Solving differential system with matrices

[hatsu27]

I am solving a system X'=AX
where A=[(1,-1,1),(0,2,-1),(0,0,1)]
I have found my eigenvalues where Lamda = 2, and 1 w/ mult.2
now in finding my eigenvectors when Lamda = 1 my matrix looks
like this: [(0,1,-1),(0,0,0),(0,0,0)] and the 1st eigenvector is (0,1,1)
and I'm pretty sure from past linear algebra class the 2nd one is (1,0,0) but I can't remember why. if in this matrix there is no x₁ how come it has a value in this vector. I remember being flumoxed by this before and I just want to understand why and what I'm doing here. Thanks!

 

[lippyka]

The eigenvectors are determined by the elements of the matrix that correspond to the eigenvalue λ. If your matrix has a row where all elements have the value 0, then that row is not taken into consideration when finding the eigenvector. In this case, since row 1 only has values 0, we don't consider it while looking for the eigenvector corresponding to λ = 1. Therefore, the eigenvector (1, 0, 0) is the correct solution.