We will keep our notations in part (1). We now generalize the result proved in Problem 2.

Problem. Let A \in M_n(k) with \det A = 1. Prove that A is a product of elementary matrices.

Solution. Suppose that A = (a_{ij}), i.e. the (i,j)-entry of A is a_{ij}. The first column of A cannot be all zero because then \det A = 0, which is not true. So if a_{11}=0, then choosing a_{i1} \neq 0, we will have the matrix E_{1i}(1)A whose (1,1)-entry is a_{i1} \neq 0. Then we can replace A by E_{1i}(1)A and so we may assume that a_{11} \neq 0. Now let B_1=\prod_{i=2}^n E_{i1}(-a_{11}^{-1}a_{i1}) and C_1= \prod_{j=2}^n E_{1j}(-a_{11}^{-1}a_{1j}). Then the first row and the first column entries of B_1AC_1 are all zero except for the (1,1)-entry which is a_{11}. We can now do the same with the second row and the second column and continue this process until we get matrices B_1, \cdots , B_{n-1} and C_1, \cdots , C_{n-1} each of which is a product of elementary matrices and

D=B_{n-1}B_{n-2} \cdots B_1 A C_1 C_2 \cdots C_{n-1}

is a diagonal matrix. Note that \det D=1 because \det B_i=\det C_i=1 for all i (see the first part of Problem 1) and \det A=1. Thus by Problem 2, D is a product of elementary matrices. Hence

A=B_1^{-1} \cdots B_{n-1}^{-1}D C_{n-1}^{-1} \cdots C_1^{-1}

is also a product of elementary matrices, by the second part of Problem 1. \Box

Definition 1. The general linear group of order n over a field k is

GL(n,k)=\{A \in M_n(k): \ \det A \neq 0 \}.

Definition 2. The special linear group of order n over a field k is

SL(n,k)=\{A \in M_n(k): \ \det A=1 \}.

Remark 1. It is clear that GL(n,k) is a group with matrix multiplication and that SL(n,k) is a subgroup of GL(n,k). If, as usual, we let k^{\times}:=k \setminus \{0\} be the multiplicative group of the field k, then we can define the group homomorphism f : GL(n,k) \longrightarrow k^{\times} by f(A)=\det A. Obviously f is onto and \ker f = SL(n,k). So SL(n,k) is a normal subgroup of GL(n,k) and GL(n,k)/SL(n,k) \cong k^{\times}.

Remark 2. By the first part of Problem 1, every elementary matrix is in SL(n,k). Thus any product of elementary matrices is also in SL(n,k). On the other hand, by the above problem, every element of SL(n,k) is a product of elementary matrices. So we get the following important result:

SL(n,k), as a group, is generated by all elementary matrices.


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