# Least Squares Solution of Minimal Norm when $A^{*}b = 0$

Suppose, given a matrix $$\textbf{A} \in \mathbb{C}^{m \times n}$$ and a vector $$\textbf{b} \in \mathbb{C}^{n}$$, I want to find the minimal norm solution of

$$\min_{\textbf{x}}\|\textbf{A}\textbf{x} - \textbf{b} \|_{2}$$

with the condition that $$\textbf{A}^{*}\textbf{b} = \textbf{0}_{n}$$.

If $$rank(\textbf{A}) =n$$, then the solution is

$$\textbf{y} = (\textbf{A}^{*}\textbf{A})^{-1}\textbf{A}^{*}\textbf{b}.$$

This would give $$\textbf{y} = \textbf{0}$$ with the stated condition. However, for $$\textbf{A}$$ of arbitrary rank, the the minimal norm solution for this problem is

$$\textbf{y} = \textbf{A}^{\dagger}\textbf{b}$$

Where $$\textbf{A}^{\dagger}$$ is the pseudoinverse.

Can we say anything about $$\textbf{y}$$ with stated condition in the general case, or is it arbitrary

It is known that a vector $$\hat {\mathbf x}$$ is a least squares solution of the system $$A {\mathbf x}= {\mathbf b}$$ iff $$A^*A\hat {\mathbf x}=A^* {\mathbf b}.$$ Thus, if the condition $$A^* {\mathbf b}={\mathbf 0}$$ is satisfied, then $$\hat {\mathbf x}= {\mathbf 0}$$ is a least squares solution. Obviously, it has the minimum possible norm (zero).