# Solve equations by solving convex optimization

I have the following equations to solve simultaneously ($$y$$ is the vector to be solved) $$\begin{cases} (A^\top A+\lambda D)y=A^\top b \\ y_1^2=y_2^2+y_3^2+y_4^2 \\ y_1 \geq 0, \end{cases}$$ where $$(A^\top A+\lambda D) \in \mathbb{R}^{4 \times 4}$$ is singular whose rank may be $$1$$, $$2$$ or $$3$$, and $$y=[y_1~y_2~y_3~y_4]^\top$$ is to be solved.

Suppose that the equations alwayse have a solution. Since $$(A^\top A+\lambda D)$$ is singular, there may be infinite solutions.

I wonder if I can convert the problem into a convex optimization, and by solving the corresponding convex optimiation we can obtain the unique solution. Or anyone can offer an efficient method to obtain a unique solution.

You have to use the bordered system to solve for $$y$$. Since $$A^TA+\lambda D$$ is singular, $$0$$ is an eigenvalue; namely, there a nonzero eigenvector $$q$$ such that $$(A^TA+\lambda D)q=0.$$ Similarly $$(A^TA+\lambda D)^T=A^TA+\lambda D^T$$ has an eigenvalue $$0$$; amely, there a nonzero eigenvector $$p$$ such that $$(A^TA+\lambda D^T)p=0$$ or $$p^T(A^TA+\lambda D)=0.$$ So the solvablity of the singularity is $$p^TA^Tb=0$$ under which, the bordered system $$\bigg[\begin{matrix}A^TA+\lambda D&q\\p^T&0\end{matrix}\bigg]\binom{w}{u}=\binom{A^tb}{0}$$ has a unique solution.
• However, there are two other constraints, one is $y_1^2=y_2^2+y_3^2+y_4^2$, another is $y_1 \geq 0$. How to gaurantee the solution satisfies the two constraints? Jun 30, 2020 at 1:57