The Unapologetic Mathematician

Mathematics for the interested outsider

How Many Generators?

Okay, one last post to fill out the week.

The shears alone generate the special linear group. Can we strip them down any further? And, with this in mind, how many generators does it take to build up the whole general linear group?

It turns out that we don’t even need all the shears. We can just use neighboring shears to build all the others. Indeed:

\displaystyle\begin{aligned}\begin{pmatrix}1&-1&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&1&-1\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&1&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&1&1\\{0}&0&1\end{pmatrix}&\\=\begin{pmatrix}1&-1&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&0&0\\{0}&1&-1\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&1&1\\{0}&1&1\\{0}&0&1\end{pmatrix}&\\=\begin{pmatrix}1&-1&0\\{0}&1&0\\{0}&0&1\end{pmatrix}\begin{pmatrix}1&1&1\\{0}&1&0\\{0}&0&1\end{pmatrix}&\\=\begin{pmatrix}1&0&1\\{0}&1&0\\{0}&0&1\end{pmatrix}\end{aligned}

In terms of elementary row operations, first we add the third row to the second. Then we add the second to the first, effectively adding the third row to the first as well. Then we subtract the third row from the second, undoing that first step. Finally, we subtract the second row (alone now) from the first, undoing the extra addition of the second row to the first. At the end of the whole process we’ve added the third row to the first. We could modify this by adding a multiple of the third row to the second, and subtracting the same multiple later. Check to see what result that has. And we have similar results using neighboring lower shears

So we can generate the special linear group \mathrm{SL}(n,\mathbb{F}) using only the 2n-2 neighboring shears. If we have an matrix M in \mathrm{GL}(n\mathbb{F}) we can take its determinant \det(M). Then we can write M=C_{1,\det(M)}\tilde{M}. Here we’ve factored out a scaling by the determinant in the first row and we’re left with a matrix \tilde{M} in \mathrm{SL}(n,\mathbb{F}), which can then be written in terms of neighboring shears. So we need 2n-1 (families of) generators here.

These are the best I can do, and I don’t see a way of improving. Roughly, upper shears can’t build up lower shears, any collection of neighboring shears can only affect the rows they cover in sequence and so can’t build up a new neighboring shear, and no shears can handle that one scaling. So it seems there’s no way to pare down this collection of generators. But there might be a completely different approach that leads to fewer families of generators. If someone has one, I’d be glad to see it.

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September 11, 2009 - Posted by | Algebra, Linear Algebra

2 Comments »

  1. Perhaps an obvious question: Is the set of all invertible operators on a finite-dimensional vector space V, together with the zero operator, a subspace of L(V)? I would appreciate a proof or a counterexample.

    Comment by odelaman | October 11, 2009 | Reply

  2. The geometry of determinental varieties (which is essentially what you’re asking about here) is one of the most annoying concepts to really get a good handle on, in my experience. It’s very much not like a subspace. And it should be really simple for you to find, say, two invertible matrices whose sum is singular.

    Comment by John Armstrong | October 11, 2009 | Reply


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