If $A$ is a symmetric matrix, then is the augmented matrix $A^tA$ positive-definite?

Question

In the lecture we talked about the conjugate gradient method to deblur an image. The idea is that we need to solve the linear system Ax = b, where A is the transformation matrix resulting from the repeated application of the image kernel (that apply the blur effect), b is the vectorized blurred image, and we need to find x that represents the vectorized matrix, that is the original image before the blur effect has been applied to it.

We have the assumption that A is symmetric and positive-definite, but in the case that it is not positive-definite, we can transform it into the augmented matrix $A^tA$ that is positive-definite, and instead of solving Ax = b, we solve the augmented system $A^tAx = A^tb$.

My question is: why $A^tA$ is positive-definite? I know that it is symmetric because each matrix multiplied by its transpose is symmetric, and I know also that $A^tA$ is always positive-semidefinite, but why in this case is positive-definite? It is something regarding the symmetry of the matrix A?

Answer 1

It's not related to the symmetry of $A$.

The nullspace of $A^TA$ will always be the same as the nullspace of $A$. (Try to see if you can prove this). So $A^TA$ is positive definite if and only if $\text{Nul } A = \{0\}$

Answer 2

The matrix is guaranteed to be positive semi-definite, since $$ \langle A^tAx,x\rangle=\langle Ax,Ax\rangle\geq 0 $$ with $0$ happening for $x$ in the kernel.