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For any $m$ by $m$ matrix, $M$, I define the following corresponding set,

$$ \mathcal{X}_M = \left\{x \in \mathbb{R}^m \,|\, x^T M\, x > 0\right\}. $$

Now for given are matrices $A$ and $B$ both in $\mathbb{R}^{2\times m}$ I define the following symmetric matrices,

$$ S_A = A^T\, \Gamma\, A, $$

$$ S_B = B^T\, \Gamma\, B, $$

with

$$ \Gamma = \begin{bmatrix} 1 & 0 \\ 0 & -1 \end{bmatrix}. $$

I am interested in finding the smallest number of (symmetric) matrices $M_n$ in $\mathbb{R}^{m\times m}$, with $n = 1,\, 2,\, \dots,\, N$, such that,

$$ \mathcal{X}_{M_1} \cup \mathcal{X}_{M_2} \cup \cdots \cup \mathcal{X}_{M_N} = \mathcal{X}_{S_A} \cap \mathcal{X}_{S_B}. $$

This problem originated from some other inequality,

$$ x^T \left(\Phi^T P\, \Phi - P\right) x < 0 \quad \forall\, x \in \mathcal{X}_A \cap \mathcal{X}_B, $$

for which I would like to find a $P$ that satisfies it. I hoped that I could rewrite it as multiple linear matrix inequalities using,

$$ \Phi^T P\, \Phi - P + M_n \prec 0 \quad \forall\, n \in \{1,\, 2,\, \dots,\, N\}. $$

In my case I actually have $m = 4$, but I just wondered if there might be some more general approach.

I am not sure how I could construct such an intersection of two sets, but here are some insight I did get when looking at the structure of the sets. Namely the matrices $S_A$ and $S_B$ should both have one positive, $\lambda_+$, and one negative, $\lambda_-$, eigenvalue and $m-2$ eigenvalues of zero. All corresponding eigenvectors are orthogonal since both matrices are symmetric. Denoting the eigenvectors corresponding to the positive, negative and the ith zero eigenvalues with $\vec{v}_+$, $\vec{v}_-$ and $\vec{v}_i$ respectively, then any $x$ can be written as a linear combination of these (normalized) eigenvectors,

$$ x = a_+\, \vec{v}_+ + a_-\, \vec{v}_- + \sum_{n=1}^{m-2} a_n\, \vec{v}_n. $$

Using all this then the quadratic inequality can therefore also be written as,

$$ \left(a_+\, \vec{v}_+ + a_-\, \vec{v}_- + \sum_{n=1}^{m-2} a_n\, \vec{v}_n\right)^T \left(a_+\, \lambda_+\, \vec{v}_+ + a_-\, \lambda_-\, \vec{v}_-\right) = a_+^2\, \lambda_+ + a_-^2\, \lambda_- > 0. $$

It can also be noted that for any $x \in \mathcal{X}_{S_A} \cap \mathcal{X}_{S_B}$ and any $\alpha \in \mathbb{R}$, then it will also be true that $\alpha\, x \in \mathcal{X}_{S_A} \cap \mathcal{X}_{S_B}$.

I am not sure if any of this helps me getting any closer to finding the matrices $M_n$.

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    You are aware that this is not always possible? For example take $m=2$ and $$A=1, B=\begin{pmatrix}0&2\\1&0\end{pmatrix}.$$ See [this](http://wolframalpha.com/input/?i=4y%5E2-x%5E2%3E0%2C+x%5E2-y%5E2%3E0&x=0&y=0) picture why. – 2017-01-07
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    @WimC I assume that you meant $A=\begin{pmatrix}1&0\\0&1\end{pmatrix}$. I also edited my question, in which I also added that I am actually interested in $m=4$. Would it that case also exist some $A$ and $B$ for which there is no solution for $M$? – 2017-01-07
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    Yes, in higher dimensions the situation gets worse. The cases where there *is* such an $M$ require a special relation between $A$ and $B$. Note that the matrix expression with $A$ (or $B$) in it is the product of two linear functionals. Having a mental image of that situation helps. – 2017-01-07
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    @WimC I have updated my question again (see last section). Namely in that case then your $m=2$ example would have a solution, but would there also be one for higher $m$? – 2017-01-08

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