How do you determine generalized eigenvectors using Jordan form?

Question

I have the following matrix

$$ A= \begin{bmatrix} 0 & 0 & 0\\ 0 & i & 1\\ 0 & 0 & i \end{bmatrix} $$

I can see that it is in Jordan normal form. I read that you can find the generalized eigenvector just by looking at the matrix but I don't know how. The answer is supposed to be $v=(0,0,1)^T$.

Answer 1

All three vectors in the standard basis are generalized eigenvectors of the matrix since this is the basis in which it takes Jordan normal form. You can read their associated eigenvalue from the corresponding diagonal element of the matrix and you can find their rank by counting their position within the respective Jordan block starting from the left.

For example, in the case of the matrix in the question, we see that

• $(1, 0, 0)^T$ is a generalized eigenvector of rank $1$ (ordinarily called an eigenvector) associated with eigenvalue $0$,
• $(0, 1, 0)^T$ is a generalized eigenvector of rank $1$ associated with eigenvalue $i$ and
• $(0, 0, 1)^T$ is a generalized eigenvector of rank $2$ associated with eigenvalue $i$.

You can confirm this directly from definition by checking that rank $r$ generalized eigenvector $v$ associated with eigenvalue $\lambda$ satisfies $(A-\lambda I)^r v = 0$ and $(A-\lambda I)^{r-1} v \ne 0$ where $A$ is the matrix in question.