How many zero elements are there in the inverse of the $n\times n$ matrix

Question

How many zero elements are there in the inverse of the $n\times n$ matrix

$A=\begin{bmatrix} 1&1&1&1&\cdots&1\\ 1&2&2&2&\cdots&2\\ 1&2&1&1&\cdots&1\\ 1&2&1&2&\cdots&2\\ \cdots&\cdots&\cdots&\cdots&\cdots&\cdots\\ 1&2&1&2&\cdots&\cdots \end{bmatrix}$

My try: Denote by $a_{ij}$ and $b_{ij}$ the elements of $A$ and $A^{-1}$,respectively then for $k\neq m$,we have $$\sum_{i=0}^{n}a_{ki}b_{im}=0$$

Then I can't.Thank you very much

Answer 1

Investigation with Maple indicates that for $n>1$, the inverse of $A$ has the following form:

(1) the main diagonal starts with $2$, finishes with $\pm1$, has zeros in between;

(2) the next diagonal each way has $1$ and $-1$ alternating;

(3) everything more than "one step away" from the main diagonal is $0$.

For example $$\pmatrix{1&1&1&1\cr 1&2&2&2\cr 1&2&1&1\cr 1&2&1&2\cr}^{-1} =\pmatrix{2&-1&0&0\cr -1&0&1&0\cr 0&1&0&-1\cr 0&0&-1&1\cr}$$ and $$\pmatrix{1&1&1&1&1\cr 1&2&2&2&2\cr 1&2&1&1&1\cr 1&2&1&2&2\cr 1&2&1&2&1\cr}^{-1} =\pmatrix{2&-1&0&0&0\cr -1&0&1&0&0\cr 0&1&0&-1&0\cr 0&0&-1&0&1\cr 0&0&0&1&-1\cr}\ .$$ I can't see a simple proof that this is true for all $n$, but if it is then the number of zeros is $n^2-2n$, provided $n>1$.

Answer 2

This isn't a complete answer, either, but note that your matrix can be written as

$$\pmatrix{ 1\cr \vdots \cr 1} \pmatrix{1 & \cdots & 1} + \pmatrix{ 0 \cr 1 \cr \vdots \cr 1} \pmatrix{0 & 1 &\cdots & 1 } - \pmatrix{ 0 \cr 0 \cr 1 \cr \vdots \cr 1}\pmatrix{ 0 & 0 & 1 & \cdots & 1}+\ldots$$

Because the matrix is symmetric all its eigenvalues are real and all its eigenvectors can be orthogonalized.

If the bottom right element equals 2 then you get something like

$$\pmatrix{0 \cr\vdots \cr 0 \cr 1} \pmatrix{0 &\cdots & 0 & 1} + \pmatrix{0 \cr \vdots \cr 0 \cr 1 \cr 1 \cr 0}\pmatrix{0 & \cdots & 0 & 1 & 1 & 0} + \dots$$

for the positive eigenvalues and something similar for the negative eigenvalues.

Hth