Q63E Consider a positive definite qua... [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 1: Q63E (page 1)

Consider a positive definite quadratic form q on $R^{n}$ with symmetric matrix. We know that there exists an orthonormal eigenbasis for ${\overset{鈫�}{v}}_{1}, . . ., {\overset{鈫�}{v}}_{n}$ for A, with associated positive eigenvalues $位_{1}, . . ., 位_{n}$ . Now consider the orthogonal Eigen basis ${\overset{鈫�}{w}}_{1}, 鈥�, {\overset{鈫�}{w}}_{n}$ , where ${\overset{鈫�}{w}}_{1} = \frac{1}{\sqrt{位}} {\overset{鈫�}{v}}_{1}$ .
Show that $q (c_{1} {\overset{鈫�}{w}}_{1} + . . . . . . + c_{n} {\overset{鈫�}{w}}_{n}) = c_{n}^{2}$ .

Short Answer

Expert verified

The diagonalizability of quadratic form is used here to prove.

Step by step solution

Given information

Let q(x) > 0 be a positive definite quadratic form which is defined by symmetric matrix $q (x) = x^{T} Ax > 0$ for all $x 鈮� 0$ .
The orthonormal eigenbasis $B = {\overset{鈫�}{v_{1}} \overset{鈫�}{v_{2}} 鈥�, \overset{鈫�}{v_{n}}}$ is and the corresponding positive eigenvalues are $位_{1}, 位_{2}, . . ., 位_{n}$ of A, and te quadratic form is diagonalizable as $q (x) = 位_{1} c_{1}^{2} + 位_{2} c_{2}^{2} + . . . + 位_{n} c_{n}^{2} . . . . . . (1)$ .
Here, are the coordinates of with regard to the eigenbasis B.

Application

Defining $C = \{{\overset{鈫�}{w}}_{1}, {\overset{鈫�}{w}}_{2}, . . ., {\overset{鈫�}{w}}_{n}\}$ , where C is the orthogonal basis of .

$q (c_{1} {\overset{鈫�}{w}}_{1} + c_{2} \overset{鈫�}{w} + . . . . + c_{n} {\overset{鈫�}{w}}_{n}) = q (c_{1} \frac{\overset{鈫�}{V_{1}}}{\sqrt{位_{1}}} + c_{2} \frac{\overset{鈫�}{V_{2}}}{\sqrt{位_{2}}} + . . . c_{n} \frac{\overset{鈫�}{V_{n}}}{\sqrt{位_{n}}}) = q (c_{1} \frac{c_{1}}{\sqrt{位_{1}}} \overset{鈫�}{v_{1}} + \frac{c_{2}}{\sqrt{位_{2}}} \overset{鈫�}{V_{2}} + . . . \frac{c_{n}}{\sqrt{位_{n}}} \overset{鈫�}{V_{n}}) = 位_{1} {(\frac{c_{1}}{\sqrt{位_{1}}})}^{2} + 位_{2} {(\frac{c_{2}}{\sqrt{位_{2}}})}^{2} + . . . + 位_{n} {(\frac{c_{n}}{\sqrt{位_{n}}})}^{2} = c_{1}^{2} + c_{2}^{2} . . . + c_{n}^{2}$

Result

It is proved using the diagonalizability of quadratic form.

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