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Let $\{X_1,X_2,...,X_n\} \subseteq GL_d$ be a subgroup of commutating matrices then show this matrices are simultaneously diagonalizable (using some argument from representation theory)

dragoboy
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    What have you done so far? For example, do you know that two diagonalizable matrices are simultaneously diagonalizable iff they commute? Are your matrices diagonalizable? Why? Why not? – Timbuc Feb 17 '15 at 12:53
  • Yes I noticed before that its enough to show diagonalizability of one matrix, but how to show that ? I was trying to show that like, setting up a homomorphism from an abelian group. Where abelian group is ${1,2,...,n,}$ and operation is defined as, ij is indices of $X_i*X_j$ and send each $i \to X_i$ is that correct ? – dragoboy Feb 17 '15 at 13:19
  • Well, I think I understand the claim when working over an algebraically closed field, say $;\Bbb C;$ , and using Schur's Lemma and stuff. This isn't given here, though. – Timbuc Feb 17 '15 at 14:25
  • Yeah I'm working on $\mathbb{C}$ and I don't see any problem in my construction ... but without that is there any way to show diagolazability using some representation th argument ? – dragoboy Feb 17 '15 at 16:28
  • Would the argument I give here be something that fits the bill? The basic idea is that as the matrices commute, they must stabilize each others' eigenspaces. The rest is by induction on $d$. – Jyrki Lahtonen Feb 17 '15 at 22:56
  • @Timbuc Even over an algebraically closed field, this will fail without further assumptions (namely that each matrix is diagonalizable). – Tobias Kildetoft Feb 18 '15 at 11:16
  • @TobiasKildetoft I'd have to check, but I think that over an alg. closed field one can prove that a begin a group then each matrix there is diagonalizable. – Timbuc Feb 18 '15 at 11:20
  • @Timbuc No, for example any upper triangular unipotent $2\times 2$ matrix will not be diagonalizable, regardless of the field (I actually had to think to find an example, but as far as I recall, being diagonalizable is a closed condition and since the invertible matrices are dense in the full set of matrices, all invertible ones being diagonalizable would imply that they would all be). – Tobias Kildetoft Feb 18 '15 at 11:26
  • @TobiasKildetoft How can an unipotent matrix be part of a finite abelian group of matrices? – Timbuc Feb 18 '15 at 11:42
  • @Timbuc Ahh, I somehow missed the "finite" part. – Tobias Kildetoft Feb 18 '15 at 13:25
  • @Timbuc Actually, in positive characteristic my example does work. – Tobias Kildetoft Feb 18 '15 at 15:57

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