The Unapologetic Mathematician

Mathematics for the interested outsider

Orbits of the General Linear Action on Matrices

We’ve previously considered the representation of the general linear group \mathrm{GL}_n(\mathbb{F}) on the vector space of n\times n matrices over \mathbb{F} by conjugation. What we want to consider now are the orbits of this group action. That is, given two matrices S and T, we will consider them equivalent if there is an invertible matrix G so that S=G^{-1}TG. We want to describe these equivalence classes explicitly, and have a test to determine whether two matrices are conjugate or not.

This isn’t just a toy problem. Remember that any matrix is a linear transformation S:\mathbb{F}^n\rightarrow\mathbb{F}^n, and conjugation by an automorphism is how we render a change of basis into the language of matrices. That is, what we’re really asking is, “given two matrices, do they represent the same linear transformation with respect to two different bases?” This will be true if and only if they are conjugate by a change-of-basis matrix G\in\mathrm{GL}_n(\mathbb{F}).

To get an idea of how to solve this problem, let’s consider a couple other orbit-finding problems for a moment.

First, let’s let the integers act on the real numbers by addition. That is, given a real number x and an integer n we set n(x)=n+x. This just slides the real line along itself by integer steps. We consider two real numbers to be equivalent if they differ by an integer. How do we classify the orbits? A common way is to pick one point in each orbit to represent the whole. In general, making such a choice can be difficult (or impossible!), but it turns out we have a way of picking out such a representative without having to do much work: pick the one point in the interval \left[0,1\right). Notice that every real number is between two integers (using the standard inclusion), and if we subtract the lower of the two we get a number between {0} and {1}. Further, no two of these numbers are equivalent to each other. This interval contains exactly one representative for each orbit.

Next, let’s let the real numbers act on the plane with an identified point. We haven’t really talked much about the plane, but an intuitive notion will suffice here. The action of the real number x will be to rotate the plane by an angle x around the identified point (again, a detailed understanding of “angle” doesn’t really matter here). The orbits here are circles around the common center. To classify them, we again want to pick out a “canonical” representative point in each orbit. To do this, we can draw a ray from the center out towards infinity. Then each circle meets this ray exactly once, and no two points on the ray are equivalent.

So let’s try to attack the problem at hand in a similar manner. We have the general linear group throwing matrices around the orbits. We want to pick exactly one matrix in each orbit as a representative of the whole.

But we know how to do this! Use a matrix in Jordan normal form! We know that within a given conjugacy class, the Jordan normal form is unique — up to rearrangement of the Jordan blocks. So we don’t quite have a unique representative, but it’s a lot easier to check if two Jordan matrices are equivalent than to search by hand for a conjugating matrix.

We have an answer to the problem of determining whether two matrices S and T are conjugate. Put each one into Jordan normal form, and compare the resulting matrices. If they differ only by a rearrangement of the blocks, then S and T are equivalent, and if not they’re not.

March 6, 2009 Posted by | Algebra, Linear Algebra | 3 Comments

   

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