Show invertibility when not full column rank? Given a $2\times 3$ matrix $A$, how can we show that $A^TA$ is not invertible? (where $A^T$ is $A$ transpose).
Hint: There are more columns than rows, thus the third column must be linearly dependent.
I tried to set it up using random values (cannot use numbers), so I let:
$A=\begin{bmatrix}a & b & c \\ d & e & f\end{bmatrix}$
Setting up $A^T  A$ give me a $3\times 2$ matrix times a $2\times 3$, giving me a $3\times 3$.
I tried to solve this by multiplying out using these general terms, and I get squared terms down the diagnoal in a $3\times 3$ matrix, however I am unsure to show how it is not invertible?

Hint: If we row-reduce $A$, there will be a free variable.  So, there will be a non-zero vector $x$ such that $Ax = 0$.
Hint: By the rank nullity theorem, the nullity of $A$ is at least $1$, so there is an $x$ as above.
In either case, $(A^TA)x = A^T(Ax)$.
