Q6E 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: Q6E (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
$6 . (\begin{matrix} 2 & 3 \\ 4 & 5 \end{matrix})$

Short Answer

Expert verified

The diagonalization of A this eigenbasis is $(\begin{matrix} \frac{7 + + \sqrt{57}}{2} & 0 \\ 0 & \frac{7 - \sqrt{57}}{2} \end{matrix})$

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).

$d e t (A - 位 l) = 0 |\begin{matrix} 2 - & 位 3 \\ 45 & - 位 \end{matrix}| = 0 (2 - 位) (5 - 位) - 12 = 0 位^{2} - 7 位 - 2 = 0$

Now, we need to find the square root

$位^{2} - 7 位 - 2 = 0 位^{2} = \frac{- 7 卤 \sqrt{49 + 8}}{2} 位_{1, 2} = \frac{7 卤 \sqrt{57}}{2}$

To solve the positive value.

To find $位 = \frac{7 卤 \sqrt{57}}{2}$ we can solve,

$(A - \frac{7 + \sqrt{57}}{2} I) = 0 (\begin{matrix} \frac{- 3 - \sqrt{57}}{2} & 3 \\ 4 & \frac{3 - \sqrt{57}}{2} \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix})$

$\frac{- 3 - \sqrt{57}}{2} x_{1} + 3 x_{2} = 0 4 x_{1} + \frac{3 - \sqrt{57}}{2} x_{2} = 0$

The basic of the eigenspace is $\{\begin{matrix} [\frac{- 3 + \sqrt{57}}{8}] \\ 1 \end{matrix}\} = \{v_{1}\}$

To solve the negative values.

To find $位 = \frac{7 - \sqrt{57}}{2}$ we solve,

$(A - \frac{7 + \sqrt{57}}{2} I) = 0 (\begin{matrix} \frac{- 3 + \sqrt{57}}{2} & 3 \\ 4 & \frac{3 + \sqrt{57}}{2} \end{matrix}) (\begin{matrix} x_{1} \\ x_{2} \end{matrix}) = (\begin{matrix} 0 \\ 0 \end{matrix}) 4_{x_{1}} + \frac{3 + \sqrt{57}}{2} x_{2} = 0$

The basic of the eigenspace is $\{\begin{matrix} [\frac{- 3 - \sqrt{57}}{8}] \\ 1 \end{matrix}\} = \{v_{2}\}$

Therefore, now $\{V_{1}, V_{2}\}$ is an eigenbasis of ${鈩�}^{2}$ .So, the diagonalization of A int this eigenbasis is

$(\begin{matrix} \frac{7 + \sqrt{57}}{2} & 0 \\ 0 & \frac{7 - \sqrt{57}}{2} \end{matrix})$

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Short Answer

Step by step solution

Algebraic Versus.

To solve the positive value.

To solve the negative values.

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