How to show that lipshitz constant for $||Ax - b||$, where A is a matrix and x, b are vectors, is given by $2 \lambda_{max} (A^T A)$. , where $\lambda_{max}$ the maximum eigen value.
So far, I have solved till
$$|| \nabla f(x) - \nabla f(y) ||_2\leq L ||x - y||_2$$
$2|| A^T (Ax - b) - A^T (Ay - b)||_2 = 2||A^T A (x - y)||_2 \leq 2\mathbf{||A^T A||} . ||x - y||_2$
In the last inequality , I am doubtful. Should the $||A^T A||_F$ occur or $||A^T A||_2$ should be there.
Any help is appreciated.
First of all, you need the $2$ norm of the matrix, see definition of this norm.
The last thing you need is that the norm of a matrix is equal to the maximum of all absolute values of all the eigenvalues. Since $A^T A$ is symmetric and positive definite all eigenvalues are real positive numbers. Hence the norm is equal to the maximal eigenvalue (witout absolute stripes)