and it follows that ∑ q i $_{=1}$α$_{i}$ − ¯ a ≤ 0 . If ∑ q i $_{=1}$α$_{i}$ − ¯ a = 0 then ¯ G$_{l}$ ( y ) is full rank and so is ¯ G ( y ) . If ¯ G ( y ) is full rank then there is a square submatrix ¯ G$_{l}$ ( y ) of full rank, for the corresponding submatrix in SG ( y ) and ∑ q i $_{=1}$α$_{i}$ − ¯ a ≥ 0. The equality follows. Proof of Lemma 4.2. Proof of Lemma 4.2. By the property (11) some a for which (13) holds exists and by the Lemma 4.1 either CLDR holds or a is such that q ∑ i =1 α$_{i}$ = ¯ α. Then by the condition on a ′ if CLDR does not hold then for some i we have α ′ i > α$_{i}$, or α ′ i < α$_{i}$. Then for any { i, j } matrix entry in the matrix SG ( y/λ ) , given by S ′ i $_{·}$G$_{·}$$_{j}$ ( y/λ ) (where for a matrix A, A$_{i}$$_{·}$ denotes the i th row and A$_{·}$$_{j}$ - the j th column) In the first case α ′ i > α$_{i}$ , this matrix entry diverges to infinity. In the second case α ′ i < α$_{i}$ and as λ → ∞ this matrix entry converges to zero for every j, thus the limit matrix lim$_{λ}$$_{→}$$_{+}$$_{∞}$ [ diag ( λ α ′ i ) SG ( y/λ ) ] has deficient rank. Proof of Lemma 4.3. Start with a q$_{v}$ × p matrix of polynomials G v ( y ) . For each matrix element, { G v ( y ) }$_{ij}$ , which is a polynomial, define the low- est order of homogeneous polynomial, ¯ k$_{{}$$_{G}$ $^{v}$( y ) $_{}$_{ij}$}$. Then select ¯ k$_{v}$ = min i,j { ¯ k$_{{}$ G $^{v}$( y ) } $_{ij}$} . 40 λ α ′ i S ′ i $_{·}$G$_{·}$$_{j}$ ( y/λ ) = λ α ′ $_{i}$− $^{α$_{i}$}$λ$^{α$_{i}$}$S ′ i $_{·}$G$_{·}$$_{j}$ ( y/λ ) lim λ → + $_{$_{∞}$λ}$∑ q i $_{=1}$α$_{i}$ det( ˜ G$_{l}$ ( y/λ )) > 0