Suppose that $A = M_n(\mathbb{C})$ be the algebra of $n*n$ matrices over $\mathbb{C}$.
If com(A) = {$B \in M_n(\mathbb{C}); AB = BA$}, then what is the $dim(com(A))?$
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2$\begingroup$ If an $n\times n$ matrix B commutes with every other $n\times n$ matrices, then B is a constant multiple of the identity matrix. // If a subgroup G of $n\times n$ matrices is irreducible (in the sense that if $\forall A\in G\hspace{2mm} AV\subseteq V $ then $V=\mathbb{C}^n$, where $V\neq\{0\}$ is a vector subspace of $\mathbb{C}^n$) then $BA=AB$ for every $A\in G$ implies that $A=cI$ for some $c\in\mathbb{C}$.. $\endgroup$– Onur OktayCommented Jul 2, 2021 at 16:18
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1 Answer
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This is known for a general field by a theorem of Frobenius:
Let $F$ be a field and $V$ a finite dimensional $F$-vector sapce with a linear operator A. When $f_i(X)$ denote the invariant factors of $A$ (such that $f_i(X)$ divides $f_{i+1}(X)$, then the dimension is equal to $\sum\limits_{i=1}^{k}{(2k-2i+1)deg(f_i(X))}$.
See Theorem 5.15 in the book "Algebra: An approach via module theory" by Adkins and Weintraub.
