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Chapter 6: Q14E (page 331)

Let \(A = \left( {\begin{aligned}{{}{}}{\bf{2}}&{\bf{1}}\\{ - {\bf{3}}}&{ - {\bf{4}}}\\{\bf{3}}&{\bf{2}}\end{aligned}} \right)\), \({\bf{b}} = \left( {\begin{aligned}{{}{}}{\bf{5}}\\{\bf{4}}\\{\bf{4}}\end{aligned}} \right)\), \({\bf{u}} = \left( {\begin{aligned}{{}{}}{\bf{4}}\\{ - {\bf{5}}}\end{aligned}} \right)\), and \({\bf{v}} = \left( {\begin{aligned}{{}{}}{\bf{6}}\\{ - {\bf{5}}}\end{aligned}} \right)\). Compute \(A{\bf{u}}\) and \(A{\bf{v}}\), and compare them with b. Could u possibly be a least-squares solution of \(A{\bf{x}} = {\bf{b}}\)?

(Answer this without computing a least-squares solution.)

Short Answer

Expert verified

None of them can be the least square solution of \(A{\bf{x}} = {\bf{b}}\).

Step by step solution

01

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\) 

The value of \({\bf{b}} - A{\bf{u}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}2&1\\{ - 3}&{ - 4}\\3&2\end{aligned}} \right)\left( {\begin{aligned}{{}{}}4\\{ - 5}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}3\\8\\2\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}2\\{ - 4}\\2\end{aligned}} \right)\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{u}}} \right\| &= \sqrt {{{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2} + {{\left( 2 \right)}^2}} \\ &= \sqrt {4 + 16 + 4} \\ &= \sqrt {24} \\ &= 2\sqrt 6 \end{aligned}\)

02

Find the value of \(\left\| {{\bf{b}} - A{\bf{v}}} \right\|\)

The value of \({\bf{b}} - A{\bf{v}}\) can be calculated as follows:

\(\begin{aligned}{}{\bf{b}} - A{\bf{u}} &= \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}2&1\\{ - 3}&{ - 4}\\3&2\end{aligned}} \right)\left( {\begin{aligned}{{}{}}6\\{ - 5}\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}5\\4\\4\end{aligned}} \right) - \left( {\begin{aligned}{{}{}}7\\2\\8\end{aligned}} \right)\\ & = \left( {\begin{aligned}{{}{}}{ - 2}\\2\\{ - 4}\end{aligned}} \right)\end{aligned}\)

Find the value of \(\left\| {{\bf{b}} - A{\bf{u}}} \right\|\).

\(\begin{aligned}{}\left\| {{\bf{b}} - A{\bf{v}}} \right\| & = \sqrt {{{\left( { - 2} \right)}^2} + {{\left( 2 \right)}^2} + {{\left( { - 4} \right)}^2}} \\ & = \sqrt {4 + 4 + 16} \\ & = \sqrt {24} \\ & = 2\sqrt 6 \end{aligned}\)

It can be observed that \(A{\bf{v}}\)and \(A{\bf{u}}\) are at an equal distance fromb.So, they are equally close to b.

Thus, none of them can be the least square solution of\(A{\bf{x}} = {\bf{b}}\).

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Most popular questions from this chapter

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set in \({\mathbb{R}^n}\). Verify the following inequality, called Bessel鈥檚 inequality, which is true for each x in \({\mathbb{R}^n}\):

\({\left\| {\bf{x}} \right\|^2} \ge {\left| {{\bf{x}} \cdot {{\bf{v}}_1}} \right|^2} + {\left| {{\bf{x}} \cdot {{\bf{v}}_2}} \right|^2} + \ldots + {\left| {{\bf{x}} \cdot {{\bf{v}}_p}} \right|^2}\)

Compute the quantities in Exercises 1-8 using the vectors

\({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 1}\\2\end{aligned}} \right),{\rm{ }}{\mathop{\rm v}\nolimits} = \left( {\begin{aligned}{*{20}{c}}4\\6\end{aligned}} \right),{\rm{ }}{\mathop{\rm w}\nolimits} = \left( {\begin{aligned}{*{20}{c}}3\\{ - 1}\\{ - 5}\end{aligned}} \right),{\rm{ }}{\mathop{\rm x}\nolimits} = \left( {\begin{aligned}{*{20}{c}}6\\{ - 2}\\3\end{aligned}} \right)\)

6. \(\left( {\frac{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm w}\nolimits} }}{{{\mathop{\rm x}\nolimits} \cdot {\mathop{\rm x}\nolimits} }}} \right){\mathop{\rm x}\nolimits} \)

Find the distance between \({\mathop{\rm u}\nolimits} = \left( {\begin{aligned}{*{20}{c}}0\\{ - 5}\\2\end{aligned}} \right)\) and \({\mathop{\rm z}\nolimits} = \left( {\begin{aligned}{*{20}{c}}{ - 4}\\{ - 1}\\8\end{aligned}} \right)\).

Let \(\left\{ {{{\bf{v}}_1}, \ldots ,{{\bf{v}}_p}} \right\}\) be an orthonormal set. Verify the following equality by induction, beginning with \(p = 2\). If \({\bf{x}} = {c_1}{{\bf{v}}_1} + \ldots + {c_p}{{\bf{v}}_p}\), then

\({\left\| {\bf{x}} \right\|^2} = {\left| {{c_1}} \right|^2} + {\left| {{c_2}} \right|^2} + \ldots + {\left| {{c_p}} \right|^2}\)

Find a \(QR\) factorization of the matrix in Exercise 11.
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