1
$\begingroup$

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 $dim(com(A))?$

$\endgroup$
1
  • 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 Oktay
    Jul 2, 2021 at 16:18

1 Answer 1

6
$\begingroup$

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.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.