$() = | − |^2$ Compute the Hessian of f and show it is positive definite.

Question

$() = \| − \|^2$ , where $$ is an $ \times $ matrix with zero null space, $$ is an $$ dimensional vector, and the solution $x$ is an $$ dimensional vector.

$f'(x) = 2A|Ax-b|$, $f''(x)=2A^2$, the square is positive definite. Is this right? How to compute the derivative for a matrix form?