# Schur's complement of a matrix with no zero entries

Let $A$ be an $n\times n$ symmetric positive-definite matrix so that itself and its inverse $A^{-1}$ both have no entry equal to $0$ ($A_{i,j} \neq 0$ and $(A^{-1})_{i,j} \neq 0$ for all $i,j \in \{1,2, ..., n\}$).

Is it true (or false) that the Schur's complement $S$ of block $A_{n,n}$ of matrix $A$ also does not have a zero entry? ($S = A_{1:n-1,1:n-1} - A_{1:n-1,n}A_{n,n}^{-1} A_{n,1:n-1}$)

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When the order of $S$ is at least $3$, the answer is "false", otherwise the answer is "true". Here is a counterexample when $S$ is 3-by-3: \begin{align} A&=\begin{pmatrix}3&3&2&1\\3&6&5&2\\2&5&6&2\\1&2&2&1\end{pmatrix}, \quad A^{-1}=\frac14\begin{pmatrix}3&-2&1&-1\\-2&4&-2&-2\\1&-2&3&-3\\-1&-2&-3&15\end{pmatrix},\\ S&=\begin{pmatrix}3&3&2\\3&6&5\\2&5&6\end{pmatrix} -\begin{pmatrix}1\\2\\2\end{pmatrix} (1)^{-1} \begin{pmatrix}1&2&2\end{pmatrix} =\begin{pmatrix}2&1&0\\1&2&1\\0&1&2\end{pmatrix}. \end{align} In general, let $A=\begin{pmatrix}X&Y\\Y^T&Z\end{pmatrix}$ ($Y$ may be a larger block, not necessary a column vector). If the order of $S$ (or $X$) is at least 3, $Z$ is positive definite and $X=YZ^{-1}Y^T+D$, where $D$ is a strictly diagonally dominant real symmetric matrix with a positive diagonal and only a pair of zero entries, then the Schur complement $X-YZ^{-1}Y^T$ would be equal to $D$ (and hence has a zero entry) and $A$ is positive definite. If you draw $Y$ at random and draw $Z$ from the set of positive definite matrices at random, the probability that $A$ or $A^{-1}$ have zero entries should be practically zero. So, it is easy to construct a counterexample in this case.
Yet, when $S$ is 1-by-1, the answer to your question is "true". Note that regardless of the dimension of $S$, in general $A$ is congruent to $S\oplus Z$. When $S$ is 1-by-1, that $S$ has a zero entry implies that $S=0$. Hence $S$ and in turn $A$ are not positive definite, which is a contradiction.
When the order of $S$ is 2, the answer is also "true". If $S$ has a zero entry, this entry must be off-diagonal, or else $S$ and in turn $A$ are not positive definite. So, $S$ must be a diagonal matrix because it is symmetric and its size is 2-by-2. But then $A^{-1}=\begin{pmatrix}S^{-1}&\ast\\ \ast&\ast\end{pmatrix}$ will have zero entries, which is a contradiction.