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?
Best Answer
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.