Q11E For a given eigenvalue, find a b... [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: Q11E (page 345)

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
$11 . (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix})$

Short Answer

Expert verified

The given matrix is an eigenbasis for ${鈩�}^{3}$ so the diagonalization of A in the eigenbasis is $(\begin{matrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 0 & 0 & 3 \end{matrix})$

Step by step solution

Algebraic Versus.

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

then $gemu (1) < almu (1)$

$\det (A - 位濒) = 0 |\begin{matrix} 1 - 位 11 \\ 11 - 位 1 \\ 111 - 位 \end{matrix}| = 0 {(1 - 位)}^{3} + 1 + 1 - 3 (1 - 位) = 0$

$- 位^{3} + 3 位^{2} = 0 位^{2} (- 位 + 3) = 0 位_{1, 2} = 0, 位_{3} = 3$

To find the value.

For $位 = 0$ , we solve

role="math" localid="1659591018014" $A_{x} = 0 (\begin{matrix} 1 & 1 & 1 \\ 1 & 1 & 1 \\ 1 & 1 & 1 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) x_{1} = x_{2} + x_{3} = 0$

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

To solve the matrix.

Similarly for $位 = 3$ we solve

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

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

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

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91影视

Short Answer

Step by step solution

Algebraic Versus.

To find the value.

To solve the matrix.

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