Least squares problem with a quadratic constraint

Question

I have to minimize $e_1^2+e_2^2+e_3^2$ subject to $\mathbf{e}^\top \mathbf{A} \mathbf{e} + \mathbf{e}^\top \mathbf{b} +c =0$

with $\mathbf{e} = [e_1, e_2, e_3]^\top$

I know that matrix $\mathbf{A}$ is positive semidefinite.

Answer 1

Suppose you have a generic quadratic program $$\min_{\mathbf{x}}\, \frac{1}{2}\mathbf{x}^TA\mathbf{x} + \mathbf{b}^T\mathbf{x} \quad \mathrm{s.t.}\quad \mathbf{x}^TC\mathbf{x} + \mathbf{d}^T\mathbf{x}+e = 0,\tag{1}$$ with $A$ positive-definite and $C$ positive-semidefinite. First, you can complete the square in the objective function, and drop the constant term: $$\min_{\mathbf{x}}\, \frac{1}{2}(\mathbf{x} + A^{-1}\mathbf{b})^TA(\mathbf{x} + A^{-1}\mathbf{b}) \quad \mathrm{s.t.}\quad \mathbf{x}^TC\mathbf{x} + \mathbf{d}^T\mathbf{x}+e = 0.$$ Now make the substitution $\mathbf{y} = A^{1/2}(\mathbf{x}+A^{-1}\mathbf{b})$ (note here we implicitly use the fact that $A$ is positive-definite). Solving for $\mathbf{x}$ gives $$\mathbf{x} = A^{-1/2}\mathbf{y} - A^{-1}\mathbf{b},$$ and \begin{align*} \mathbf{x}^TC\mathbf{x} + \mathbf{d}^T\mathbf{x} + e &= \mathbf{y}^T\left[A^{-1/2}CA^{-1/2}\right]\mathbf{y} + \left[-2\mathbf{b}^TA^{-1}CA^{-1/2} + \mathbf{d}^TA^{-1/2}\right]\mathbf{y} + \mathbf{b}^TA^{-1}CA^{-1}+e\\ &= \mathbf{y}^T\tilde{C}\mathbf{y} + \tilde{\mathbf{d}}^T\mathbf{y} + \tilde{e}, \end{align*} where $\tilde{C}$ is clearly still positive-semidefinite. Therefore any problem of the form (1) is equivalent to $$\min_{\mathbf{y}} \frac{1}{2}\|\mathbf{y}\|^2\quad\mathrm{s.t.}\quad \mathbf{y}^T\tilde{C}\mathbf{y} + \tilde{\mathbf{d}}^T\mathbf{y} + \tilde{e} = 0,$$ which is exactly your problem. So your formulation is no easier than solving an arbitrary QCQP with one equality constraint, which is not convex and does not have a closed-form solution. You can try to solve it numerically using packages such as MOSEK.

In order to have a closed-form solution you need more restrictions on the problem, for instance, the linear term in the constraint vanishing.

EDIT: Initially I missed that the quadratic constraint is an equality rather inequality, so the problem is not even convex.