Q64E For the quadratic form q(x1x2)=8... [FREE SOLUTION]

91影视

Linear Algebra With Applications

Otto Bretscher

$Math Studyset 91影视 Explanations$ Math

5th Edition 路 442 pages

ISBN: 9780321796974

Chapter 8: Q64E (page 400)

For the quadratic form $q (x_{1} x_{2}) = 8 x_{1}^{2} - 4 x_{1} x_{2} + 5 x_{2}^{2}$ , find an orthogonal basis ${\overset{鈫�}{w}}_{1}, {\overset{鈫�}{w}}_{2}$ of $R^{2}$ such that $q (c_{1} {\overset{鈫�}{w}}_{1} + c_{2} {\overset{鈫�}{w}}_{2}) = c_{1}^{2} + c_{2}^{2}$ . Use your answer to sketch the level curve $q (\tilde{x}) = 1$ . Compare with Example 4 and Figure 4 in this section. Exercise 63 is helpful.

Short Answer

Expert verified

The vectors ${\overset{鈫�}{w}}_{2} = \frac{1}{3 \sqrt{5}} [\begin{matrix} 2 \\ - 1 \end{matrix}]$ and ${\overset{鈫�}{w}}_{1} = \frac{1}{2 \sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}]$ forms $q (c_{1} {\overset{鈫�}{w}}_{1} + c_{2} {\overset{鈫�}{w}}_{2}) = c_{1}^{2} + c_{2}^{2}$

Step by step solution

of 2: Given information

Here, $q (x) = q (x_{1}, x_{2}) = 8 x_{1}^{2} - 4 x_{1} x_{2} + 5 x_{2}^{2} = [x_{1} x_{2}] [\begin{matrix} 8 & - 2 \\ - 2 & 5 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \end{matrix}] = x^{T} Ax A = [\begin{matrix} 8 & - 2 \\ - 2 & 5 \end{matrix}]$

It is positive definite as $|A^{(1)}| = 8 > 0$ and role="math" localid="1659674077363" $|A^{(2)}| = 6 > 0$
The eigenvalues of are represented by $|A - 位濒| = 0$ , where

$(8 - 位) (5 - 位) - 4 = 0 鈬� (位 - 9) (位 - 4) = 0 鈬� 位_{1} = 4 > 0, 位_{2} = 9 > 0$

of 2: Application

For $位_{1} = 4$ , we have $(A - 4 l) {\overset{鈫�}{v}}_{1} = \overset{鈫�}{0} 鈬� [\begin{matrix} 4 & - 2 \\ - 2 & 1 \end{matrix}] [\begin{matrix} v_{11} \\ v_{12} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] 鈬� 2 v_{11} - v_{12} = 0$ .

Hence, one normalized eigenvector corresponding to $位_{1} = 4$ is ${\overset{鈫�}{v}}_{1} = \frac{1}{\sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}]$

For $位_{2} = 9$ , we have $(A - 9 l) \overset{鈫�}{v_{2}} = \overset{鈫�}{0} 鈬� [\begin{matrix} - 1 & - 2 \\ - 2 & - 4 \end{matrix}] [\begin{matrix} v_{21} \\ v_{22} \end{matrix}] = [\begin{matrix} 0 \\ 0 \end{matrix}] 鈬� v_{21} + 2 v_{22} = 0$ .

Hence, one normalized eigenvector corresponding to $位_{2} = 9$ is $\overset{鈫�}{v_{2}} = \frac{1}{\sqrt{5}} [\begin{matrix} 2 \\ - 1 \end{matrix}]$
Hence, the orthonormal eigenbasis of A is formed by $B = \{\overset{鈫�}{v_{1}}, \overset{鈫�}{v_{2}}\}$ corresponding to the eigen values $位_{1} = 4$ and $位_{2} = 9$ .

With the solution from Exercise 63, the orthogonal basis of A is formed by the vectors $\overset{鈫�}{w_{1}} = \frac{\overset{鈫�}{V_{1}}}{\sqrt{位_{1}}} = \frac{1}{2 \sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}]$ and $\overset{鈫�}{w_{2}} = \frac{\overset{鈫�}{V_{2}}}{\sqrt{位_{2}}} = \frac{1}{3 \sqrt{5}} [\begin{matrix} 2 \\ - 1 \end{matrix}]$ which gives $q (c_{1} {\overset{鈫�}{w}}_{1} + c_{2} {\overset{鈫�}{w}}_{2}) = c_{1}^{2} + c_{2}^{2}$

Here comes the level curve $q (x) = q (x, y) = 8 x^{2} - 4 xy + 5 y^{2} = 1$ which is a rotated eclipse.

Result:

$q (c_{1} {\overset{鈫�}{w}}_{1} + c_{2} {\overset{鈫�}{w}}_{2}) = c_{1}^{2} + c_{2}^{2}$ is formed by the vectors ${\overset{鈫�}{w}}_{1} = \frac{1}{2 \sqrt{5}} [\begin{matrix} 1 \\ 2 \end{matrix}]$ and ${\overset{鈫�}{w}}_{2} = \frac{1}{3 \sqrt{5}} [\begin{matrix} 2 \\ - 1 \end{matrix}]$

Unlock Step-by-Step Solutions & Ace Your Exams!

Full Textbook Solutions
Get detailed explanations and key concepts
Unlimited Al creation
Al flashcards, explanations, exams and more...
Ads-free access
To over 500 millions flashcards
Money-back guarantee
We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91影视!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Recommended explanations on Math Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

91影视

Short Answer

Step by step solution

of 2: Given information

of 2: Application

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Decision Maths

Probability and Statistics

Pure Maths

Calculus

Study anywhere. Anytime. Across all devices.

91影视

For the quadratic form q(x1x2)=8x12-4x1x2+5x22, find an orthogonal basis w鈫�1,w鈫�2of R2such that q(c1w鈫�1+c2w鈫�2)=c12+c22. Use your answer to sketch the level curve q(x~)=1. Compare with Example 4 and Figure 4 in this section. Exercise 63 is helpful.

Short Answer

Step by step solution

of 2: Given information

of 2: Application

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Discrete Mathematics

Decision Maths

Probability and Statistics

Pure Maths

Calculus

Study anywhere. Anytime. Across all devices.