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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.

Afternote. Not long ago, someone mentioned that Olga Tausky-Todd had once said that "If an assertion about matrices is false, there is usually a 2x2 matrix that reveals this." Since these are the words of the great Olga Tausky-Todd, I didn't dare to disagree. But deep in my heart, I thought that in general 2x2 matrices do not suffice, but 3x3 matrices should be sophisticated enough for the building of counterexamples in most cases. But you know what? There are always exceptional cases where even 3x3 matrices are not good enough. Here we encounter one such case.

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