1

Let $A,B$ be commuting matrices such that some positive power of each matrix is the identity. Prove that there is an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both diagonal.

Similar questions to this have been asked on here a number of times, and I understand the solution that one can prove that $A$ and $B$ have the same eigenvectors forming a basis for the vector space, so if we change to that basis both $A$ and $B$ will be diagonal.

My confusion comes from where I found this question. It is an exercise from chapter 10 of Algebra, by Artin, on group representations. I'm finding it difficult to see the connection between this problem and group representations. Some thoughts I had were that there are one dimensional representations of $A$, $B$ and $AB$, but I failed to find any connection between that and diagonalization.

What is the connection between this problem and group representations? Is there a version of a proof for this problem that mainly uses the theory behind group representations?

Miles Johnson
  • 505
  • I'd generalize the problem, even if it means losing the connection to group representations. The generalization is: Let $A$ and $B$ be two commuting diagonalizable matrices; then there is an invertible matrix $P$ such that $PAP^{-1}$ and $PBP^{-1}$ are both diagonal. (This generalizes your problem, because a square matrix over $\mathbb{C}$ that has an identity power must always be diagonalizable. I am hoping that you are working over $\mathbb{C}$, since otherwise your problem is false.) – darij grinberg Feb 23 '19 at 21:54
  • The generalization is proven in https://kconrad.math.uconn.edu/blurbs/linmultialg/simulcomm.pdf (Theorem 5) and in https://math.stackexchange.com/questions/56307/simultaneous-diagonalization . – darij grinberg Feb 23 '19 at 21:55
  • I already saw and understood the proof of the generalization. I just want to know why the question was in this chapter – Miles Johnson Feb 23 '19 at 23:32
  • 2
    Maybe because (in the specific situation of the question) you can define an abelian group $G := Z_p \times Z_q$ (where $Z_m$ means the cyclic group of order $m$, and where $p$ and $q$ are positive integers satisfying $A^p = I$ and $B^q = I$), and then this group $G$ acts on your vector space by $\left(i, j\right) \mapsto A^i B^j$. Maschke's theorem then shows that every representation is a direct sum of irreducibles, while the abelianity of $G$ yields that irreducibles are $1$-dimensional. Thus the diagonalization. – darij grinberg Feb 23 '19 at 23:51
  • @darijgrinberg Thanks! I think that's definitely the answer this question was looking for. It seems I was trying to look for a representation for the group generated by A and B, but A and B are already a suitable representation for a simple group. Definitely makes me think about the different ways representation theory might be applied haha – Miles Johnson Feb 24 '19 at 02:40

0 Answers0