The currying map of each inner product is injective. Indeed, if \(b(z,u)=0\) for every \(z\), take \(z=u\); positive definiteness gives \(u=0\). Finite dimension makes this map onto the dual, which has the same dimension. Thus these forms are chosen perfect pairings, and the paired-congruence identity applies whenever \(C\) is invertible.

Assume \(Q>0\). For \(y\ne 0\), \(\iota _Yy\ne 0\) and \[b_Y(y,Cy)=b_M(\iota _Yy,Q\iota _Yy)>0.\] Hence \(C>0\). If \(Cy=0\), positivity forces \(y=0\); finite dimension then turns injectivity into surjectivity, so \(C\) is invertible. For \(x\ne 0\), the vector \((x,-C^{-1}Ex)\) is nonzero and congruence gives \[b_X(x,Sx) =b_M((x,-C^{-1}Ex),Q(x,-C^{-1}Ex))>0.\] Thus \(S>0\), and the same injectivity and finite-dimension argument makes \(S\) invertible.

Conversely, suppose \(C>0\) and \(S>0\). For any nonzero \(z=(x,y)\), the invertible map \(V(x,y)=(x,y+C^{-1}Ex)\) has \(Vz\ne 0\). At least one of its two components is nonzero. The corresponding term of \[b_M(z,Qz)=b_X(x,Sx)+b_Y(y+C^{-1}Ex,C(y+C^{-1}Ex))\] is strictly positive, and the other is nonnegative. Hence \(Q>0\). If a summand is zero, its positivity assertion is vacuous and its unique endomorphism is invertible, so the same argument includes it.