5
$\begingroup$

I sort of intuitively see why we care about similar matrices, i.e., when $A=S^{-1}BS$ for some invertible matrix $S$. But I want to make this intuition more precise and abstract.

Matrices: First of all, as mentioned here,

Because matrices are similar if and only if they represent the same linear operator with respect to (possibly) different bases, similar matrices share all properties of their shared underlying operator.

This is followed by a long list of shared properties. However, I feel this doesn't give the full story. For example, we also care about when two different operators are similar, in which case (conversely), they can be represented by the same matrix with appropriate bases. In this case, the long list of properties is still shared by both operators. How can one precisely say what type of properties are shared by similar operators, and what's the most abstract way to understand this?

Generalizations: If $A, B, S$ are elements of a group $G$, then $A = S^{-1}BS$ is described by saying $A$ is the conjugation of $B$ by $S$. For any $S \in G$, conjugation by $S$ is an endomorphism of $G$, and hence preserves all group-theoretic properties of any element. Matrix similarity is an extension of this idea, where we conjugate elements of an algebra $L(V)$ (operators) with the group of units in the algebra $GL(V)$ (invertible matrices). This type of conjugation then provides an algebra endomorphism, so we should expect the properties of some $T \in L(V)$ as an element of the algebra $L(V)$ to be preserved by conjugation—but some of the properties in the list linked above (e.g., determinant) are specific to $L(V)$ and cannot be generalized to an arbitrary algebra.

The most general framework I can think of is as follows: We have some group $G$ which acts on some structure $X$ from the left and right. Thus conjugation makes sense. Can we say anything about what properties of an arbitrary $x \in X$ are preserved under conjugation, in a way that includes matrix similarity as a special case?

$\endgroup$
7
  • 1
    $\begingroup$ Would it be possible for you to elaborate more on your "However, I feel this doesn't give the full story" comment? It seems to me that the long list of shared properties are a consequence of the fact that A and B represent the same linear operator. I'm having a bit of difficulty interpreting what kind of properties you're thinking of in the generalized case. $\endgroup$
    – mhum
    Mar 29 at 2:24
  • $\begingroup$ @mhum My point was, if one only considered similarity to be interesting because it indicates when two matrices can represent the same operator, then one could not explain why we care about when two (distinct) operators are similar, and why similar operators share the same properties. $\endgroup$
    – WillG
    Mar 29 at 2:33
  • $\begingroup$ Admittedly, my question is a bit loose. I'm basically looking for more abstract insight into matrix similarity. As for what kind of properties I'm thinking of in the generalized case, see my last paragraph: It seems like a two-sided group action is most general setting for similarity, and I'd like to understand what properties are preserved in this case (in the language of group actions), and whether matrix similarity can be understood as a special case. $\endgroup$
    – WillG
    Mar 29 at 2:36
  • $\begingroup$ If $F$ is any field, $n$ is any natural number, and $M_n(F)$ is the ring of $n\times n$ matrices over $F$ then for each invertible matrix $S\in M_n(F)$, the conjugation $X\mapsto S^{-1}XS$ is an automorphism of $M_n(F)$, so the conjugation has the properties of automorphisms of rings. $\endgroup$ Mar 29 at 12:35
  • $\begingroup$ @AlexRavsky Indeed. But note that some properties of elements of $M_n(F)$ are not only ring properties, but "linear algebraic" properties, e.g., eigenvalues and determinants. But these latter properties are also preserved under conjugation. $\endgroup$
    – WillG
    Mar 29 at 16:31

1 Answer 1

3
+50
$\begingroup$

Here's one way of approaching matrix similarity, from an abstract POV. Instead of an $n\times n$ matrix $M$, let's consider an arbitrary endomorphism $\phi$ of an $n$ dimensional vector space $V$. An endomorphism $\phi$ of $V$ is equivalent to a $k[x]$ module structure on $V$, where $x\cdot v:=\phi(v)$. From this perspective, similarity of endomorphisms is when the associated modules are isomorphic. So any properties of finite dimensional $k[x]$ modules (of course invariant under isomorphism) are what similar matrices share.

For instance, the structure theory of modules over a PID gives normal form theorems, generalised eigenvalues are the isomorphism classes of simple modules in the composition series, etc.

The category of finite dimensional $k[x]$ modules also has duals, and (symmetric) tensor products, so we can understand transpose and exterior powers without bases, giving characteristic polynomials categorically.

As a concrete example, from this perspective, we can prove easily that a matrix and it's transpose are similar, by interpreting all the pieces of this statement module theoretically.

Understanding when the linear algebra is describing properties of finite dimensional $k[x]$ modules is very helpful, and helps delineate matrix properties from abstract linear endomorphism properties (which may generalised to other settings).

$\endgroup$
3
  • $\begingroup$ This is great—it's exactly the kind of abstract perspective I was looking for. I had never considered how linear maps induce a $k[x]$ module structure on $V$. Now I just need to study more module theory in order to understand it :) $\endgroup$
    – WillG
    Apr 1 at 0:41
  • $\begingroup$ Out of curiosity, is every $k[x]$ module structure on $V$ induced by some operator in this way? $\endgroup$
    – WillG
    Apr 1 at 0:42
  • $\begingroup$ Yea modules really clarify the situation. It is true that every $k[x]$ module structure is induced by a linear operator, spelling this out is a good exercise to familiarize yourself with the definitions. $\endgroup$
    – Chris H
    Apr 1 at 9:10

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .