$A^{T}A$ positive definite then A is invertible? Say if $A$ is an $n \times  n$ matrix, why is it that if $A^{T}A$ is positive definite, the matrix $A$ is then invertible? All I know is $A^{T}A$ gives a symmetric matrix but what does $A^{T}A$ is positive definite tell or imply or hint about the matrix $A$ itself that leads to the fact that it will be invertible?
 A: Hint: Show that $A$ is injective as a linear transformation on $\mathbb R^n$.  This implies invertibility by the rank-nullity theorem.
A: $A^TA$ is always non-negative and I think you don't have any problem with it. (otherwise see that $x^TA^TAx$ is nothing but $\|Ax\|^2_2 \geq 0$) 
Suppose A is square and not invertible hence rank deficient. Due to the homework tag, I will just add that when two matrices multiplied they cannot form a matrix that has a higher rank than the individual matrices, for example if you multiply a vector $x x^T$ the resulting matrix cannot be of rank 2.
A: Actually $\ker(A^T A) = ker(A)$ so one is invertible if and only if the other is invertible. Proof:
If $v \in \ker(A)$ then $Av=0$ so $A^T A v = 0$, hence $v \in \ker(A^T A)$.
On the other hand, if $v \in \ker (A^T A)$ then $\left\langle w,A^{T}Av\right\rangle =0$ for all $w \in \mathbb{R}^n$, so $\left\langle Aw,Av\right\rangle =0$ for all $w$, and in particular for $w=v$ you get $Av=0$.
A positive definite matrix is invertible (for instance, because it has positive eigenvalues) so you're done.
