Let $T = D + N$ be the $D−N$ decomposition of a linear transformation $T \in L(V)$. Prove that a linear transformation $E \in L(V)$ commutes with $T$ if and only if $E$ commutes with $D$ and $N$.
$D-N$ decomposition: Let $T \in L(V)$ & $V(F)$ contains all eigen values of $T$. Then there exist linear transformations $D$ and $N$ on $V$ such that, $T=D+N$ where $D$ is diagnolizable and $N$ is nilpotent.
If $E$ commutes with $D$ and $N$ then $E$ clearly commutes with $T$. However, if $E$ commutes with $T$, I don't understand how it commutes with $N$ and $D$. It is not necessary that a diagnol matrix commutes with every matrix. For example, $
\left( {\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array} } \right) $
$
\left( {\begin{array}{cc}
0 & 1 \\
0 & 0 \\
\end{array} } \right) \neq $
$
\left( {\begin{array}{cc}
0 & 1 \\
0 & 0 \\
\end{array} } \right) $
$
\left( {\begin{array}{cc}
1 & 0 \\
0 & -1 \\
\end{array} } \right) $. Am I missing something?
Best Answer
You’re right that in general a diagonal matrix doesn’t compute with a nilpotent matrix.
However in the $D-N$ décomposition, an important hypothesis is that... $D,N$ commute.
In the proof of the $D-N$ decomposition, it can also be proven that $D,N$ can be written as polynomials of $T$. Then, if $E$ commutes with $T$, it is obvious that it commutes with $D,N$.
Note: I refer above to French articles as I’m French and $D-N$ decomposition is named « décomposition de Dunford » in French. It seems that in English it is named Jordan-Chevalley decomposition. The article provides an English proof of the arguments I mention above. See in particular proposition 4.1