How to prove that $ A^TA$ is singular for $2\times 3$ matrix $A$

Question

I was trying to find the determinant for $A^TA$ where $$ A = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ \end{array} \right) $$ I tried out with some numbers in place of $a, b, c, \dots$ and saw that the determinant is always $0$. Is there any way to prove this?

One thing I thought of, was writing a matrix $B$

$$ B = \left( \begin{array}{ccc} a & b & c \\ d & e & f \\ 0 & 0 & 0 \\ \end{array} \right) $$ $B^TB$ is always the same as $A^TA$ because the $0$ elements don't contribute to anything, whereas both $B$ and $B^T$ are clearly singular, but I don't have a rigorous way to show that $B^TB = A^TA$ other than multiplying out the matrices which is the very thing I'm trying to avoid.

So, how do I show that the product of a matrix and it's transpose, when they are not square, is singular?

Answer 1

Note that $A$ (dimension $2\times 3$) has rank at most 2 and so cannot have full column rank. That is, there is $v$ ($3\times 1$) not identically $0$ such that $$ Av=0. $$ But then $(A'A)v=A'(Av)=0$, implying that $A'A$ cannot be invertible. Alternatively, note that $(A'A)v=0$ and $v\neq 0$ expose the fact that $A'A$ (a $3\times 3$ matrix) does not have full (column) rank.

Answer 2

In general $\text{rank}(AB) \le \max(\text{rank}(A), \text{rank}(B))$. In particular, if $A$ and $B$ are $m \times n$ and $n \times m$ with $n < m$, $AB$ must be singular. However, it's incorrect to say "the product of a matrix and its transpose, when they are not square, is singular": if $A$ is $m \times n$ with $n < m$ , $A A^T$ (which is $m \times m$) is singular, but $A^T A$ (which is $n \times n$) may well be nonsingular.