Systems Of Linear D.E's, Complex Eigenvalues

[EnragedToilet]

1. Find the General Solution of the given system

[ -1 -1 2 ] X = X'
[ -1 1 0 ]
[ -1 0 1 ]

det(A-lambda*Identity matrix) = 0, solve for eigenvalues/values of lambda
(A-lambda*Identity matrix|0)



The eigenvalues we got are 1 and 1 +/- i. The matrix generated for (A-(1-i)I|0) is

[ i -1 2 | 0 ]
[ -1 i 0 | 0 ]
[ -1 0 i | 0 ]

Where do we go from here??

 

[Mark44]

I think you have made a mistake on your eigenvalue calculations. The eigenvalues I get are \lambda = 0 (of multiplicity 2) and \lambda = 1.

After you have found the eigenvectors, you need to do this for each one:
Substitute the \lambda in the matrix A - \lambdaI.
Row reduce this matrix. You are guaranteed that the row-reduced matrix will have at least one row of zeroes.
Of the nonzero rows that remain, solve for x₁ in terms of the other variable(s). Solve for x_x in terms of the other variables.

As an example, if you ended up with this:
[1 2 0]
[0 1 -1]
[0 0 0]

This says that
x₁ = -2x₂
x₂ = x₃
x₃ = x₃

The three variables on the left side are given in terms of x₂ and x₃ on the right side. If you choose x₂ = 1 and x₃ = 0, that gives you one solution. If you choose x₂ = 0 and x₃ = 1, that gives you another solution. Every possible solution is a linear combination of values for you choose x₂ and x₃.

 

[EnragedToilet]

[ 1 -1 2 ] X = X' ...that first one in the upper left is not negative :( sorry!
[ -1 1 0 ]
[ -1 0 1 ]