5

I try to solve it by KKT conditions.

The Lagrangian is $L=x^TAy-\lambda x^Ty$.

Its KKT conditions are given by $$ \begin{align} Ay-\lambda y=&0\quad (1)\\ A^Tx-\lambda x=&0\quad (2)\\ \lambda\geq &0\quad (3)\\ x^Ty\geq &0 \quad (4)\\ \lambda x^Ty=&0\quad (5)\\ \end{align} $$ From 1 and 2 we see that $\lambda$ is an eigenvalue of $A$, $\lambda\geq 0$, and $y=cx$ where $c\in\mathbf{R}$. From 3, 4 and 5, we know if $x^Ty>0$, $\lambda=0$, $\min x^TAy=\min \lambda x^Ty=0$. Otherwise, if $x^Ty=0$, we can also deduce that $\min x^TAy=\min \lambda x^Ty=0$. So $\min x^TAy=0$. Is it correct? Thx.

Rodrigo de Azevedo
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Hao WANG
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1 Answers1

1

Counterexample: Consider $x=[1,-1]^T$, $y=[1,1]^T$, $A=\begin{bmatrix}1&0\\0&2\end{bmatrix}$
$\Rightarrow$ $x^Ty=0$ and A is PD matrix obviously. However $x^TAy=-1$ which is feasible and smaller than 0(your optimal objective function value).

Ducheng Peng
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