# The Unapologetic Mathematician

## Another Carnival

The Secret Blogging Seminar is hosting the latest installment of the Carnival of Mathematics. They seem to have done a fair job of collecting links, with more (and more varied) entries than I remember seeing before. Enjoy.

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Posted by John Armstrong | Uncategorized

## 3 Comments	»

1. John, I have appreciated your help in the past and I would like to get your insight into yet another question I have concerning linear algebra.

What is the relationship between the set of all linear transformations on a vector space over a field F of dimension n and the set of all nxn matrices with entries from the same field F. Clearly, it is not a one-to-one relation. There may be many, possibly an infinite number of matrices that can represent the same linear transformation with respect to different basis of the space. What exactly is the relation?

Jonathan Greenberg

Comment by Jonathan Grennberg | December 11, 2007 | Reply

2. When you’re dealing with $R$-linear functions between free $R$-modules (all vector spaces — modules over a field — are free), you can pick a basis for each of the domain and range as I discuss here. Once you do, then any such linear transformation can be written down uniquely as a matrix with respect to those bases, as I discuss in that link. Also, given a choice of bases any matrix corresponds to exactly one linear transformation.

But as you notice, there’s that choice of basis. Changing the basis of a free module is the same as applying an invertible linear transformation from that module to itself. That is, $\hom(R^m,R^n)$ is the space of linear transformations from a rank $m$ module to a rank $n$ module. Then there is an action of the group $\mathrm{GL}(m,R)$ (invertible $m\times m$ matrices with entries in $R$) on the one side, and an action of $\mathrm{GL}(n,R)$ on the other. What you’re really asking about is what the set of transformations looks like once we quotient out by these group actions.

And that’s where the story gets complicated. Over an algebraically closed field (like the complex numbers), the answer is contained in the Jordan normal form, which is a standard part of any good senior-level linear algebra course. It gives a canonical representative of each orbit of the group action. Beyond that it’s a huge mess. Sorry I can’t be more accurate.

Comment by | December 11, 2007 | Reply

3. John, thank you for your insight into my question. I will have to think about your answer as I study linear algebra in more depth.

JG

Comment by Jonathan Greenberg | December 11, 2007 | Reply

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## About this weblog

This is mainly an expository blath, with occasional high-level excursions, humorous observations, rants, and musings. The main-line exposition should be accessible to the “Generally Interested Lay Audience”, as long as you trace the links back towards the basics. Check the sidebar for specific topics (under “Categories”).

I’m in the process of tweaking some aspects of the site to make it easier to refer back to older topics, so try to make the best of it for now.

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