Second Order Eigenvector In my differential equations class, my professor calls solutions of the following matrix equation second order eigenvectors:

$$(A-LI)^2 v = 0.$$

A vector $v$ that satisfies this equation where $L$ is an eigenvalue, $A$ is a matrix and $I$ is the identity matrix is called a second order eigenvector.

I can't find any information on this topic so I suspect it goes by a different name. If so, what is the name?

Also in particular, I need to know how to find second order eigenvectors.

Thanks.

P.S.: To add context, we are solving linear differential equations of the form $v' = Av$.
The particular case we are looking at is when we have a double real eigenvalue and only one eigenvector.

My professor said the solution is of this form where $L$ is the eigenvalue, $w$ is a eigenvector, and $u$ is a second order eigenvector:

$$e^{Lt}w + te^{Lt}u.$$

Here is an example:
$$A= \begin{bmatrix} 1 &amp; -1 &amp; 1 \\ 0 &amp; 0 &amp; 1 \\ 0 &amp; -1 &amp; 2 \end{bmatrix}.$$

After finding $L=1$ to be the only eigenvalue, I get this:
$$A - LI = \begin{bmatrix} 0 &amp; -1 &amp; 1 \\ 0 &amp; -1 &amp; 1 \\ 0 &amp; -1 &amp; 1 \end{bmatrix}.$$

If I square this though to solve $(A-LI)^2 v = 0$, I get the zero matrix, which means I don't get a unique second order eigenvector.
If 
$$
A =
\left(
\begin{array}{rrr}
1 &amp; -1 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
0 &amp; -1 &amp; 2
\end{array}
\right)
$$
take
$$
R =
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 1 \\
\end{array}
\right)
$$

$$
R^{-1} =
\left(
\begin{array}{rrr}
1 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; -1 &amp; 1 \\
\end{array}
\right)
$$
then
$$
(A-I)R =
\left(
\begin{array}{rrr}
0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
\end{array}
\right)
$$
and
$$
AR =
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; 1 \\
0 &amp; 1 &amp; 1 \\
0 &amp; 1 &amp; 2 \\
\end{array}
\right)
$$
and we arrive at the Jordan Normal Form:
$$
R^{-1}AR =
\left(
\begin{array}{rrr}
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
\end{array}
\right)
$$
If 
$$
A =
\left(
\begin{array}{rrr}
1 &amp; -1 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
0 &amp; -1 &amp; 2
\end{array}
\right)
$$
take
$$
R =
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; 1 &amp; 1 \\
\end{array}
\right)
$$

$$
R^{-1} =
\left(
\begin{array}{rrr}
1 &amp; -1 &amp; 0 \\
0 &amp; 1 &amp; 0 \\
0 &amp; -1 &amp; 1 \\
\end{array}
\right)
$$
then
$$
(A-I)R =
\left(
\begin{array}{rrr}
0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
\end{array}
\right)
$$
and
$$
AR =
\left(
\begin{array}{rrr}
1 &amp; 1 &amp; 1 \\
0 &amp; 1 &amp; 1 \\
0 &amp; 1 &amp; 2 \\
\end{array}
\right)
$$
and we arrive at the Jordan Normal Form:
$$
R^{-1}AR =
\left(
\begin{array}{rrr}
1 &amp; 0 &amp; 0 \\
0 &amp; 1 &amp; 1 \\
0 &amp; 0 &amp; 1 \\
\end{array}
\right)
$$