## Elementary matrices (2)

Posted: January 29, 2011 in Elementary Algebra; Problems & Solutions, Linear Algebra
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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.