Maximum index of a generalized eigenvector of $A$ associated to $\lambda$

Question

My textbook says the following:

Let $\lambda$ be an eigenvalue of a matrix $A$. Then max-ind($λ$), the maximum index of a generalized eigenvector of $A$ associated to $\lambda$, is the largest value of $j$ such that $A$ has a generalized eigenvector of index $j$ associated to the eigenvalue $\lambda$.

However, the index of the generalized eigenvector $v$ is defined as the smallest $j$ with $(A − \lambda I)^j(v)=0$.

Is it just me or does the definition contradict itself with regard to "largest/smallest" value of $j$?

Answer 1

Consider an easy example, like $$A=\begin{pmatrix}1&1\\0&1\end{pmatrix}$$.

Consider $\lambda =1$. It has ordinary eigenvector $\begin{pmatrix} 1\\0\end{pmatrix}$ and generalized eigenvectors of the form $\begin{pmatrix}a\\1\end{pmatrix}$ (easy to see).

The index of $ \begin{pmatrix} 1\\0\end{pmatrix}$ is the min over all $n$ such that $(A-1I)^n\begin{pmatrix} 1\\0\end{pmatrix}=0$. So it's $1$.

For $\begin{pmatrix}1\\1\end{pmatrix}$, say, $(A-1I) \begin{pmatrix}1\\1\end{pmatrix}=\begin{pmatrix}1\\0\end{pmatrix}$. So the index is $2$.

Finally since the $\operatorname{max-ind}(1)$ of $\lambda =1$ is defined to be the max of these two indices, it is $2$.