What's wrong with this proof that all square matrices are invertible?

Question

For any square matrix A,

1. $A^tA$ is definite positive, since $x^tA^tAx=y^ty>0, \forall x\in\mathbb{R}^n≠0$ with $y=Ax$.
2. Definite positive matrices are invertible. (If a matrix B was singular, there would exist $x\in\mathbb{R}^n ≠ 0$ such that $Bx = 0$. But then $Bx = 0 \Rightarrow x^tBx = 0$, which is a contradiction. So B must be invertible.) So $det(A^t A) ≠ 0$.
3. $det(A^t A)=det(A^t) det(A)$ since $A^t$ and $A$ are square matrices of the same size. So if $det(A^t A)≠0$ then $det(A)≠0.$
4. Therefore, A is invertible.

I just can't see it...

Answer 1

It is your point 1 which is wrong. $A^t A$ is generally only positive semi-definite, i.e. it may happen that $x^tA^tAx=0,$ i.e. $Ax=0,$ for some non-zero vectors $x.$