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.

Thanks for your insightful comments.


1 Answer 1


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.

  • $\begingroup$ 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? $\endgroup$ Jun 30, 2020 at 1:57

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