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Let $A \in \mathbb{R}^{n\times n}$ be a positive definite matrix, $x \in \mathbb{R}^n$ and $c \in \mathbb{R} \setminus \{0\}$. Determine the maximum $$ \max_{y \in \mathbb{R}^n} \left\{ c^T y : y \in \mathcal{E} (A,x) \right\} $$ where $\mathcal{E} (A,x)$ is an ellipsoid defined by $A$ and $x$.

How can I determine this maximum? Can anyone give me a hint?

Rodrigo de Azevedo
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Thesinus
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1 Answers1

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The goal is to solve the following maximization problem:

\begin{align*} & \max c^Ty \\ & \text{s.t. } (y-x)^TA^{-1}(y-x) \leq 1 \end{align*}

Consider the Langrangian approach

\begin{align*} & \max c^Ty+\lambda(1-(y-x)^TA^{-1}(y-x))\\ & \text{s.t. } \lambda \geq 0 \end{align*}

Differentiating w.r.t. $y$,

$$c+2\lambda \left[ A^{-1}x-A^{-1}y\right]=0$$

If $\lambda=0$, the equation above will give us a contradiction, hence $\lambda>0$.

Also, $$y=x+\frac{1}{2\lambda}Ac$$

Hence, if I can solve for $\lambda$, I have solve the whole thing.

Substitute that the equation above to the equation of the ellipsoid(Since $\lambda>0$, the optimal solution must occur at the boundary by complementary slackness condition.)

$$\left(\frac{1}{2\lambda}Ac\right)^TA^{-1}\left(\frac{1}{2\lambda}Ac\right)=1$$

and hence,

$$\lambda^2=\frac{1}{4}c^TAc$$

Hence $$y=x+\frac{1}{\sqrt{c^TAc}}Ac$$

Siong Thye Goh
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  • That looks great. Is it possible to extend this result for $\text{maximize}\quad x^T x$ with the same ellipsoidal constraint? – Jacob Aug 02 '18 at 18:16
  • I think it should work. – Siong Thye Goh Aug 03 '18 at 02:06
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    Refer to https://math.stackexchange.com/a/3496110/269050 for a solution that doesn't use the Lagrangian approach. – Leandro Caniglia Jan 05 '20 at 13:58
  • @LeandroCaniglia, calling $B=A^{-1}$ and reapeating Siong answer forgetting that $B=A^{-1}$, but assuming $B$ is symmetric, this answer only requires $B$ to have left inverse, which is a quite weak assumption. – R. W. Prado Aug 15 '23 at 08:50