# Easy linear matrix inequality (LMI) problem

I'm trying to find a matrix $P$ such that: \begin{equation} x^T P x \leq -1\quad \forall x\neq 0. \end{equation}

I'm not sure if I can use Schur's complement, which states that: \begin{equation} \left[ \begin{array}{cc} Q(x) & S(X)\\ S(x)^T & R(X) \end{array} \right] \succ0 \end{equation} with $Q(x) = Q(x)^T$, $R(x) = R(x)^T$, $S(x)$ affine in $x$, is equivalent to: \begin{equation} R(x) \succ 0 \quad \mathrm{and} \quad Q(x) - S(x) R^{-1} S(x) \succ 0. \end{equation} The reason being that I do not want $P \succ 0$. (This would be nonsensical, because then $0\leq x^T P x$ and $x^T P x\leq -1\quad \forall x \neq 0$.)

Any hints?

• There is no matrix $P$ such that $$x^T P x \leq -1\qquad \forall x \neq 0_n$$Did you mean the following? $$x^T P x \leq -x^T x$$ – Rodrigo de Azevedo Jun 2 '16 at 11:56
• Ah you're right! No, I meant what I wrote... $x^T P x \leq -1$. Thanks, I'll need to rethink what I'm doing. :) – MathsStudent Jun 2 '16 at 12:08