Now that one week has lapsed, here is my solution:

If $\lambda$ is a multiplicity 3 eigenvalue of a 3x3 matrix [imath]\mathbf A[/imath], then we can bring it to Jordan normal form by forming matrix [imath]\mathbf P[/imath] with column vectors $\mathbf B^2 \mathbf v, \mathbf B\mathbf v, \mathbf v$, where [imath]\mathbf B=\mathbf A-\mathbf I[/imath], and [imath]\mathbf v[/imath] is any (column) vector for which [imath]\mathbf B^2 \mathbf v \neq \mathbf 0[/imath]:

[math]\mathbf P =
\left(
\mathbf B^2 \mathbf v ,
\mathbf B \mathbf v ,
\mathbf v
\right)[/math]In the basis [imath](\mathbf B^2 \mathbf v, \mathbf B\mathbf v, \mathbf v)[/imath] matrix [imath]\mathbf B[/imath] is represented by:

[math]\mathbf B^\prime = \left(
\begin{array}{ccc}
0 & 1 & 0\\
0 & 0 & 1\\
0 & 0 & 0
\end{array}\right)[/math]i.e.

[math]\mathbf B^\prime = \mathbf P^{-1}\mathbf B \mathbf P[/math]But then

[math]\mathbf A^\prime =\;\;
\mathbf P^{-1}\mathbf A \mathbf P =\;\;
\mathbf P^{-1} (\mathbf B+\lambda \mathbf I) \mathbf P =\;\;
\mathbf P^{-1}\mathbf B \mathbf P + \mathbf P^{-1}\mathbf \lambda \mathbf I \mathbf P =\;\;
\mathbf P^\prime + \lambda \mathbf I =\;\;[/math][math]=\left( \begin{array}{ccc}
\lambda & 1 & 0\\
0 & \lambda & 1\\
0 & 0 & \lambda
\end{array}\right)[/math]Note: I actually don't know how to prove that columns of [imath]\mathbf P[/imath] form a basis, i.e., are linearly independent, but it might not be too difficult in the case of a specific [imath]\mathbf A[/imath].