Q13E 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: Q13E (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
$13 . (\begin{matrix} 3 & 0 & - 2 \\ - 7 & 0 & 4 \\ 4 & 0 & 3 \end{matrix})$

Short Answer

Expert verified

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

Step by step solution

Algebraic Versus.

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

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

$\det (A - 位濒) = 0 (\begin{matrix} 3 - 位 & 0 & - 2 \\ - 7 & - 位 & 4 \\ 4 & 0 & - 3 - 位 \end{matrix}) = 0 - 位 (3 - 位) (3 - 位) - 8 位 = 0 - 位 (位^{2} - 9 + 8) = 0 - 位 (位^{2} - 1) = 0 - 位 (位 - 1) (位 + 1) = 0 位_{1} = - 1, 位_{2} = 0, 位_{3} = 1$

To find the values.

For $位 = 1$ ,

$(A - l) x = 0 (\begin{matrix} 4 & 0 & - 2 \\ - 7 & 1 & 4 \\ 4 & 0 & - 2 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) 2 x_{1} - x_{3} = 0, - 7 x_{1} + x_{2} + 4 x_{3} = 0$

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

Now, we need to solve $位 = 0$ , we get

$A x = 0 (\begin{matrix} 3 & 0 & - 2 \\ - 7 & 0 & 4 \\ 4 & 0 & 3 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) - 3 x_{1} + 2 x_{3} = 0, - 7 x_{1} + 4 x_{1} - 3 x_{3} = 0 x_{1} = x_{2} = 0$

The basis of this eigenspace is $\{[\begin{matrix} 0 \\ 1 \\ 0 \end{matrix}]\} = \{v_{2}\}$ value.

To solve the value.

Finally for $位 = 1$ , we get

$(A - l) x = 0 (\begin{matrix} 2 & 0 & - 2 \\ - 7 & - 1 & 4 \\ 4 & 0 & - 4 \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \\ x_{3} \end{matrix}) = (\begin{matrix} 0 \\ 0 \\ 0 \end{matrix}) x_{1} - x_{3} = 0, - 7 x_{1} + x_{2} + 4 x_{3} = 0$

The basis of this eigenbasis is $\{[\begin{matrix} 1 \\ - 3 \\ 1 \end{matrix}]\} = \{v_{3}\}$

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

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91影视

Short Answer

Step by step solution

Algebraic Versus.

To find the values.

To solve the value.

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