Q54E 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: Q54E (page 325)

find an eigenbasis for the given matrice and diagonalize:
$A = [\begin{matrix} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix}]$

Short Answer

Expert verified

The eigenbasis for the given matrice is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{matrix}]$ .

Step by step solution

Solving the given matrices

We solve:

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

Solving by different values of λ

For $位 = 0$ , we solve:

$\begin{array}{l} [\begin{matrix} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix}] [\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] \\ x_{1} - y_{1} + z_{1} = 0 \end{array}$

We can see all the eigenvector for $位 = 0$ are spanned by vectors:

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

And

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

For $位 = 3$ , we solve:

$\begin{array}{l} [\begin{matrix} - 2 & - 1 & 1 \\ - 1 & - 2 & - 1 \\ 1 & - 1 & - 2 \end{matrix}] [\begin{matrix} x_{3} \\ y_{3} \\ z_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] \\ - 2 x_{3} - y_{3} + z_{3} = 0, - x_{3} - 2 y_{3} - z_{3} = 0, x_{3} - y_{3} + z_{3} = 0 \end{array}$

We can choose an eigenvector:

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

Now, $\{v_{1}, v_{2}, v_{3}\}$ is an eigenbasis for $R^{3}$ , therefore the diagonalization of A in the eigenbasis is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{matrix}]$ .

Hence the final answer is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{matrix}]$ .

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91影视

find an eigenbasis for the given matrice and diagonalize:
$A = [\begin{matrix} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix}]$

Short Answer

Step by step solution

Solving the given matrices

Solving by different values of λ

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91影视

find an eigenbasis for the given matrice and diagonalize:A=[1-11-11-11-11]

Short Answer

Step by step solution

Solving the given matrices

Solving by different values of λ

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Math Textbooks

Applied Mathematics

Pure Maths

Discrete Mathematics

Decision Maths

Mechanics Maths

Logic and Functions

Study anywhere. Anytime. Across all devices.

find an eigenbasis for the given matrice and diagonalize:
$A = [\begin{matrix} 1 & - 1 & 1 \\ - 1 & 1 & - 1 \\ 1 & - 1 & 1 \end{matrix}]$