Finding best phase in least-squares manner

Question

I have the following problem: $$argmin_{\vec{x},\phi}||A\vec{x}-\vec{y}e^{j\phi}||_2^2$$ Here, $x$ and $y$ are vectors and $\phi$ is a constant phase factor that applies to the all entries of $y$. I could not make any progress in this problem. If anyone can point me to a possible closed form solution that would be great.

Answer 1

It seems to me that you can't do any better than taking the classical least-squares solution $x = A^+y$. To see that, replay the solution of the least-squares problem via SVD: take an SVD $A=U\Sigma V^*$, with $\Sigma\in\mathbb{R}^{m\times n}$ diagonal, and set $z = V^*x$ and $w = U^* y$. Then, $z$ solves $$ \arg\min_{z,\phi} \|\Sigma z - w e^{j\phi}\|, $$ and this problem is a separable minimization problem in the entries of $z$ that is simple enough to solve by hand.

In particular, the objective function is the 2-norm of a vector with entries $\sigma_i z_i - w_i e^{j\phi}$ for $i=1,2,\dots, r$, $r$ being the rank of $A$, and $-w_i e^{j\phi}$ for $i>r$. The first $r$ entries become zero if one sets $$z_i = \frac{w_i}{\sigma_i e^{j\phi}}, \quad i=1,2,\dots,r \tag{*}$$ and the entries after that have constant absolute value independently of $z$ and $\phi$. So any pair $(z,\phi)$ that satisfies (*) is a minimizer.

In particular, one of the minimizers is $x=A^+y$, obtained by taking $\phi=0$ and $z_i=0$ for each $i>r$ in addition to the conditions (*). This is also a minimum-norm solution.