## 3 Properties of the Set of Normalized Immobile Indices
Hence, according to Proposition 2 in [20], the following inequalities hold true: \(\mathcal{A}(x)\tau(i)\geq 0\ \forall i\in I\), for all \(x\in X\). However, this contradicts the assumption that \(\Delta X=\emptyset\). The proposition is proved. **Proposition 4**. : _Given a finite set \(\{\tau(i),i\in I\}\subset T_{im}(\mathcal{Z})\), suppose that there are numbers and vectors \(\gamma_{i}>0,\ \tau(i)\in T,\ i\in\Delta I,\ \ \lambda(i)\in\mathbb{R}_{+}^{p},\ i\in I,\) such that_  
\[ \begin{array}{l}\sum_{i\in\Delta I}\gamma_{i}(\tau(i))^{\top}A_{j}\tau(i)+ \sum_{i\in I}(\lambda(i))^{\top}A_{j}\tau(i)=0\ \forall j=1,...,n;\\ \sum_{i\in\Delta I}\gamma_{i}(\tau(i))^{\top}A_{0}\tau(i)+\sum_{i\in I}( \lambda(i))^{\top}A_{0}\tau(i)=:\eta<0.\end{array}  \tag{20}\]  
_Then, \(X=\emptyset\)._  
Proof. : Suppose the contrary: relations (20) hold true and there exists \(x^{*}\in X.\) Then from Proposition 2 it follows that \(\tau(i)\in T_{im}(X)\ \forall i\in I,\) and hence  
\[\mathcal{A}(x^{*})\tau(i)\geq 0\ \forall i\in I,\ t^{\top}\mathcal{A}(x^{*})t \geq 0\ \forall t\in T. \tag{21}\]  
Denote  
\[\beta_{*}:=\sum_{i\in\Delta I}\gamma_{i}(\tau(i))^{\top}\mathcal{A}(x^{*}) \tau(i)+\sum_{i\in I}(\lambda(i))^{\top}\mathcal{A}(x^{*})\tau(i).\]  
It follows from relations (21) that \(\beta_{*}\geq 0.\) On the other hand, from (20) we can conclude that \(\beta_{*}=\eta<0.\) The resulting contradiction implies that \(X=\emptyset.\) The proposition is proved. **Proposition 5**. : _Given a set \(W:=\{\tau(i),i\in I\}\subset T_{im}(\mathcal{Z})\), the corresponding number \(\sigma(W)\), and the sets \(\Omega(W)\), \(\mathcal{X}(W)\) defined in (12) -(14), suppose that the set \(\Delta X\) defined in (19) is not empty and for some vector \((\tilde{y},\tilde{y}_{0})\in\mathbb{R}^{n+1}\), it holds_  
\[\tilde{y}_{0}\geq 0,\ \mathcal{B}(\tilde{y},\tilde{y}_{0})\tau(i)\geq 0,\ i\in I ;\ t^{\top}\mathcal{B}(\tilde{y},\tilde{y}_{0})t\geq\mu>0\ \forall t\in\Omega(W). \tag{22}\]  
_Then \(\mathcal{X}(W)=X\
eq\emptyset\) and there exists \(\tilde{x}\in X\) such that_  
\[t^{\top}\mathcal{A}(\tilde{x})t>0\ \forall t\in\Omega(W). \tag{23}\]  
Proof. : First, let us show that \(X\
eq\emptyset\) and there exists \(\tilde{x}\in X\) such that inequalities (23) hold true. From (22) and Theorem 1, it follows that \((\tilde{y},\tilde{y}_{0})\in\mathcal{Z}.\) Consider the following cases: (a) \(\tilde{y}_{0}>0\) and (b) \(\tilde{y}_{0}=0.\)  
In the case (a), set \(\tilde{x}:=\tilde{y}/\tilde{y}_{0}.\) It is easy to see that, by construction, \(\tilde{x}\in X\) and inequalities (23) hold true. Now consider the case (b). Here relations (22) take the form  
\[\Big{(}\sum_{j=1}^{n}A_{j}\tilde{y}_{j}\Big{)}\tau(i)\geq 0\ \forall i\in I;\ t^{\top} \Big{(}\sum_{j=1}^{n}A_{j}\tilde{y}_{j}\Big{)}t\geq\mu>0\ \forall t\in\Omega(W). \tag{24}\]  
For a fixed \(x^{*}\in\Delta X,\) set:  
\[\alpha_{*}:=\min_{t\in\Omega(W)}t^{\top}\mathcal{A}(x^{*})t\ \ \ \text{and}\ \ \theta:=\left\{ \begin{array}{cl}1,&\text{if}\ \alpha_{*}\geq 0,\\ -2\alpha_{*}/\mu>0,&\text{if}\ \alpha_{*}<0,\end{array} \right.\]  
and consider vector \(\tilde{x}:=x^{*}+\theta\tilde{y}.\) Taking into account relations (24) and the inclusion \(x^{*}\in\Delta X,\) we obtain  
\[\mathcal{A}(\tilde{x})\tau(i)\geq 0,\ i\in I,\ t^{\top}\mathcal{A}(\tilde{x})t \geq\alpha_{*}+\theta\mu=\left\{ \begin{array}{cl}\alpha_{*}+\mu>0,&\text{ if}\ \alpha_{*}\geq 0,\\ -\alpha_{*}>0,&\text{if}\ \alpha_{*}<0,\end{array} \right.\ \forall t\in\Omega(W).\]  
From the relations above and from Theorem 1 in [18], it follows that \(\tilde{x}\in X\) and inequalities (23) hold true. Since \(X\
eq\varnothing\), from Proposition 2 we deduce that \(T_{im}(X)=T_{im}(\mathcal{Z})\). This means that \(W\subset T_{im}(X)\), and hence according to Theorem 1, we conclude that \(X=\mathcal{X}(W)\). The proposition is proved.