let $A$ be a complex matrix. Denote by $J(A)$ the Jordan Canonical Form of $A$. Let $C[J(A)]$ be the centralizer of $J(A)$ in $M_n(\mathbb C)$. Can we construct a real matrix $B$, that is, $B$ has only real entries, verifying the equality $C[J(A)]=C(B)$, in $M_n(\mathbb C)$?
Asked
Active
Viewed 747 times
1 Answers
3
$J(A)$ can be written as $S+N$, where $S$ is diagonal and $N$ is nilpotent (and has only entries $0$ and $1$). Moreover, $S$ and $N$ are polynomials in $J(A)$.
It follows that $C$ centralizes $J(A)$ if and only if it centralizes $S$ and $N$.
The recipe for $B$ is now clear, $B = T+N$, where $T$ is obtained from $S$ by replacing the distinct eigenvalues with distinct real numbers.
This is because $S$ and $T$ have the same centralizer, which consists of block diagonal matrices, with arbitrary blocks on each eigenspace.
Andreas Caranti
- 68,873
-
+1 Nice argument. By the way, should $B$ be $T+kN (k\in\mathbb{C})$ instead of $T+N$? – user1551 Feb 22 '13 at 14:28
-
@user1551, why do that? Moreover, $B$ must be real. – Andreas Caranti Feb 22 '13 at 18:24
-
Oops, sorry, I meant $k\in\mathbb{R}$. You see, if $C$ centralizes $S$ and $N$, it also centralizes $kN$ for every real $k$. – user1551 Feb 22 '13 at 18:27