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Q15E

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Linear Algebra With Applications

Found in: Page 345

Book edition 5th

Author(s) Otto Bretscher

Pages 442 pages

ISBN 9780321796974

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Short Answer

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
$15 . (\begin{matrix} - 1 & 0 & 1 \\ - 3 & 0 & 1 \\ 4 & 0 & 3 \end{matrix})$

The given matrix A is not diagonalizable.

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Step by Step Solution

TABLE OF CONTENTS :

TABLE OF CONTENTS

Step 1: Algebraic Versus.

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

then

$gemu (1) < almu (1)$

$\det (A - λ I) = 0 (\begin{matrix} - 1 - λ & 0 & 1 \\ - 3 & - λ & 1 \\ - 4 & 0 & 3 - λ \end{matrix}) = 0 - λ (1 - λ) (3 - λ) - 4 λ = 0 λ = (λ^{2} - 2 λ - 3 + 4) = 0$

$- λ (λ^{2} - 2 λ - 3 + 4) = 0 - λ (λ^{2} - 2 λ + 1) = 0 - λ (λ - 1)^{2} = 0 λ_{1} = 0, λ_{2, 3} = 1$

Step 2: Solve the values

For $λ = 0$

$A x = 0 (\begin{matrix} - 1 & 0 & 1 \\ - 3 & 0 & 1 \\ - 4 & 0 & 3 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) x_{1} + x_{3} = 0, - 3 x_{1} + x_{3} = 0, - 4 x_{1} + 3 x_{3} = 0 x_{1} = x_{2} = 0$

The basic of the eigenspace is $\{[\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]\} = \{v_{1}\}$

Step 3: Find the value.

For $λ = 1$ , we get

$(A - I) x = 0 (\begin{matrix} - 2 & 0 & 1 \\ - 3 & - 1 & 1 \\ - 4 & 0 & 2 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) - 2 x_{1} + x_{3} = 0, - 3 x_{1} - x_{2} + x_{3} = 0$

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

Therefore, the $gemu (1) < almu (1)$ . So A is not diagonalizable.

Most popular questions for Math Textbooks

Consider the matrix $A = | \begin{matrix} a & b \\ b & c \end{matrix} |$ where a, b, and c are nonzero constants. For which values of a, b, and c does A have two distinct eigenvalues?

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

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Consider the matrix $A = [\begin{matrix} 2 & 0 \\ 3 & 4 \end{matrix}]$ Show that 2 and 4 are eigenvalues ofand find all corresponding eigenvectors. Find an eigen basis for A and thus diagonalize A.

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