equation \({\bf{y}} = X{\bf{\hat \beta }} + \in \). Denote the first column of \(X\) by 1. Use the fact that the residual vector \( \in \) is orthogonal to the column space of \(X\) and hence is orthogonal to 1.)

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

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

\(\left[ {\begin{align}{ 2}\\{ - 7}\\{-1}\end{align}} \right]\), \(\left[ {\begin{align}{ - 6}\\{ - 3}\\9\end{align}} \right]\), \(\left[ {\begin{align}{ 3}\\{ 1}\\{-1}\end{align}} \right]\)

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

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