$\det(AA^{T})$: order of operations?

Question

Let: $A \in \mathcal{M}_{2 \times 3}\big( \mathbb{R} \big)$

Question: does the $\det(AA^{T})$ exist or not?

So I have two different approaches to this problem.

First approach:

Find $A^T \rightarrow$ calculate $AA^{T} \rightarrow$ calculate determinant of the result.

Second approach is what interests me:

Incomplete theorem: $$\det(AB) = \det(A) \circ \det(B)$$ hence: $$\det(AA^{T}) = \det(A) \circ \det(A^{T})$$

Because $A$ is not a square matrix, $\det(A)$ cannot be calculated. Same about $A^{T}$.

So this comes down to this:

$$\det(AA^{T}) = \text{<undefined>} \circ \text{<undefined>}$$

How can I make this theorem work in my case? Why does it not work? What is the correct answer to my main question, after all?

Answer 1

You cannot make it work, since the theorem that states that $\det(AB)=\det(A)\det(B)$ is about square matrices (of the same size).

Answer 2

Let $A$ be an $m\times n$ matrix. Then note that $AA^\top$ is the product of the $m\times n$ matrix $A$ with the $n\times m$ matrix $A^\top$. This product is well-defined and produces a $m\times m$ matrix. Since every square matrix has a determinant, it follows that $\det(AA^\top)$ exists.

Unless we know more about $A$, there isn't much more we can say about $\det(AA^\top)$. As observed in another answer, the formula $\det(AA^\top)=\det(A)\det(A^\top)$ does not make sense unless $A$ is square.

Answer 3

As others have already told you, the formula $\det(AB)=\det A \cdot \det B$ only for square matrices of the same size.
I would like to add the fact that we may prove by the Cauchy-Binet formula that $$\det(AA^t)\ge0,\forall A \in \mathcal{M_{n,m}(\mathbb{C})}$$