Q63E find an eigenbasis for the given... [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 7: Q63E (page 325)

find an eigenbasis for the given matrice and diagonalize:
$A = \frac{1}{14} [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{matrix}]$

Short Answer

Expert verified

The eigenbasis for the given matrice is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}]$ .

Step by step solution

Solving the given matrices

We solve:

$d e t (A - 位 I) = 0 |\begin{matrix} 1 - 14 位 & 2 & 3 \\ 2 & 4 - 14 位 & 6 \\ 3 & 6 & 9 - 14 位 \end{matrix}| = 0 {(\frac{1}{14})}^{3} ((1 - 14 位) (4 - 14 位) (9 - 14 位)) + 36 + 36 - 36 (1 - 14 位) - 9 (4 - 14 位) - 4 (9 - 14 位) = 0 {(\frac{1}{14})}^{3} (位^{2} (14 - 14 位)) = 0 位_{1, 2} = 0, 位 = 1$

Solving by different values of

For $位 = 0$ , we solve:

$\begin{aligned} [\begin{array}{lll} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{array}] [\begin{matrix} x_{1} \\ y_{1} \\ z_{1} \end{matrix}] = 14 [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] \\ x_{1} + 2 y_{1} + 3 z_{1} = 0 \end{aligned}$

We can see two non-colinear eigenvector:

$V_{1} = [\begin{matrix} 3 \\ 0 \\ - 1 \end{matrix}]$

And

$v_{2} = [\begin{matrix} 2 \\ - 1 \\ 0 \end{matrix}]$

For $位 = 14$ , we solve:

$[\begin{array}{lll} 0 & 2 & 3 \\ 2 & 3 & 6 \\ 3 & 6 & 8 \end{array}] [\begin{matrix} x_{3} \\ y_{3} \\ z_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] 2 y_{3} + 3 z_{3} = 0, 2 x_{3} + 3 y_{3} + 6 z_{3} = 0, 3 x_{3} + 6 y_{3} + 8 z_{3} = 0$

We can choose an eigenvector:

$v_{3} = [\begin{matrix} 1 \\ 2 \\ 3 \end{matrix}]$

Now, $\{V_{1}, V_{2}, V_{3}\}$ is an eigenbasis for $R^{3}$ , therefore the diagonalization of A in the

eigenbasis is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}]$ .

Hence the final answer is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 1 \end{matrix}]$ .

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91影视

find an eigenbasis for the given matrice and diagonalize:
$A = \frac{1}{14} [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{matrix}]$

Short Answer

Step by step solution

Solving the given matrices

Solving by different values of

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91影视

find an eigenbasis for the given matrice and diagonalize:A=114[123246369]

Short Answer

Step by step solution

Solving the given matrices

Solving by different values of

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Theoretical and Mathematical Physics

Mechanics Maths

Logic and Functions

Discrete Mathematics

Decision Maths

Applied Mathematics

Study anywhere. Anytime. Across all devices.

find an eigenbasis for the given matrice and diagonalize:
$A = \frac{1}{14} [\begin{matrix} 1 & 2 & 3 \\ 2 & 4 & 6 \\ 3 & 6 & 9 \end{matrix}]$