$$

Matrix multiplication and transpose

Problem

Let

$A$ and

$B$ be

$n\times n$ matrices. Show the following.

1. $det(AB)=det(A)det(B)$.
2. $det(A)=det(A^{\mathrm{\top }})$.

Thought

These are well-known properties of the determinant. The hard part is to show it by the definition . Therefore, it is critical to think again which one implies which one and avoid circular arguments. In our case, we defined the determinant algorithmically by

$det(I)=1$ and three rules. Equivalently, we knew there are two cases:

1. If
   $A$ is invertible, then
   $A$ can be written as a sequence of elementary matrices
   $F_1\cdots F_s$, whose determinants are defined by the three rules, and
   $det(A)=det(F_1)\cdots det(F_s)$.
2. If
   $A$ is singular, then
   $det(A)=0$.

Sample answer

For Problem 1, consider two cases.

Case 1: Both

$A$ and

$B$ are invertible.

In this case, we may write

$A$ and

$B$ as products of elementary matrices. That is,

$$\begin{array}{rl}A& =F_1\cdots F_a\text{ and }\\ B& =G_1\cdots G_b.\end{array}$$

By definition,

$det(A)=det(F_1)\cdots det(F_a)$ and

$det(B)=det(G_1)\cdots det(G_b)$. Now it is easy to check that

$$AB=F_1\cdots F_a{G}_1\cdots G_b,$$ which is a product of elementary matrices. Therefore,

$$det(AB)=det(F_1)\cdots det(F_a)det(G_1)\cdots det(G_b)=det(A)det(B).$$

Case 2: One of

$A$ or

$B$ is singular.

Recall that for a square matrix

$M$, the following are equivalent.

1. $M$ is singular.
2. There is a nonzero vector
   $\mathbf{x}$ such that
   $M\mathbf{x}=\mathbf{0}$. (Right kernel contains a nonzero vector.)
3. There is a nonzero vector
   $\mathbf{y}$ such that
   ${\mathbf{y}}^{\mathrm{\top }}M={\mathbf{0}}^{\mathrm{\top }}$. (Left kernel contains a nonzero vector.)

If

$A$ is singular, then there is a vector

$\mathbf{y}$ such that

${\mathbf{y}}^{\mathrm{\top }}A={\mathbf{0}}^{\mathrm{\top }}$. Thus,

$AB$ is a square matrix such that

${\mathbf{y}}^{\mathrm{\top }}AB={\mathbf{0}}^{\mathrm{\top }}$ with

$\mathbf{y}$ nonzero, so

$AB$ is singular. Similarly, if

$B$ is singular, then there is a vector

$\mathbf{x}$ such that

$B\mathbf{x}=\mathbf{0}$. Thus,

$AB$ is a square matrix such that

$AB\mathbf{x}=\mathbf{0}$ with

$\mathbf{x}$ nonzero, so

$AB$ is singular.

Therefore, if

$A$ or

$B$ is singular, then

$AB$ is singular. Consequently,

$$det(AB)=0=det(A)det(B).$$

For Problem 2, consider two cases.

Case 1:

$A$ is invertible.

If

$A$ is invertible, then

$A$ can be written as a product of elementary matrices. That is,

$A=F_1\cdots F_s$.

Now we examine the transpose of an elementary matrix. If

$E$ is an elementary matrix corresponding to swapping, then

$E=E^{\mathrm{\top }}$ and

$det(E)=det(E^{\mathrm{\top }})=-1$. If

$E$ is an elementary matrix corresponding to rescaling by

$k$, then

$E=E^{\mathrm{\top }}$ and

$det(E)=det(E^{\mathrm{\top }})=k$. Finally, if

$E$ is the elementary matrix corresponding to a row combination

${\rho }_i:+k{\rho }_j$, then

$E^{\mathrm{\top }}$ is the elementary matrix corresponding to the row combination

${\rho }_j:+k{\rho }_i$. In this case,

$det(E)=det(E^{\mathrm{\top }})=1$. In summary,

$det(E)=det(E^{\mathrm{\top }})$ if

$E$ is an elementary matrix.

Now it is easy to see that

$$A^{\mathrm{\top }}=F_n^{\mathrm{\top }}\cdots F_1^{\mathrm{\top }},$$

which is again a product of elementary matrices. Therefore,

$$det(A^{\mathrm{\top }})=det(F_n^{\mathrm{\top }})\cdots det(F_1^{\mathrm{\top }})=det(F_n)\cdots det(F_1)=det(A).$$

Case 2:

$A$ is singular.

If

$A$ is singular, then

$A^{\mathrm{\top }}$ is also singular, so

$det(A)=0=det(A^{\mathrm{\top }})$.

This note can be found at Course website > Learning resources.