Finding geneal solution to X'=AX

[borkborkmath]

How do I find the general solution to X'=AX where

A =
[0 1 0]
[1 0 0]
[1 1 1]

 

[galactus]

First, find your eigenvalues and eigenvectors.

Note, this one has repeated eigenvalues: 1 and -1.

1 is repeated.

 

[borkborkmath]

do i have to find the adjoint eigenvector for when the eigenvalue is 1?
then the answer comes in the form

c1e^t(V) + c2te^t(W) + c3e^-t(K) where V,W,K are the eigenvectors?

 

[yus310]

The solution to X'=AX, lies in the fact that X=e^(At). You must find the the eigenvalues of the matrix, and then find the associated eigenvectors (A-lambda*I). It is much easier if you look at the system as KZ=K*D, where D is the diagonal matrix that contains e^(At), thus Z=K*D*K^-1, so the solution is finding the K matrix, which is the eigenvectors, D matrix is the eigenvalues in a diagonal fashion.