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3.3. Invertible matrices as a product of elementary matrices

In this subsection, we will prove a fundamental result: Any invertible matrix is the product of elementary matrices (Theorem 3.3.4). For an introduction to elementary matrices, see Section 2. First we will look more closely at how the elementary matrices multiply with each other. The proof of Lemma 3.3.1 is an exercise in matrix multiplication, and was given as Exercise 2.4.6.

Lemma 3.3.1.

Let $n\geq 2$ be an integer, and consider square matrices of size $n$ . For any non-zero scalars $\lambda,\mu\in{\mathbb{R}}$ and any integers $i$ and $j$ with $1\leq i,j\leq n$ and $i\neq j$ , we have:

(i)

$E_{ij}^{2}=I_{n}$ .
(ii)

$E_{i}(\lambda)E_{j}(\mu)=E_{j}(\mu)E_{i}(\lambda)$ and $E_{i}(\lambda)E_{i}(\mu)=E_{i}(\lambda\mu)$ .
In particular, $E_{i}(\lambda)E_{i}({\lambda}^{-1})=I_{n}$ .
(iii)

$E_{ij}(\lambda)E_{ij}(\mu)=E_{ij}(\mu)E_{ij}(\lambda)=E_{ij}(\lambda+\mu)$ . In particular, $E_{ij}(\lambda)E_{ij}(-\lambda)=I_{n}.$

In other words, every elementary matrix is invertible, and the inverse of each elementary matrix is itself an elementary matrix.

Example 3.3.2.

It is easiest to view Lemma 3.3.1 for $2\times 2$ matrices. For example:

$\begin{pmatrix}0&1\\ 1&0\end{pmatrix}^{2}=I_{2},\quad\quad\begin{pmatrix}1&\lambda\\ 0&1\end{pmatrix}\begin{pmatrix}1&\mu\\ 0&1\end{pmatrix}=\begin{pmatrix}1&\lambda+\mu\\ 0&1\end{pmatrix}.$

Lemma 3.3.3.

A square matrix is invertible if and only if its reduced echelon form is the identity matrix.

Proof.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ and write $A_{r}$ for its reduced echelon form. Thus, $A_{r}$ is a square upper-triangular matrix, with leading coefficients (a.k.a. pivots) all equal to $1$ . By inspection of the possibilities for $A_{r}$ , we distinguish two cases:

(i)

$A_{r}$ is the identity matrix, or
(ii)

$A_{r}$ has a zero row.

By the algorithm in this section, in the case (i), we have that $A$ is invertible. In the case (ii), $A_{r}$ has its last row filled with zeros. Thus, $A_{r}B$ also has its last row filled with zeros, for any matrix $B\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ . In particular, there is no matrix $B$ such that $A_{r}B=I_{n}$ which shows that $A_{r}$ is not invertible. Since $A_{r}=L_{k}\cdots L_{1}A$ , which is a product of elementary matrices times $A$ , by Theorem 3.1.5 we see that $A$ cannot be invertible either. ∎

Theorem 3.3.4.

Every invertible matrix can be factorised into a product of elementary matrices.

Proof.

Let $A\in\operatorname{M}_{{n}}({{\mathbb{R}}})$ be an invertible matrix. Let $A_{r}$ be the reduced echelon form of $A$ , and say that $L_{1},\dots,L_{k}$ are elementary matrices used to put $A$ into reduced echelon form. That is,

A_{r}=L_{k}\dots L_{1}A.

By Lemma 3.3.3, $A_{r}$ must be equal to $I_{n}$ . So, by Theorem 3.1.5, (notice the order reversal):

A={(L_{k}\dots L_{1})}^{-1}={L}^{-1}_{1}\dots{L}^{-1}_{k}.

By Lemma 3.3.1, every ${L}^{-1}_{i}$ is itself an elementary matrix, so we have proved the result.

∎