{%hackmd 5xqeIJ7VRCGBfLtfMi0_IQ %}

# 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. $\det(A) = \det(A\trans)$.

## 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,

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

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

$$
AB = F_1\cdots F_aG_1\cdots G_b,
$$
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 $\bx$ such that $M\bx = \bzero$. (Right kernel contains a nonzero vector.)
3. There is a nonzero vector $\by$ such that $\by\trans M = \bzero\trans$. (Left kernel contains a nonzero vector.)

If $A$ is singular, then there is a vector $\by$ such that $\by\trans A = \bzero\trans$. Thus, $AB$ is a square matrix such that $\by\trans AB = \bzero\trans$ with $\by$ nonzero, so $AB$ is singular. Similarly, if $B$ is singular, then there is a vector $\bx$ such that $B\bx = \bzero$. Thus, $AB$ is a square matrix such that $AB\bx = \bzero$ with $\bx$ 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 $E = E\trans$ and $\det(E) = \det(E\trans) = -1$. If $E$ is an elementary matrix corresponding to rescaling by $k$, then $E = E\trans$ and $\det(E) = \det(E\trans) = k$. Finally, if $E$ is the elementary matrix corresponding to a row combination $\rho_i:+k\rho_j$, then $E\trans$ is the elementary matrix corresponding to the row combination $\rho_j:+k\rho_i$. In this case, $\det(E) = \det(E\trans) = 1$. In summary, $\det(E) = \det(E\trans)$ if $E$ is an elementary matrix.

Now it is easy to see that

$$
A\trans = F_n\trans \cdots F_1\trans,
$$

which is again a product of elementary matrices. Therefore,

$$
\det(A\trans) = \det(F_n\trans) \cdots \det(F_1\trans) = \det(F_n) \cdots \det(F_1) = \det(A).
$$

**Case 2**: $A$ is singular.

If $A$ is singular, then $A\trans$ is also singular, so $\det(A) = 0 = \det(A\trans)$.

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