October 26, 2011 / indefiniteintegirl

Scales and Shears are (Sort of) Commutative and the Latter Generate the Special Linear Group

For a first post I am going to post a response to this article that states without formal proof (“easy to verify”) that you can pull any scale matrix to the left of a shear matrix. This is a relatively easy first problem but it has interesting implications, namely that with only simple modifications, we can write any invertible matrix in the form

$ABX = I$

where $A$ is the product of elementary matrices of the third kind (scales) and $B$ is the product of elementary matrices of the first kind (shears).
Since all invertible matrices can be written as a product of elementary matrices of the first and third kind, we just have to show that we can row-reduced $X$ into the identity using shears before scales, namely, that we can write any row-reduction in the form above.

The article gives examples of this type of reorganization, but in general form we want to be able to produce a matrix $M_3$ such that

$M_1M_2 =M_2M_3$

where $M1$ , $M2$ , and $M3$ are matrices defined as such:

$M_1 =\begin{bmatrix} 1 & & & & & & 0\\ &\ddots & & & & &\\ & & 1 & & a & & \\ & & &\ddots & & &\\ & & & & 1 & &\\ & & & & & \ddots &\\ 0 & & & & & & 1\\ \end{bmatrix}$ , $M_2 = \begin{bmatrix} 1 & & & & & & 0\\ &\ddots & & & & &\\ & & 1 & & & & \\ & & & b & & &\\ & & & & 1 & &\\ & & & & & \ddots &\\ 0 & & & & & & 1\\ \end{bmatrix}$ ,

$M_3 = \begin{bmatrix} 1 & & & & & & 0\\ &\ddots & & & & &\\ & & 1 & & c & & \\ & & &\ddots & & &\\ & & & & 1 & &\\ & & & & & \ddots &\\ 0 & & & & & & 1\\ \end{bmatrix}$ $M_2 = \begin{bmatrix} 1 & & & & & & 0\\ &\ddots & & & & &\\ & & 1 & & & & \\ & & & b & & &\\ & & & & 1 & &\\ & & & & & \ddots &\\ 0 & & & & & & 1\\ \end{bmatrix}$

and $c$ is some function of $a$ and $b$ .

There are three cases:

1. Where $a$ is in position $(n,j)$ of $M_1$ and $b$ is in position $(i,i)$ of $M_2$ , and $n,j \neq i$ .

Thus the only non-zero and non-one entries of the product $M_2M_1$ are in $(n,j)$ and $(i,i)$ and they are $a$ and $b$ respectively. Similarly, if we switch $M_1$ and $M_2$ , the product $M_1M_2$ has the values $a$ and $b$ in $(n,j)$ and $(i,i)$ respectively as well, and so $M_1M_2 = M_2M_1$ is commutative and we can “pull” the scale $M2$ to the left of $M1$ without modifying either. Thus, $c=a$ .

2. In our second case, we explore when $n=i$ . Thus, when multiplying $M_1M_2$ , $b$ is placed in $(i,i)$ while $a$ is placed in $(i,j)$ , but when multiplying $M_2M_1$ , $b$ is placed in $(i,i)$ once again but row $i$ is multiplied by column $j$ and the result is $ab$ in position $(i,j)$ . Thus, $c = a/b$ .

3. Similarly, if $j = i$ then when row $j$ is multiplied by column $i$ , the result is $ab$ in $(i,j)$ but in the switch, $n \neq i$ so $a$ is in position $(i,j)$ and once again, $b$ is in $(i,i)$ . Thus, $c= ab$ .

For the points 2. and 3. above, we assume that only one of $n,j$ is equal to $i$ . This is obvious because otherwise $M_1$ is not an elementary matrix of the first kind. In addition, $a$ and $b$ are not equal to $0$ since otherwise $M_1$ is an identity matrix (and by the definition of the identity matrix, $IX = XI$ ).

The rest of the proof is provided in the article, which uses the above fact in proving that shears generate the Special Linear Group. I’m not going to reiterate that proof but when I first did this problem, my first instinct was to use induction. I’m not sure if this method is harder or easier but it is probably more straightforward. So as a digression, I’m going to prove that $SL_n \mathbb{R}$ can be generated by shears using induction instead.

We start with a $2 \times 2$ matrix. We want to create an upper-triangular matrix using only shears, which we do just by subtracting the proper multiple of the first row from the second so that the term $a_{2,1}$ is $0$ . For a $2 \times 2$ matrix, this makes it in upper-triangular row echelon form.

However, since the determinant of any $2 \times 2$ matrix is $ad - dc$ , and since $b=0$ , we must have $ad=1$ . Using only shears, we can reduce $a$ to be 1, and in the process $d$ must also equal 1 since $a$ and $d$ are reciprocals. Now we have a $(2 \times 2)$ matrix with only 1′s in the diagonal.

Using induction, we can assume that we have a $(m-1 \times m-1)$ matrix that is in upper-triangular with only 1′s on its diagonal, and use this to prove that the $(m,m)$ case is the same. We can subtract a multiple equal to the value of each of element $1, 2, ... m-1$ of the $m^{th}$ row, so all elements before the $(m,m)$ element in the $m^{th}$ row are zero. Since all elements on the diagonal before the $(m,m)^{th}$ element are 1′s, and the diagonal elements of an upper-triangular matrix multiplies to the determinant, it is obvious that the $(m,m)^{th}$ element must be 1 as well. Thus, using induction, we have proved the base case and the inductive case. All $n \times n$ matrices of determinant 1 can be reduced to an upper-triangular matrix with only 1′s on its diagonal.

We can continue to reduce this upper-triangular matrix to the identity using a series of shears. Starting with the last row, where the only non-zero element is the $n^{th}$ element, equal to 1, we can subtract a multiple of that row from every row above it where the $n^{th}$ element is not a 0, until they are all 0′s. Now we take the $(n-1)^{th}$ row, where the only non-zero element if the $(n-1)^{th}$ 1, and subtract a multiple of that from every row above it. Eventually we will have constructed the identity using only shears.

Thus,

$I = AXB$

and

$B^{-1}A^{-1} = X$

and since the inverse of an elementary matrix is the same type of elementary matrix, we can conclude that all elements of $SL_n \mathbb{R}$ can be written as a product of shears and thus $SL_n \mathbb{R}$ can be generated by shears.

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