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Chapter 6: Q5E (page 331)

Let \(X\) be the design matrix used to find the least square line of fit data \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\). Use a theorem in Section 6.5 to show that the normal equations have a unique solution if and only if the data include at least two data points with different \(x\)-coordinates.

Short Answer

Expert verified

It is verified that the normal equations have a unique solution if the data set includes at least two data points with different \(x\)-coordinates.

Step by step solution

01

Write the Theorem

Consider an \(m \times n\) matrix \(A\), and the following statements will be logically the same as:

a. For each \({\bf{b}}\) in \({\mathbb{R}^m}\), \(A{\bf{x}} = {\bf{b}}\) has a unique least-squares solution.

b. Columns of matrix \(A\) are linearly independent.

c. \({A^T}A\) is an invertible matrix.

02

Necessary condition

Let the normal equation have a unique solution.

Then, by using the theorem stated in step 1, if a normal equation has a unique solution, the matrix \({X^T}X\) is invertible, and if a matrix is invertible, then columns of that matrix are linearly independent, so there cannot be the same -coordinates, which implies the data will have at least two data points with different -coordinates.

03

Sufficient condition

Assume that the two data points have different \(x\)-coordinates, so the columns of the matrix \(X\) cannot be multiple of each other, which implies columns are linearly independent, and by the theorem in step 1, if the columns of the matrix are linearly independent, then the normal equation will have a unique solution.

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Most popular questions from this chapter

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

In Exercises 11 and 12, find the closest point to\[{\bf{y}}\]in the subspace\[W\]spanned by\[{{\bf{v}}_1}\], and\[{{\bf{v}}_2}\].

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

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

10. \(\left( {\begin{aligned}{{}{}}{ - 1} & 6 & 6 \\ 3 & { - 8}&3\\1&{ - 2}&6\\1&{ - 4}&{ - 3}\end{aligned}} \right)\)

In Exercises 7鈥�10, let\[W\]be the subspace spanned by the\[{\bf{u}}\]鈥檚, and write y as the sum of a vector in\[W\]and a vector orthogonal to\[W\].

7.\[y = \left[ {\begin{aligned}1\\3\\5\end{aligned}} \right]\],\[{{\bf{u}}_1} = \left[ {\begin{aligned}1\\3\\{ - 2}\end{aligned}} \right]\],\[{{\bf{u}}_2} = \left[ {\begin{aligned}5\\1\\4\end{aligned}} \right]\]

Question 16: Let \({\mathop{\rm y}\nolimits} = \left( {\begin{array}{*{20}{c}}{ - 3}\\9\end{array}} \right)\), and \({\mathop{\rm u}\nolimits} = \left( {\begin{array}{*{20}{c}}1\\2\end{array}} \right)\). Compute the distance from y to the line through u and the origin.
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