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Sometimes a determinant identity is established by first proving the special case where the matrix involved is nonsingular and then making an appeal to continuity for the general case. The following curiosity (due to M. J. Pelling) shows you shouldn't be too cavalier about doing so.
Proposition Suppose A,B,C,D are n by n matrices satisfying CDT = – DCT with D nonsingular. Then the determinant of the 2 by 2 block matrix with blocks A,B,C,D is equal to det(ADT + BCT).
Can you remove the restriction that D be nonsingular by an appeal to ''continuity''? No—the proposition is false without this restriction! The full story is that if CDT = – DCT, then the block determinant is equal to (–1)r det(ADT + BCT) where r is the rank (= total geometric multiplicity of nonzero eigenvalues) of C, and this is true whether or not D is singular. The point is that D nonsingular forces rank(C) to be even and so the (–1)r sign factor disappears when D is nonsingular.
In this paper we drop the ''transposes'', consider the general hypothesis CD = kDC with k a scalar, and prove that kr times the block determinant is equal to (–1)r det(AD – kBC) where the exponent r is now the total algebraic multiplicity of nonzero eigenvalues of C. The proof is somewhat complicated because it requires use of the Jordan canonical form PAP –1, whereas the simpler Smith canonical form PAQ suffices for the ''transpose'' case.