# The Unapologetic Mathematician

## Subgroups Generated by Shears

Okay, when I introduced elementary matrices I was a bit vague on the subgroup that the shears generate. I mean to partially rectify that now that we’ve got elementary row operations to work with.

I assert that the upper shears — those elementary matrices with one nonzero entry above the diagonal — generate the group of “upper unipotent” matrices. A matrix is unipotent if it is the identity plus a nilpotent transformation, so all of its eigenvalues are ${1}$. Specifically, we want those that are also upper-triangular. Thus the matrices we’re talking about have ${0}$ everywhere below the diagonal and all ${1}$ on the diagonal, as

$\displaystyle A=\begin{pmatrix}1&a_{1,2}&\cdots&&a_{1,n}\\{0}&1&&&\\\vdots&&\ddots&&\vdots\\&&&1&a_{n-1,n}\\{0}&&\cdots&0&1\end{pmatrix}$

What I’m going to do, now, is turn this matrix into the identity by using elementary row operations. Specifically, I’ll only ever add a multiple of one row to another row, which can be effected by multiplying the matrix on the left by shear matrices. And I’ll only add multiples of one row to rows above it, which can be effected by using upper shears.

So first, let’s add $-a_{n-1,n}$ times the $n$th row to the $n-1$th.

$\displaystyle\left(H_{n-1,n,-a_{n-1,n}}\right)A=\begin{pmatrix}1&a_{1,2}&&\cdots&&a_{1,n}\\{0}&1&&&&\\&&\ddots&&&\vdots\\\vdots&&&1&a_{n-2,n-1}&a_{n-2,n}\\&&&&1&0\\{0}&&&\cdots&0&1\end{pmatrix}$

We’ve cleared out the last entry in the next-to-last row in the matrix. Keep going, clearing out all the rest of the last column

$\displaystyle\left(H_{1,n,-a_{1,n}}\dots H_{n-1,n,-a_{n-1,n}}\right)A=\begin{pmatrix}1&a_{1,2}&&\cdots&&0\\{0}&1&&&&\\&&\ddots&&&\vdots\\\vdots&&&1&a_{n-2,n-1}&0\\&&&&1&0\\{0}&&&\cdots&0&1\end{pmatrix}$

Now we can use the next-to-last row — which has only a single nonzero entry left — to clear out the rest of the next-to-last column

\displaystyle\begin{aligned}\left(\left(H_{1,n-1,-a_{1,n-1}}\dots H_{n-2,n-1,-a_{n-2,n-1}}\right)\left(H_{1,n,-a_{1,n}}\dots H_{n-1,n,-a_{n-1,n}}\right)\right)A&\\=\begin{pmatrix}1&a_{1,2}&&\cdots&&0\\{0}&1&&&&\\&&\ddots&&&\vdots\\\vdots&&&1&0&0\\&&&&1&0\\{0}&&&\cdots&0&1\end{pmatrix}&\end{aligned}

Keep going, clearing out the columns from right to left

$\displaystyle\left(\left(H_{1,2,-a_{1,2}}\right)\dots\left(H_{1,n,-a_{1,n}}\dots H_{n-1,n,-a_{n-1,n}}\right)\right)A=\begin{pmatrix}1&0&&\cdots&&0\\{0}&1&&&&\\&&\ddots&&&\vdots\\\vdots&&&1&0&0\\&&&&1&0\\{0}&&&\cdots&0&1\end{pmatrix}$

and we’ve got the identity matrix! So that means that this big product of all these upper shears on the left is actually the inverse to the matrix $A$ that we started with. Now we just multiply the inverse of each of these shears together in reverse order to find

$\displaystyle A=\left(\left(H_{n-1,n,a_{n-1,n}}\dots H_{1,n,a_{1,n}}\right)\dots\left(H_{1,2,a_{1,2}}\right)\right)$

So any upper-unipotent matrix can be written as a product of upper shears. Similarly, any lower-unipotent matrix (all ${0}$ above the diagonal and all ${1}$ on the diagonal) can be written as a product of lower shears. If we add in scalings, we can adjust the diagonal entries too. Given an invertible upper-triangular matrix, first factor it into the product of a diagonal matrix and an upper-unipotent matrix by dividing each row by its diagonal entry. Then the diagonal part can be built from scalings, while the upper-unipotent part can be built from shears. And, of course, scalings and lower shears together generate the subgroup of invertible lower-triangular matrices.

August 28, 2009 Posted by | Algebra, Linear Algebra | 1 Comment