Q52E 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: Q52E (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:

$d e t (A - 位 l) = 0$

localid="1659532000702" $|\begin{matrix} 1 - 位 & 1 & 1 \\ 1 & 1 - 位 & 1 \\ 1 & 1 & 1 - 位 \end{matrix}| = 0 {(1 - 位)}^{3} + 1 + 1 - 3 (1 - 位) = 0 - 位^{3} + 3 位^{3} = 0 位^{2} (- 位^{3} + 3) = 0 位_{1, 2} = 0, 位_{3} = 3$

Solving by different values of

For $位 = 0$ , we solve:

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

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

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

For $位 = 0$ , we solve:

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

We can choose an eigenvector:

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

Now, localid="1659533326963" $\{v_{1}, v_{2}, v_{3}\}$ 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

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91影视

find an eigenbasis for the given matrice and diagonalize:A=[111111111]

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

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