Check that
| $$\rho$$ | $$ \det E_{\rho}$$ |
|---|---|
| $$R_i \leftrightarrow R_j$$ | $$(-1)$$ |
| $$\lambda R_i \to R_i\quad (\lambda \neq 0)$$ | $$\lambda$$ |
| $$\lambda R_i+R_j \to R_j$$ | $$1$$ |
where $E_\rho$ is the elementary matrix correponding to the ERO $\rho$.
Proposition 1. $\det(EM) = \det(E)\det(M)$ for any matrix $M \in M_n(K)$ and any elementary matrix $E \in M_n(K)$.
Proof. Suppose $E$ is an elementary matrix corresponds to swapping two rows $R_i \leftrightarrow R_j$. Then $\det(E) = (-1)$ (see the table above) and $EM$ is the matrix obtained from $M$ by swapping its $i$-th row and its $j$-th row. Thus, by Property (1) of determinant (see the previous lecture notes),
$$ \det(EM) = (-1)\det(M) = \det(E)\det(M). $$The proof that $\det(EM) = \det(E)\det(M)$ for elementary matrices of the remaining two types are similar.
Checkpoint. Verify that $\det(EM) = \det(E)\det(M)$ for the remaining two types of elementary matrices.
Theorem 1. $M \in M_n(K)$ is invertible if and only if $\det(M) \neq 0$.
Proof. By Gauss-Jordan elimination, $$ \text{rref}(M) = E_k\cdots E_1 M \tag{1} $$ for some elementary matrices $E_1, \ldots, E_k$. Thus it follows by apply Proposition 1 successively that:
\begin{align*} \det(\text{rref}(M)) &= \det(E_k \cdots E_1M) = \det(E_k)\det(E_{k-1} \cdots E_1M)\\ & = ... = \det(E_k) \cdots \det(E_1)\det(M). \tag{2} \end{align*}We have also proved that $M$ is invertible if and only if rref($M$) = $I_n$ (Theorem 1 in Notes 7). Thus, if $M$ is invertible then $$ 1 = \det(I_n) = \det(\text{rref}(M)) \stackrel{(2)}{=} \det(E_k)\cdots \det(E_1)\det(M) $$ Therefore, $\det(M) \neq 0$.
Conversely, suppose $\det(M) \neq 0$. Since the determinant of an elementary matrix is nonzero (either -1, $\lambda \neq 0$, or $1$), it follows from Equation (2) that $\det(\text{rref}(M)) \neq 0$ as well. Therefore, rref($M$) cannot have any zero rows and hence must be $I_n$. That means $M$ is invertible.
Theorem 2. $\det(AB) = \det(A)\det(B)$ for any $A,B \in M_n(K)$.
Proof.
Case 1 $A$ is not invertible. Then $AB$ is not invertible, for otherwise there is $C$ such that $I_n = (AB)C = A(BC)$ but then $BC$ would be an inverse of $A$, contradiction. Thus, by Theorem 1 $\det(A) = 0 = \det(AB)$ and so $\det(AB) =\det(A)\det(B)$ (both sides being 0).
Case 2 $A$ is invertible. So, $A = E_k \ldots E_1$ is a product of elementary matrices. By applying Proposition (1) successively, we see that
$$ \det(AB) = \det(E_k\cdots E_1 B) = \det(E_k)\cdots \det(E_1)\det(B) = \det(E_k\cdots E_1)\det(B) = \det(A)\det(B). $$Proposition 2. $\det(A^T) = \det(A)$.
Proof. $A$ is invertible if and only if $A^T$ is. In the case when $A$ is not invertible, then both $A$ and $A^T$ have determinant 0.
Since $\det(E^T) = \det(E)$ for any elementary matrix $E$ and any invertible matrix $A$ is a product of some elementary matrices $E_k \cdots E_1$, $$ \begin{align*} \det(A^T) &= \det((E_k\cdots E_1)^T) = \det(E_1^T\cdots E_k^T)\\ &= \det(E_1)\cdots \det(E_k) = \det(E_k)\cdots \det(E_1) \\ & = \det(E_k\cdots E_1) =\det(A). \end{align*} $$
These few results demonstrate an important idea: to establish a property for the elements of a group (in this case the group of invertible matricecs GL$_n(K)$), it is often sufficient and easier to establish the same property for its set of generators (in this case the set of elementary matrices).