# 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

where is the product of elementary matrices of the third kind (scales) and 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 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 such that

and is some function of and .

There are three cases:

1. Where is in position of and is in position of , and .

Thus the only non-zero and non-one entries of the product are in and and they are and respectively. Similarly, if we switch and , the product has the values and in and respectively as well, and so is commutative and we can “pull” the scale to the left of without modifying either. Thus, .

2. In our second case, we explore when . Thus, when multiplying , is placed in while is placed in , but when multiplying , is placed in once again but row is multiplied by column and the result is in position . Thus, .

3. Similarly, if then when row is multiplied by column , the result is in but in the switch, so is in position and once again, is in . Thus, .

For the points 2. and 3. above, we assume that only one of is equal to . This is obvious because otherwise is not an elementary matrix of the first kind. In addition, and are not equal to since otherwise is an identity matrix (and by the definition of the identity matrix, ).

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 can be generated by shears using induction instead.

We start with a 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 is . For a matrix, this makes it in upper-triangular row echelon form.

However, since the determinant of any matrix is , and since , we must have . Using only shears, we can reduce to be 1, and in the process must also equal 1 since and are reciprocals. Now we have a matrix with only 1′s in the diagonal.

Using induction, we can assume that we have a matrix that is in upper-triangular with only 1′s on its diagonal, and use this to prove that the case is the same. We can subtract a multiple equal to the value of each of element of the row, so all elements before the element in the row are zero. Since all elements on the diagonal before the element are 1′s, and the diagonal elements of an upper-triangular matrix multiplies to the determinant, it is obvious that the element must be 1 as well. Thus, using induction, we have proved the base case and the inductive case. All 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 element, equal to 1, we can subtract a multiple of that row from every row above it where the element is not a 0, until they are all 0′s. Now we take the row, where the only non-zero element if the 1, and subtract a multiple of that from every row above it. Eventually we will have constructed the identity using only shears.

Thus,

and

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

Interesting. Makes me miss actual math classes.