Steps to finding a generalised eigenspace, given a linear transformation.

Question

I'm currently taking a linear algebra course and we are reading Sheldon Axler's "Linear Algebra Done Right". On page 254, he gives an example:

Suppose $T\in {\cal L}(V)$ is defined by $T(z_1, z_2, z_3)=(6z_1+3z_2+4z_3, 6z_2 + 2z_3, 7z_3)$. The eigenvalues are 6 and 7. You can verify that the generalized eigenspaces of T are as follows:

$G(6, T)=$ span$((1, 0, 0), (0, 1, 0))$ and $G(7, T)=$ span$((10, 2, 1))$

I was able to find the eigenvalues (as above), but I don't know how Axler found these generalized eigenspaces. Is there a general procedure/steps that I could take to find the generalised eigenspace of any linear transformation, given that I have already found the eigenvalues?

Also, we aren't covering determinants in this course (hence I don't know anything about determinants), which is something I've seen mentioned on similar StackExchange posts, so I wouldn't be able to use determinants on an exam, technically.

Answer 1

If you know $\lambda$ is an eigenvalue, you can check the nullspaces of $T - \lambda I, (T-\lambda I)^2, (T - \lambda I)^3, \ldots$ to get the generalized eigenvectors. You can stop once you find that the nullspaces of $(T-\lambda I)^k$ and $(T-\lambda I)^{k+1}$ are the same for some $k$.

Answer 2

I have just realised that theorem 8.11 along with @angryavian's answer would be helpful.