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() = | − |^2 , where is an × 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?

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    According to your definition, $x$ must be a $n-$vector, so $f'$ does not make sense, since $f$ has $n$ variables. Try writing $x=(x_1,\ldots, x_n)$ to get a clear image of what is going on. – 2017-02-13
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    You presumably mean $\| Ax -b \|_{2}$ rather than $| Ax-b |$. You can express $f(x)$ as $f(x)=(Ax-b)^{T}(Ax-b)=x^{T}(A^{T}A)x-2b^{T}A^{T}x+b^{T}b$. From there you can find the gradient and Hessian of f. The gradient is $\nabla f(x)=2A^{T}Ax$ and the Hessian is $\nabla^{2}f(x)=2A^{T}A$. – 2017-02-13

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