maximum number of zero elements (entries) in $A^{-1}?$

Question

Let $A $ be an ${n \times n}$ symmetric invertible matrix with real positive elements. Then, find the maximum number of zero elements in $A^{-1}$ ?

Answer 1

Consider, for different positive numbers $a,b$, the matrix

$$M=\begin{bmatrix} a & a & a\\ a & a & b\\ a&b&b \end{bmatrix}$$

If $C_{ij}$ denotes the cofactor of the entry in the $i$th row and the $j$th column, then $$C_{12}=C_{21}=C_{33}=0$$

Consequently, $A^{-1}$ has at-least three zeros.

More generally, if $$M=\begin{bmatrix} a&a&\cdots&a\\ a&a&\cdots&b\\ \vdots\\ a&b&\cdots&b \end{bmatrix}$$ then the inverse of $A$ has at-least $n^2-2n$ zeros.

Finally, we prove that there cannot be more than $n^2-2n$ zeros in $A^{-1}$. To prove that, observe that if there are more than $n^2-2n$ zeros, then there are at-least $n^2-2n+1=(n-1)^2$ zeros. So, there are at-most $n^2-(n-1)^2=2n-1$ non-zero entries in $A^{-1}$. Using the pigeonhole principle, there must be a row in $A^{-1}$ that has at-most $1$ non-zero entry. That is, there is a row that has either no non-zero entry or exactly one non-zero entry. In the former case, matrix $A^{-1}$ is singular, a clear contradiction. In the latter case, without loss of generality, let the only non-zero entry in a row (if any) be the $ij$th entry in $A^{-1}$. The cofactor of any other entry in the same column in $A^{-1}$ is zero. Since a zero cofactor in $A^{-1}$ means a zero entry in $A$, This means that an entry in $A$ is zero, a contradiction. Thus, the maximum number of zeros in $A^{-1}$ is $n^2-2n$.

For $n=3$, it is $3^2-2\times 3=9-6=3$.