Am trying to prove the following proposition in which all matrices are real:
Proposition. Let $A$ be an $n\times n$ symmetric matrix and $B$ be an $n\times k$ matrix with $\text{rank}(B)=k$. Then
\begin{bmatrix}A & B \\ B^T & 0 \end{bmatrix}
is invertible if and only if $C'AC$ is invertible for some $n\times (n-k)$ matrix $C$ with $\text{rank}(C)=n-k$ and $C'B=0$. In this case we have
$$\begin{bmatrix}A & B \\ B^T & 0 \end{bmatrix}^{-1}=
\begin{bmatrix}C(C'AC)^{-1}C' & [I-C(C'AC)^{-1}C'A]B(B'B)^{-1} \\ (B'B)^{-1}B'[I-AC(C'AC)^{-1}C'] & -(B'B)^{-1}B'[A-AC(C'AC)^{-1}C'A]B(B'B)^{-1} \end{bmatrix}$$
Note that for a given $n\times k$ matrix $B$ with $\text{rank}(B)=k$ we can always find an $n\times (n-k)$ matrix $C$ with $\text{rank}(C)=n-k$ and $C'B=0$; It suffices to choose as columns for $C$ a basis of the orthogonal complement of the column space of $B$. Such $C$ is not unique, but it always satisfies
$$I_n=B(B'B)^{-1}B'+C(C'C)^{-1}C'$$
To see this, let $D=[B: C]^T$ and note that $D^{-1}=[B(B'B)^{-1} : C(C'C)^{-1}]$. Hence $I_n=D^{-1}D=B(B'B)^{-1}B'+C(C'C)^{-1}C'$.
Now suppose $C'AC$ is invertible. Direct computations using the above relationship shows that the inverse is indeed as stated in the proposition.
Any ideas how to prove the converse?
Thanks a lot for your help.
Best Answer
Let $P=\pmatrix{B&C}$ and $$ P^{-1}A(P^T)^{-1}=\pmatrix{X&Y\\ Y^T&Z} $$ where $Z$ is $(n-k)\times(n-k)$. Then $$ \begin{aligned} Z&=\pmatrix{0&I_{n-k}}\pmatrix{X&Y\\ Y^T&Z}\pmatrix{0\\ I_{n-k}}\\ &=(C^TC)^{-1}C^TP\pmatrix{X&Y\\ Y^T&Z}P^TC(C^TC)^{-1}\\ &=(C^TC)^{-1}C^TAC(C^TC)^{-1}. \end{aligned} $$ Hence the invertibility of your block matrix, which is equivalent to the invertibility of $$ \pmatrix{P^{-1}\\ &I_k}\pmatrix{A&B\\ B^T&0}\pmatrix{P^{-T}\\ &I_k} =\pmatrix{X&Y&I_k\\ Y^T&Z&0\\ I_k&0&0}, $$ is equivalent to the invertibility of $Z$, and in turn, of $C^TAC$.