Suggested languages for you:

Americas

Europe

Q29E

Expert-verified
Found in: Page 1

### Linear Algebra With Applications

Book edition 5th
Author(s) Otto Bretscher
Pages 442 pages
ISBN 9780321796974

# Find the least square solutions of the system ${\mathbf{A}}\stackrel{\mathbf{⇀}}{\mathbf{x}}{\mathbf{=}}\stackrel{\mathbf{⇀}}{\mathbf{b}}$ where ${\mathbf{A}}{\mathbf{=}}\left[\begin{array}{cc}1& 1\\ {10}^{-10}& 0\\ 0& {10}^{-10}\end{array}\right]{\mathbf{}}{\mathbf{and}}{\mathbf{}}\stackrel{\mathbf{⇀}}{\mathbf{b}}{\mathbf{=}}\left[\begin{array}{c}1\\ {10}^{-10}\\ {10}^{-10}\end{array}\right]$.

The solution is $a=\left(\frac{1-{10}^{-40}}{{10}^{-40}+{10}^{-20}-1}\right),andb=\frac{{10}^{-20}\left(1+{10}^{-20}\right)}{\left({10}^{-40}+{10}^{-20}-1\right)}$.

See the step by step solution

## Step:1 Definition of least square

Consider a linear system as $\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\stackrel{⇀}{\mathrm{b}}$.

Here A is an matrix, a vector $\stackrel{⇀}{\mathrm{x}}$ in ${\mathrm{R}}^{\mathrm{n}}$ called a least square solution of this system if role="math" localid="1659687531916" $||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}\stackrel{⇀}{{\mathrm{x}}^{\mathrm{k}}}||\le ||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}\stackrel{⇀}{{\mathrm{x}}^{\mathrm{k}}}||\mathrm{for}\mathrm{all}\stackrel{⇀}{\mathrm{x}}\mathrm{in}{\mathrm{R}}^{\mathrm{m}}$.

## Step:2 Explanation of the solution

Consider a linear system as $\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\stackrel{⇀}{\mathrm{b}}$ where $\mathrm{A}=\left[\begin{array}{cc}1& 1\\ {10}^{-10}& 0\\ 0& {10}^{-10}\end{array}\right]\mathrm{and}\stackrel{⇀}{\mathrm{b}}=\left[\begin{array}{c}1\\ {10}^{-10}\\ {10}^{-10}\end{array}\right]$and A also A is A is an matrix, a vector $\stackrel{⇀}{\mathrm{x}}$ in ${\mathrm{R}}^{\mathrm{n}}$ called a least square solution of this system if

localid="1659688022040" $||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}{\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}||\le ||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}{\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}||\mathrm{for}\mathrm{all}\stackrel{⇀}{\mathrm{x}}\mathrm{in}{\mathrm{R}}^{\mathrm{m}}$

The term least square solution reflects the fact that minimizing the sum of the squares of the component of the vector $\left(\stackrel{⇀}{\mathrm{b}}-\mathrm{A}\stackrel{⇀}{\mathrm{x}}\right)$.

To show ${\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}$ of a linear system $\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\stackrel{⇀}{\mathrm{b}}$.

Consider the following string of equivalent statements.

The vector ${\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}$ is a least square solution of the system $\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\stackrel{⇀}{\mathrm{b}}$.

Then by the definition as follows.

$⇔||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}{\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}||\le ||\stackrel{⇀}{\mathrm{b}}-\mathrm{A}{\stackrel{⇀}{\mathrm{x}}}^{\mathrm{k}}||\mathrm{for}\mathrm{all}\stackrel{⇀}{\mathrm{x}}\mathrm{in}{\mathrm{R}}^{\mathrm{m}}$

Also by the theorem 5.4.3 as follows.

$⇔\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\mathrm{proj}\left(\stackrel{⇀}{\mathrm{b}}\right)\mathrm{where}\mathrm{V}=\mathrm{im}\left(\mathrm{A}\right)$

Similarly by the theorem 5.1.4 and 5.4.1 as follows.

$⇔\stackrel{⇀}{\mathrm{b}}-\mathrm{A}\stackrel{⇀}{\mathrm{x}}\mathrm{is}\mathrm{in}\mathrm{V}-={\left(\mathrm{imA}\right)}^{\perp }=\mathrm{ker}\left({\mathrm{A}}^{\mathrm{T}}\right)\phantom{\rule{0ex}{0ex}}⇕\phantom{\rule{0ex}{0ex}}{\mathrm{A}}^{\mathrm{T}}\left(\stackrel{⇀}{\mathrm{b}}-\mathrm{A}\stackrel{⇀}{\mathrm{x}}\right)=\stackrel{⇀}{0}\phantom{\rule{0ex}{0ex}}⇕\phantom{\rule{0ex}{0ex}}{\mathrm{A}}^{\mathrm{T}}\mathrm{A}\stackrel{⇀}{\mathrm{x}}={\mathrm{A}}^{\mathrm{T}}\stackrel{⇀}{\mathrm{b}}$.

Therefore, the least square of the system $\mathrm{A}\stackrel{⇀}{\mathrm{x}}=\stackrel{⇀}{\mathrm{b}}$ are the exact solution of the system .

${\mathrm{A}}^{\mathrm{T}}\mathrm{A}\stackrel{⇀}{\mathrm{x}}={\mathrm{A}}^{\mathrm{T}}\stackrel{⇀}{\mathrm{b}}$

The system ${\mathrm{A}}^{\mathrm{T}}\mathrm{A}\stackrel{⇀}{\mathrm{x}}={\mathrm{A}}^{\mathrm{T}}\stackrel{⇀}{\mathrm{b}}$ is called the normal equation of .

Now, simplify as follows.

$\left[\begin{array}{ccc}1& {10}^{-10}& 0\\ 1& 0& {10}^{-10}\end{array}\right]\left[\begin{array}{cc}1& 1\\ {10}^{-10}& 0\\ 0& {10}^{-10}\end{array}\right]=\left[\begin{array}{ccc}1& {10}^{-10}& 0\\ 1& 0& {10}^{-10}\end{array}\right]\left[\begin{array}{c}1\\ {10}^{-10}\\ {10}^{-10}\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{cc}1+{10}^{10}& 1\\ 1& {10}^{-20}\end{array}\right]\stackrel{⇀}{\mathrm{x}}=\left[\begin{array}{c}1+{10}^{-20}\\ 1+{10}^{-20}\end{array}\right]\phantom{\rule{0ex}{0ex}}\left[\begin{array}{c}{\mathrm{x}}_{1}\\ {\mathrm{x}}_{2}\end{array}\right]=\left[\begin{array}{c}\mathrm{a}\\ \mathrm{b}\end{array}\right]\phantom{\rule{0ex}{0ex}}\mathrm{where}{\mathrm{x}}_{1}=\mathrm{a}\left(\frac{1+{10}^{-40}}{{10}^{-40}+{10}^{-20}-1}\right)\mathrm{and}{\mathrm{x}}_{2}=\mathrm{b}=\frac{{10}^{-20}\left(1+{10}^{-20}\right)}{\left({10}^{-40}+{10}^{-20}-1\right)}\phantom{\rule{0ex}{0ex}}\mathrm{Hence},\mathrm{the}\mathrm{solution}\mathrm{is}\mathrm{a}=\left(\frac{1+{10}^{-40}}{{10}^{-40}+{10}^{-20}-1}\right)\mathrm{and}\mathrm{b}=\frac{{10}^{-20}\left(1+{10}^{-20}\right)}{\left({10}^{-40}+{10}^{-20}-1\right)}.$