TITLE: Additive Jordan decomposition of a 3x3 matrix
QUESTION [0 upvotes]: 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?

REPLY [0 votes]: 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.
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