Let k be a field, V a finite-dimensional k-vectorspace and

Question

Let k be a field, V a finite-dimensional k-vectorspace and

tobeint39r 2022-04-30 Answered

M \in E n d (V)

. How can I determine Z, the centralizer of

M \otimes M

in

E n d (V) \otimes E n d (V)

?

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Answers (1)

Penelope Carson

Answered 2022-05-01 Author has 16 answers

Step 1
This looks like it can get complicated. Generically, at least over an algebraically closed field, a matrix M will have distinct eigenvalues $m_{1}, \dots, m_{n}$ , and generically $M \otimes M$ will have distinct eigenvalues $m_{1}^{2}, \dots, m_{n}^{2}$ with multiplicity one, and $m_{1} m_{2}, m_{1} m_{3}, \dots, m_{n - 1} m_{n}$ with multiplicity two. Thus the centralizer will have dimension $n + 4 (\binom{n}{2}) = 2 n^{2} - n$
But there are many degenerate cases: for instance if M has eigenvalues 1 $1, a, \dots, a^{n - 1}$ then $M \otimes M$ will have eigenvalues $1, a, \dots, a^{2 n - 2}$ with multiplicities $1, 2, \dots n - 1, n, n - 1, \dots, 1$ . Things can get more complicated still.
Then M might have non-trivial Jordan blocks, and then the real fun starts!

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Let G be a finite group with $c a r d (G) = p^{2} q$ with $p < q$ two ' numbers. We denote $s_{q}$ the number of q-Sylow subgroups of G and similarly for p. I have just shown that $s_{q} \in {1, p^{2}}$ . Now I want to show that
$\underset{S \in S y l_{q} (G)}{\cup} S ∖ {1} = {\dot{\cup}}_{S \in S y l_{q} (G)} S ∖ {1}$
i.e. that for $S, T \in S y l_{q} (G)$ with $S \neq T$ we that $S ∖ {1} \cap T ∖ {1} = \emptyset$