Q8E 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: Q8E (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
$8 ((\begin{matrix} 1 & 1 & 0 \\ 0 & 2 & 2 \\ 0 & 0 & 3 \end{matrix})$

Short Answer

Expert verified

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

Step by step solution

Algebraic Versus.

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

then gemu(位) 鈮� almu(位).

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

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

For $位 = 1$ , 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

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

To find the values.

First, $位 = 2$ ,

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

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

To find the values.

Now, $位 = 3$ , 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 ${鈩�}^{3}$ , 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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91影视

Short Answer

Step by step solution

Algebraic Versus.

To find the values.

To find the values.

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