Q20E 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: Q20E (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
$20 . A = (\begin{matrix} 1 & 0 & 1 \\ 1 & 1 & 1 \\ 1 & 0 & 1 \end{matrix})$

Short Answer

Expert verified

The matrix of the diagonalization of A is $(\begin{matrix} 0 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 2 \end{matrix})$

Step by step solution

Algebraic Versus.

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

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

$\det (A - 位滨) = 0 (\begin{matrix} 1 - 位 & 1 & 0 \\ 0 & 1 - 位 & 0 \\ 0 & 0 & 1 - 位 \end{matrix}) = 0 (1 - 位)^{3} - (1 - 位) = 0 (1 - 位) (位^{2} - 2 位) = 0$

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

To find the value.

For $位 = 0$

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

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

To find the eigenvalues.

For, $位 = 1$ we get

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

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

To find the eigenvalues.

For, $位 = 2$ we get

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

The basic for this eigenvalues is $\{[\begin{matrix} 1 \\ 2 \\ 1 \end{matrix}]\} = \{v_{3}\}$

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

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91影视

Short Answer

Step by step solution

Algebraic Versus.

To find the value.

To find the eigenvalues.

To find the eigenvalues.

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