Suppose that $A$ is a real $m \times n$ matrix, $m <n$, of full rank (this means that $A$ has full row rank).
Let $\phi(x)=\frac{1}{2}||Ax-b||_2^2$.
Show that Newton's method solves $\nabla \phi(x)=0$ in one step.
I am confused because I am not sure how to even take the gradient of $\phi(x)$. If someone could demonstrate, I would really appreciate it. Thank you.
Noting that $\phi(x) = {1 \over 2} \langle Ax-b, Ax-b\rangle = {1 \over 2} (\langle x, A^T A x\rangle -2\langle x, A^T b\rangle + \|b\|^2 )$, expand $\phi(x+h)-\phi(x)$.
This will have a linear term and a term bounded by $\|h\|^2$. This shows that $\phi$ is differentiable and the linear part will be the derivative.