Rank of a block matrix.

Question

Let $M$ be a matrix in the following block form

$$M = \pmatrix{A & C \\ C^t & B}$$

where blocks $A$ and $B$ are symmetric and have full rank. Note that $A \neq C \ne B$. From here, can we conclude that $M$ is also full rank?

Answer 1

The formula for the determinant of a block matrix gives $$ \det(M)=\det (A) \cdot\det(B-C^tA^{-1}C). $$ Nothing prevents $B-C^tA^{-1}C$ from being non-invertible. For example, just take $B=C^tA^{-1}C$ for invertible $A,B,C$; for instance, take $A=\frac 12 I$, $C$ any orthogonal matrix, $B=2I$.