Give an explicit solution of the following QCLP $$\begin{split} & c^\top x \to \min\limits_{x \in \mathbb{R}^n }\\ \text{s.t. } & x^\top A x \leq 1\end{split}$$ where $A \in \mathbb{S}^n_{++}$ and $c \neq 0$. What is the solution if the problem is not convex ($A \notin \mathbb{S}^n_{++}$)?
Source: Robert M. Freund, Quadratic Functions, Optimization, and Quadratic Forms [PDF], February 2004.
My idea is to make a Lagrangian function
$$L = c^Tx + \lambda x^TAx - \lambda$$
find its extra value
$$c + 2\lambda Ax^*=0,\qquad x^* = -\frac{1}{2}(\lambda A)^+c$$
and, finally, make a dual function
$$ g(\lambda) = -\frac{1}{4}c^T(\lambda A)^+-\lambda, \text{ if } 0 \preceq \lambda A $$
And maybe if $A \notin \mathbb{S}^n_{++}$ the optimal solution is $-\infty.$
However, I found the answer similar to this question here on page 15, number 7, and it does not coincide with mine. Where am I wrong?
The objective is linear, therefore the minimum is on the boundary and the inequality can be exchanged to equality. Let us first consider that the matrix $\mathbf{A}$ is positive definite. \begin{align} &\min_{\mathbf{x}}~ \mathbf{c}^{\text{T}}\mathbf{x}\\ &\text{s.t.}~~\mathbf{x}^{\text{T}}\mathbf{Ax}=1 \end{align} Then \begin{align} \mathcal{L}&=\mathbf{c}^{\text{T}}\mathbf{x}+\lambda(\mathbf{x}^{\text{T}}\mathbf{Ax}-1),\\ \frac{\partial \mathcal{L}}{\partial \mathbf{x}}&=\mathbf{c}+2\lambda\mathbf{Ax}=\mathbf{0} \implies \mathbf{x}=-\frac{1}{2\lambda}\mathbf{A}^{-1}\mathbf{c}\\ \frac{\partial \mathcal{L}}{\partial \lambda}&=\mathbf{x}^{\text{T}}\mathbf{Ax}-1=0 \end{align} By inserting the solution the second Eq. we get $2\lambda=\sqrt{\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}}$ and finally $\mathbf{x}=\frac{-\mathbf{A}^{-1}\mathbf{c}}{\sqrt{\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}}}$, where $\mathbf{A}^{-1}$ is also positive definite. The question now is how the problem changes if we use general matrix $\mathbf{A}$. One problem is that $\mathbf{c}^{\text{T}}\mathbf{A}^{-1}\mathbf{c}$ is not guaranteed to be nonnegative, therefore $\lambda$ cannot be determined in this way (if we use pseudoinverse). We can find an example, in which the minimum is $-\infty$, because some components of $\mathbf{x}$ can be out of control of $\mathbf{x}^{\text{T}}\mathbf{Ax}=1$, for example by $n=2$ and $\mathbf{A}=\left[\begin{array}{cc}1 & 0\\0 & 0\end{array}\right]$.