Consider a $3 \times 3$ matrix over the real numbers of the form
$$\begin{pmatrix} \lambda_1 & 1 & 1 \\ 0 & \lambda_2 & 1 \\ 0 & 0 & \lambda_3 \end{pmatrix},$$
where the $\lambda_i$’s are nonzero. How to decompose this matrix as a sum of a semisimple and nilpotent matrices that commute with each other?
I can do the case where all the $\lambda_i$’s are equal. In this case we simply have the diagonal matrix with the $\lambda_i$ on the diagonal as the semisimple part. How does one get the general case?
We denote the given matrix by $$ A := \begin{pmatrix} \lambda_1 & 1 & 1 \\ & \lambda_2 & 1 \\ & & \lambda_2 \end{pmatrix}. $$ We determine the Jordan decomposition of $A = S + N$ with $S$ semisimle and $N$ nilpotent by case distinction:
If $\lambda_1 = \lambda_2 = \lambda_3 =: \lambda$ then the Jordan decomposition is given by $$ S = \begin{pmatrix} \lambda & & \\ & \lambda & \\ & & \lambda \end{pmatrix} \qquad\text{and}\qquad N = \begin{pmatrix} 0 & 1 & 1 \\ & 0 & 1 \\ & & 0 \end{pmatrix} $$ If $\lambda_1, \lambda_2, \lambda_3$ are pairwise different then $A$ is diagonalizable, and we find that $$ S = A \qquad\text{and}\qquad N = 0. $$
In the case that we only have two distinct eigenvalues, recall that $S$ and $A$ have the same eigenvalues, and that the $\lambda_i$-eigenspace of $S$ coincide with the generalized $\lambda_i$-eigenspace of $A$. We get three subcases:
Suppose that $\lambda_1 = \lambda_2 =: \lambda$ and $\lambda_3 =: \mu$, so that $$ A = \begin{pmatrix} \lambda & 1 & 1 \\ & \lambda & 1 \\ & & \mu \end{pmatrix}. $$ with $\lambda \neq \mu$. Then the generalized $\lambda$-eigenspace of $A$ is given by $\langle e_1, e_2 \rangle$, and for $c := \lambda - \mu \neq 0$ the generalized $\mu$-eigenspace of $A$ is given by $\langle (1-c, -c, c^2) \rangle$. With respect to the basis $\mathcal{B} = (e_1, e_2, (1-c, -c, c^2))$ of $\mathbb{R}^3$ the semisimple part $S$ is therefore given by the diagonal matrix $$ \begin{pmatrix} \lambda & & \\ & \lambda & \\ & & \mu \end{pmatrix}. $$ Thus we find that \begin{align*} S &= \begin{pmatrix} 1 & 0 & 1-c \\ 0 & 1 & -c \\ 0 & 0 & c^2 \end{pmatrix} \begin{pmatrix} \lambda & & \\ & \lambda & \\ & & \mu \end{pmatrix} \begin{pmatrix} 1 & 0 & 1-c \\ 0 & 1 & -c \\ 0 & 0 & c^2 \end{pmatrix}^{-1} \\ &= \begin{pmatrix} 1 & 0 & 1-c \\ 0 & 1 & -c \\ 0 & 0 & c^2 \end{pmatrix} \begin{pmatrix} \lambda & & \\ & \lambda & \\ & & \mu \end{pmatrix} \begin{pmatrix} 1 & 0 & (c-1)/c^2 \\ 0 & 1 & 1/c \\ 0 & 0 & 1/c^2 \end{pmatrix} \\ &= \begin{pmatrix} 1 & 0 & 1-c \\ 0 & 1 & -c \\ 0 & 0 & c^2 \end{pmatrix} \begin{pmatrix} \lambda & 0 & \lambda (c-1)/c^2 \\ 0 & \lambda & \lambda/c \\ 0 & 0 & \mu/c^2 \end{pmatrix} \\ &= \begin{pmatrix} \lambda & 0 & \lambda(c-1)/c^2 + \mu(1-c)/c^2 \\ 0 & \lambda & \lambda/c - \mu/c \\ 0 & 0 & \mu \end{pmatrix} = \begin{pmatrix} \lambda & 0 & (c-1)/c \\ 0 & \lambda & 1 \\ 0 & 0 & \mu \end{pmatrix} \end{align*} and $$ N = A - S = \begin{pmatrix} 0 & 1 & 1/c \\ & 0 & 0 \\ & & 0 \end{pmatrix}. $$
The subcases $\lambda_1 = \lambda_3$ and $\lambda_2 = \lambda_3$ can be calculated similarly. The case $\lambda_2 = \lambda_3$ can actually be derived from the above result: By applying the matrix transpose to the decomposition $$ \begin{pmatrix} \lambda & 1 & 1 \\ & \lambda & 1 \\ & & \mu \end{pmatrix} = \begin{pmatrix} \lambda & 0 & (c-1)/c \\ 0 & \lambda & 1 \\ 0 & 0 & \mu \end{pmatrix} + \begin{pmatrix} 0 & 1 & 1/c \\ & 0 & 0 \\ & & 0 \end{pmatrix} $$ we arrive at the decomposition $$ \begin{pmatrix} \mu & 1 & 1 \\ & \lambda & 1 \\ & & \lambda \end{pmatrix} = \begin{pmatrix} \mu & 1 & (c-1)/c \\ 0 & \lambda & 0 \\ 0 & 0 & \lambda \end{pmatrix} + \begin{pmatrix} 0 & 0 & 1/c \\ & 0 & 1 \\ & & 0 \end{pmatrix}. $$ Here we use that the matrix transpose preserves diagonalizability, nilpotence and commutativity. By I’m not sure if there is a nice way to derive the case $\lambda_1 = \lambda_3$ in a similar way.