Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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For a given eigenvalue, find a basis of the associated eigenspace. Use the geometric multiplicities of the eigenvalues to determine whether a matrix is diagonalizable. For each of the matrices A in Exercises 1 through 20, find all (real) eigenvalues. Then find a basis of each eigenspace, and diagonalize A, if you can. Do not use technology
$8 ((\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix})$

The eigenbasis for the $ℝ^{3}$ , so the diagonalization of A in the eigenbasis is $(\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix})$

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TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Algebraic Versus.

Algebraic versus geometric multiplicity If λ is an eigenvalue of a square matrix A,

then gemu(λ) ≤ almu(λ).

This matrix is upper triangle, so its eigenvalues are the entries units main diagonal, which are

$\begin{array}{l} λ_{1} = 1 \\ λ_{2} = 2 \\ λ_{3} = 3 \end{array}$

For $λ = 1$ , we solve

role="math" localid="1659588503956" $(A - l) x = 0 (\begin{matrix} 0 & 1 & 0 \\ 0 & 1 & 2 \\ 0 & 0 & 2 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) x_{2} = 0, x_{2} + 2 x_{3} = 0, 2 x_{3} = 0 x_{2} = 0, x_{3} = 0$

The basic of this eigenspace is

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

Step 2: To find the values.

First, $λ = 2$ ,

role="math" localid="1659588665728" $(A - 2 λ) x = 0 (\begin{matrix} - 1 & 1 & 0 \\ 0 & 0 & 2 \\ 0 & 0 & 1 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) - x_{1} = x_{2} = 0, 2 x_{3} = 0, x_{3} = 0 x_{1} - x_{2} = 0, x_{3} = 0$

The basic for this eigenspace is $\{[\begin{matrix} 1 \\ 1 \\ 0 \end{matrix}]\} = \{v_{2}\}$

Step 3: To find the values.

Now, $λ = 3$ , we solve

$(A - 3 l) x = 0 (\begin{matrix} - 2 & 1 & 0 \\ 0 & - 1 & 2 \\ 0 & 0 & 0 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) - 2 x_{1} + x_{2} = 0, - x_{2} + 2 x_{3} = 0$

The basic of eigenspace $\{[\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}]\} = \{v_{3}\}$

Therefore, $\{V_{1}, V_{2}, V_{3}\}$ is an eigenbasis for $ℝ^{3}$ , so the diagonalization of A in the eigenbasis is $(\begin{matrix} 1 & 0 & 0 \\ 0 & 2 & 0 \\ 0 & 0 & 3 \end{matrix})$

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Linear Algebra With Applications

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Linear Algebra With Applications

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