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Find a formula for the least-squares solution of\(Ax = b\)when the columns of A are orthonormal.

Short Answer

Expert verified

The formula for the least-square solution is \(\hat x = {A^T}b\).

Step by step solution

01

Given information

Given that \(A\) is an \(m \times n\) matrix whose columns are orthonormal.

02

Formula for the least square solution

It is known that theleast-square solution of an \(m \times n\) matrix is given by \(\hat x = {\left( {{A^T}A} \right)^{ - 1}}{A^T}b\).

Since, the columns of\(A\)are orthonormal. So,\({A^T}A = I\).

Substitute\({A^T}A = I\)in the formula for least-squares solution to get:

\(\begin{aligned}{}\hat x &= {\left( {{A^T}A} \right)^{ - 1}}{A^T}b\\\hat x &= {\left( I \right)^{ - 1}}{A^T}b\\\hat x &= {A^T}b\end{aligned}\)

Hence, the formula for the least-square solution is \(\hat x = {A^T}b\).

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Most popular questions from this chapter

A certain experiment produce the data \(\left( {1,7.9} \right),\left( {2,5.4} \right)\) and \(\left( {3, - .9} \right)\). Describe the model that produces a least-squares fit of these points by a function of the form

\(y = A\cos x + B\sin x\)

In Exercises 13 and 14, the columns of Q were obtained by applying the Gram-Schmidt process to the columns of A. Find an upper triangular matrix R such that \(A = QR\). Check your work.

13. \(A = \left( {\begin{aligned}{{}{}}5&9\\1&7\\{ - 3}&{ - 5}\\1&5\end{aligned}} \right),{\rm{ }}Q = \left( {\begin{aligned}{{}{}}{\frac{5}{6}}&{ - \frac{1}{6}}\\{\frac{1}{6}}&{\frac{5}{6}}\\{ - \frac{3}{6}}&{\frac{1}{6}}\\{\frac{1}{6}}&{\frac{3}{6}}\end{aligned}} \right)\)

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.

Given data for a least-squares problem, \(\left( {{x_1},{y_1}} \right), \ldots ,\left( {{x_n},{y_n}} \right)\), the following abbreviations are helpful:

\(\begin{aligned}{l}\sum x = \sum\nolimits_{i = 1}^n {{x_i}} ,{\rm{ }}\sum {{x^2}} = \sum\nolimits_{i = 1}^n {x_i^2} ,\\\sum y = \sum\nolimits_{i = 1}^n {{y_i}} ,{\rm{ }}\sum {xy} = \sum\nolimits_{i = 1}^n {{x_i}{y_i}} \end{aligned}\)

The normal equations for a least-squares line \(y = {\hat \beta _0} + {\hat \beta _1}x\) may be written in the form

\(\begin{aligned}{c}{{\hat \beta }_0} + {{\hat \beta }_1}\sum x = \sum y \\{{\hat \beta }_0}\sum x + {{\hat \beta }_1}\sum {{x^2}} = \sum {xy} {\rm{ (7)}}\end{aligned}\)

Derive the normal equations (7) from the matrix form given in this section.

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

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

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