Dimension of Commutator Space [closed]

For each $n\times n$ matrix $A$ with real entries the set $$C(A)=\{X\in M_n(\mathbb{R}): AX=XA\}$$ is obviously a linear subspace of $M_n(\mathbb{R})$.

Can we recognize the dimension of this space by looking only at the Jordan form of $A$?

We know that this dimension is at least the degree of the minimal polynomial of $A$, but this bound is not sharp, since for example for the identity matrix the degree of the minimal polynomial is 1 and $\dim C(I)=n^2$.

• The answer is yes we can, though the answer can be messy particularly when $A$ is nilpotent, which is the essential case. – Geoff Robinson Aug 28 '14 at 18:07
• To see that the answer is positive, one can note that $\dim_\mathbb{R}\,C(A)=\dim_\mathbb{C}\,C(\overline{A})$, where $C(\overline{A})$ is the complex commutator of complexification of $A$. Then, it suffices to check $C(J^{-1}AJ)=J\,C(A)\,J^{-1}$. – user56203 Aug 28 '14 at 18:46
• See the answer to this question. – abx Aug 29 '14 at 6:39

Consider the conjugation representation of $GL_n$ on $M_n$. Suppose that $A$ is an $n\times n$ matrix. Then $C(A)$ is the $\mathfrak{gl}_n$-stabilizer of $A$, so that $\dim(C(A))$ is equal to the dimension of the $GL_n$-stabilizer of $A$. In other words, $\dim(C(A))$ is $n^2$ minus the dimension of the conjugacy class of $A$. If $A$ is a nilpotent matrix in Jordan canonical form, then I believe there are expressions for the dimension of the conjugacy class of $A$ in terms of the partition giving its Jordan canonical form. You might find this in Nilpotent Orbits in Semisimple Lie Algebras by Collingwood and McGovern.