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Recall that two $n\times n$ matrices over $\mathbb{C}$ are conjugate if and only if they have the same Jordan canonical form.

Question. Is there a similar classification for commuting pairs of matrices?

To be precise, a commuting pair of matrices is an ordered pair $(A,B)$ of $n\times n$ matrices over $\mathbb{C}$ for which $AB=BA$. Two commuting pairs $(A,B)$ and $(A',B\,')$ are conjugate if there exists a nonsingular matrix $P$ so that $$ PAP^{-1} = A'\qquad\text{and}\qquad PBP^{-1} = B\,'. $$ What are the equivalence classes under this relation? Can we enumerate them? Can we check whether two pairs are in the same class?

A few notes:

  1. If $A$ and $B$ are diagonalizable, they can be simultaneously diagonalized, and the resulting pair of diagonal matrices determines the conjugacy class.

  2. According to this answer, a commuting pair of matrices cannot in general be simultaneously Jordanized.

  3. According to this Wikipedia article, a commuting pair of matrices can be simultaneously triangularized, but of course the pair of triangular matrices is not uniquely determined.

By the way, this question is equivalent to asking for a classification of indecomposable modules over $\mathbb{C}[x,y]$ with finite dimension over $\mathbb{C}$.

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Jim Belk
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1 Answers1

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Gelfand and Ponomarev proved, that a category of all artinian modulus over $K[X,Y]$ contains a full subcategory isomorphic to the category $L_k$ of artinian modulus over free algebra on $k$ elements. This subcategory can be constructed explicitly.

So if such classification exists then one can also classify systems of $n$ non-commuting linear operators on up to conjugation. That seems impossible.

Here is a link to the article: http://link.springer.com/article/10.1007%2FBF01076321

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