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Chapter 6: Q24E (page 331)

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

Compute the least-squares error associated with the least square solution found in Exercise 3.

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

\(\left[ {\begin{array}{*{20}{c}}3\\{-2}\\1\\3\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}{-1}\\3\\{-3}\\4\end{array}} \right]\), \(\left[ {\begin{array}{*{20}{c}}3\\8\\7\\0\end{array}} \right]\)

Use the Gram–Schmidt process as in Example 2 to produce an orthogonal basis for the column space of

\(A = \left( {\begin{aligned}{{}{r}}{ - 10}&{13}&7&{ - 11}\\2&1&{ - 5}&3\\{ - 6}&3&{13}&{ - 3}\\{16}&{ - 16}&{ - 2}&5\\2&1&{ - 5}&{ - 7}\end{aligned}} \right)\)

Determine which pairs of vectors in Exercises 15-18 are orthogonal.

15. \({\mathop{\rm a}\nolimits} = \left( {\begin{aligned}{*{20}{c}}8\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm b}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 2}\\{ - 3}\end{aligned}} \right)\)

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

2. \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{2}}}&{\bf{0}}\\{\bf{2}} {\bf{3}}\end{aligned}} \right)\), \(b = \left( {\begin{aligned}{{}{}}{ - {\bf{5}}}\\{\bf{8}}\\{\bf{1}}\end{aligned}} \right)\)
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