Need help with eigenvector!!!?

Question

Given the matrix A =
[ 1 0 2 ]
[ 0 1 2 ]
[ 0 0 2 ]
State all the eigenvalues for matrix A and for each eigenvalue, find a basis for the corresponding eigenspace.
Ok! So i've got found the eigenvalues which are 1 and 2 but I can't seem to get the eigenspaces and the 3 linearly independent eigenvectors!!!....PLEASE HELP!!!!!

Answer 1

If v is an eigenvector with eigenvalue λ, then we would have:

Av=λv
Av-λv = 0
(A-λI)v = 0

So v is an eigenvector with eigenvalue λ iff it is in the kernel of the matrix A-λI. So first, let us find the kernel of A-I:

A-I =
[0, 0, 2]
[0, 0, 2]
[0, 0, 1]

Now, obviously the vectors (1, 0, 0) and (0, 1, 0) suffice. Since the eigenspace has dimension of at most 2 (since that was the multiplicity of the eigenvalue), these vectors span it, and there are no more linearly independent eigenvectors to be found here. Moving on to A-2I:

[-1, 0, 2]
[0, -1, 2]
[0, 0, 0]

Simple inspection reveals that the vector (2, 2, 1) is in the kernel of this matrix. And since this eigenspace has dimension at most 1, we're done.

So a basis of eigenvectors is {(1, 0, 0), (0, 1, 0), (2, 2, 1)}.