Kiara Haas

2022-04-03

Solving homogenous system with complex eigenvalues

$\frac{dx}{dt}=2x+8y$

$\frac{dy}{dt}=-x-2y$

When I solve the determinant of the matrix, I get$\lambda =\pm 2i$ . Then , I plug it in the matrix, and get the for the first eigenvalue, $\lambda =2i$ :

$0=(2-2i)x+8y$

$0=-x-(2+2i)y$

This gives

$x=\frac{-8}{2-2i}y$

$x=-(2+2i)y$

When I solve the determinant of the matrix, I get

This gives

Ireland Vaughan

Beginner2022-04-04Added 14 answers

Note that $\frac{-8}{2-2i}=\frac{-8(2+2i)}{(2-2i)(2+2i)}=\frac{-8(2+2i)}{8}=-(2+2i)$ ,

so the two equations are saying the same thing (as they must, if the eigenvalues are correctly computed). So you can simply take, for example,$y=-1\text{}\text{and}\text{}x=2+2i$ to get an eigenvector.

so the two equations are saying the same thing (as they must, if the eigenvalues are correctly computed). So you can simply take, for example,

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