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I'm trying to characterize equivalence classes of matrices over a vector space.

Specifically, let $V$ be a vector space over a field $K$, let $M \in M_n(V)$ be an $n \times n$ matrix with entries in $V$ and let $A,B \in M_n(K)$ be $n \times n$ matrices with entries in $K$. I want to characterize the equivalence class formed by all matrices formed as

$$A M B$$

i.e, equivalent matrices, but over a vector field instead of a ring, and the invertible "side" matrices are picked from the base field, not the vector field itself.

The problem is related to constructing the Jacobian variety of a curve.

So far, I've been experimenting with the $2\times2$ case:

$$\left[\matrix{v_1 & v_3 \\ v_2 & v_4}\right]$$

Let's multiply the equation out:

$$AMB = \left[\matrix{a & c \\ b & d}\right] \left[\matrix{v_1 & v_3 \\ v_2 & v_4}\right] \left[\matrix{e & g \\ f & h}\right]$$

$$=\left[\matrix{eav_1 + ebv_2 + gav_3 + gbv_4 && fav_1+fbv_2+hav_3+hbv_4 \\ ecv_1+edv_2 +gcv_3+gdv_4 && fcv_1+fdv_2+hcv_3+hdv_4}\right]$$

Obviously, all of the vectors have to be in the linear space spanned by the original four vectors. So, is picking a four-dimensional subspace of $V$ enough to specify our equivalence class?

No. Let the transformed matrix have the form:

$$\left[\matrix{ A_1v_1 + A_2v_2 + A_3v_3 + A_4v_4 && C_1v_1 + C_2v_2 + C_3v_3 + C_4v_4 \\ B_1v_1 + B_2v_2 + B_3v_3 + B_4v_4 && D_1v_1 + D_2v_2 + D_3v_3 + D_4v_4 }\right]$$

We see that the obvious equality $ea\cdot gb = eb \cdot ga$ implies that $A_1A_4 = A_2A_3$. Likewise, $B_1B_4=B_2B_3$, $C_1C_4=C_2C_3$ and $D_1D_4=D_2D_3$.

There are also similar cross-relationships between the four vectors, but the restriction I just derived is enough to show that a four-dimensional subspace alone isn't enough to specify an equivalence class. The vectors have to picked from a particular three-dimensional subvariety, and it's a subvariety, not a subspace, because the relationships $A_1A_4 = A_2A_3$, etc., are not linear.

So, specifying a equivalence class requires specifying a four-dimensional subspace, then picking a certain three-dimensional subvariety from that space. I've left out considerations like the vectors being linearly dependent, but I think this is enough to give the sense of what I'm trying to accomplish.

Has anybody seen anything like this? Any idea how to characterize these equivalence classes?

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If $\dim V = d$, then this is equivalent to the problem of simultaneous similarity of $d$-tuples of $n \times n$ matrices over $K$. I'll prove the equivalence below, but let me first discuss the consequences for your question. The latter is a famous "wild" problem (in the formal sense of representation-wildness, see, e.g. Chapter 19 of Simson & Skowronski Vol. 3). This means that classifying the equivalence classes here is universal for classifying the representations of arbitrary finite-dimensional algebras over $K$, and hence is sometimes said to be "hopeless." Nonetheless, there is some hope and quite a lot is known. For example, there is a polynomial-time algorithm to decide when two such matrices of vectors lie in the same orbit [Sergeichuk; Brooksbank-Luks; Chistov-Ivanyos-Karpinski].

Sergeichuk also gives a "normal form," although the normal form is (necessarily) quite a bit more complicated than that for single matrices, and is recursive in nature.

Proof of equivalence: choose a basis $v_1, \dotsc, v_d$ of $V$. Write the $(i,j)$ entry of your matrix as $\sum_{k=1}^d T_{ijk} v_k$. Then your transformation of $A M B$ transforms the tensor $T$ into another tensor $T'_{ijk} = \sum_{i'=1}^n \sum_{j'=1}^n A_{ii'} B_{j'j} T_{i'j'k}$. If you look at what this action is, it's the same as if we instead considered the $d$-tuple of matrices $(M_1, \dotsc, M_d)$ where $(M_k)_{ij} = T_{ijk}$ under the action $A(M_1, \dotsc, M_d)B = (AM_1 B, \dotsc, A M_d B)$. This problem is wild for any $d \geq 3$. For $d=1$ the normal form is obvious and classical. For $d=2$ it is the Kronecker normal form for pencils, which I guess is also classical, though perhaps less well-known (e.g. not often taught in first courses on linear algebra).

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  • $\begingroup$ Thank you, Joshua. This is one of the most useful answers I've ever gotten on Stack Exchange! I'd never heard of simultaneous similarity, so I just didn't know what to search for, but now I do. Stack Exchange beats Google, which is just saying that a human expert beats an AI. The first paper you cited is general enough to embrace simultaneous equivalence, which is what I'm looking for. I'll have to study these papers now; this looks like a real promising approach. Thanks again. $\endgroup$ – Brent Baccala Jun 27 '18 at 1:06

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