TITLE: $A^{T}A$ positive definite then A is invertible?
QUESTION [6 upvotes]: 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?

REPLY [4 votes]: $A^TA$ 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.