Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

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

The eigenbasis for the given matrice is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{matrix}]$ _.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Solving the given matrices

We solve:

$\det (A - λl) = 0 |\begin{matrix} 1 - λ & 2 & 3 \\ 2 & 4 - λ & 6 \\ 3 & 6 & 9 - λ \end{matrix}| = 0 (1 - λ) (4 - λ) (9 - λ) + 36 + 36 - 36 (1 - λ) - 9 (4 - λ) - 4 (9 - λ) = 0 - λ^{3} + 14 λ^{2} = 0 λ^{2} (- λ + 14 λ) = 0 λ_{1, 2} = 0, λ_{3} = 14$

Step 2: Solving by different values of λ

For $λ = 0$ , we solve:

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

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

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

For $λ = 14$ , we solve:

$[\begin{matrix} - 13 & 2 & 3 \\ 2 & - 10 & 6 \\ 3 & 6 & - 5 \end{matrix}] [\begin{matrix} x_{3} \\ y_{3} \\ z_{3} \end{matrix}] = [\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}] - 13 x_{3} + 2 y_{3} + 3 z_{3} = 0, 2 x_{3} - 10 y_{3} + 6 z_{3} + 0, 3 x_{3} + 6 y_{3} - 5 z_{3} = 0$

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

$[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{matrix}]$

Hence the final answer is $[\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 14 \end{matrix}]$ .

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Linear Algebra With Applications

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Short Answer

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

Step by Step Solution

Step 1: Solving the given matrices

Step 2: Solving by different values of λ

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Linear Algebra With Applications

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Short Answer

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

Step by Step Solution

Step 1: Solving the given matrices

Step 2: Solving by different values of λ

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Pure Maths

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