 Solution

Take the usual symmetric meaning of positive definite $Q$. To include all data allowed by the source, first check feasibility. PutFull row rank makes $B$ positive definite. Every $x$ satisfying $Ax=b$ is uniquely $x_0+z$ with $Az=0$, and $z^{T}Q{x}_0=0$. ThusIf ${\rho }^2>1$, there is no feasible point. If ${\rho }^2=1$, the only feasible point is $x_0$, and $x_0$ is optimal for every $c$. If ${\rho }^2<1$, the Slater condition holds, and the following KKT conditions are necessary and sufficient:Necessity follows from the constraint qualification proved in (ii). Sufficiency follows from convexity: the stationary optimization Lagrangian $c^{T}x+y^T(b-Ax)+\gamma (x^{T}Qx-1)$ is minimized at $x$, while its feasible values never exceed the objective, and complementary slackness makes its value at $x$ equal the objective.

The singleton case cannot be silently covered by ordinary KKT necessity. For example, let $Q=I$, $A=(10)$, $b=1$, $c=(0,1{)}^T$. The sole feasible point $(1,0)$ is optimal, but stationarity would require $(0,1)$ to lie in the span of $(1,0)$, which is impossible. The feasibility split above supplies genuinely necessary and sufficient conditions without assuming an unprinted regularity hypothesis.