. Let12A =andb =18(a) Use the Gram-Schmidt process to find an or-thonormal basis for the column space of A.(b) Factor A into a product QR, where Q has an or-thonormal set of column vectors and Ris uppertriangular.(c) Solve the least squares problem Ax = b

Answered: Let 12 A = and b = 18 (a) Use the…

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Let
12
A =
and
b =
18
(a) Use the Gram-Schmidt process to find an or-
thonormal basis for the column space of A.
(b) Factor A into a product QR, where Q has an or-
thonormal set of column vectors and Ris upper
triangular.
(c) Solve the least squares problem Ax = b

Expert Solution

Step 1

Given: $A = [\begin{matrix} 2 & 1 \\ 1 & 1 \\ 2 & 1 \end{matrix}] b = [\begin{matrix} 12 \\ 6 \\ 18 \end{matrix}]$

To find

(a) By using the Gram-Schmidt process find a basis for column space of A.

(b) A into QR

Step 2

(a) Column space of A is

$A = [\begin{matrix} 2 & 1 \\ 1 & 1 \\ 2 & 1 \end{matrix}]$

$R_{3} \leftrightarrow R_{3} - R_{1} R_{2} \leftrightarrow R_{2} - \frac{1}{2} R_{1} A = [\begin{matrix} 2 & 1 \\ 0 & \frac{1}{2} \\ 0 & 0 \end{matrix}]$

So, the column space of A is $A = [\begin{matrix} 2 & 1 \\ 0 & \frac{1}{2} \\ 0 & 0 \end{matrix}]$ .

Step 3

So basis vector of A is $\{x_{1}, x_{2}\} = \{(2, 1, 2), (1, 1, 1)\}$

By Gram-Schmidt process

Let $v_{1} = (2, 1, 2)$

$v_{2} = x_{2} - \frac{<x_{2}, v_{1}>}{<v_{1}, v_{1}>} v_{1} v_{2} = (1, 1, 1) - \frac{<(1, 1, 1), (2, 1, 2)>}{<(2, 1, 2), (2, 1, 2)>} (2, 1, 2) v_{2} = (1, 1, 1) - \frac{5}{9} (2, 1, 2) v_{2} = (- \frac{1}{9}, \frac{4}{9}, - \frac{1}{9}) ||v_{2}|| = \sqrt{{(- \frac{1}{9})}^{2} + {(\frac{4}{9})}^{2} + {(- \frac{1}{9})}^{2}} ||v_{2}|| = \sqrt{\frac{1}{81} + \frac{16}{81} + \frac{1}{81}} ||v_{2}|| = \sqrt{\frac{18}{81}} ||v_{2}|| = \sqrt{\frac{2}{9}} ||v_{2}|| = \frac{\sqrt{2}}{3}$

$\{v_{1}, v_{2}\}$ is orthogonal basis.

Now, orthogonal normal is $S = \{\frac{v_{1}}{||v_{1}||}, \frac{v_{2}}{||v_{2}||}\}$ .

So, $S = \{\frac{1}{3} (2, 1, 2), \frac{3}{\sqrt{2}} (- \frac{1}{9}, \frac{4}{9}, - \frac{1}{9})\}$ .

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