?Eigenvector Shortcut For a 2 by 2 matrix\(A=\left[\begin{array}{ll} a & b \\ c & d

Chapter 5, Problem 17

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Eigenvector Shortcut For a 2 by 2 matrix

$A=\left[\begin{array}{ll} a & b \\ c & d \end{array}\right]$

with eigenvalue λ, show that if b ≠ 0, then the corresponding eigenvector is

$\bar{v}=\left[\begin{array}{cc} -b & \\ a & -\lambda \end{array}\right]$

Equation Transcription:

$A=\left[{{\ }_{c\ \ \ \ \ \ \ \ \ \ \ \ \ d}^{a\ \ \ \ \ \ \ \ \ \ \ \ \ b}}\right]$

$\overline{v}=\left[{{\ }_{\ a\ \ \ -\lambda{}}^{\ \ -b}}\right]$

Text Transcription:

A=[a b _ c d]

bar v = [-b _ a - lambda]

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