Hey everyone, let's dig into this interesting linear algebra problem:

Let $A$ be an $n \times n$ matrix where every element comes from the set ${-1, 0, 1}$. Let $r_i$ stand for the sum of elements in the $i$-th row, and $c_i$ for the sum of elements in the $i$-th column. We want to figure out which values of $n$ allow all the numbers $r_1,...,r_n,c_1,...,c_n$ to be pairwise distinct.

Even $n$: Constructing valid matrices #

I've already worked out that when $n$ is even, we can build such matrices recursively. Let's call these matrices $A_n$:

• For $n=2$:
  $$A_2=\begin{pmatrix}1&1\-1&0\end{pmatrix}$$
  The row sums are 2 and -1, column sums are 0 and 1—all four values are distinct.
• For $n=4$:
  $$A_4=\begin{pmatrix}1&1&1&1\-1&1&1&1\-1&-1&0&1\-1&-1&-1&0\end{pmatrix}$$
  Row sums here are 4, 2, -1, -3; column sums are -2, 0, 1, 3. All 8 sums are unique.
• Recursive rule for $A_{n+2}$ from $A_n$:
  $$A_{n+2}=\begin{pmatrix}1& \begin{matrix}1&\cdots&1\end{matrix} &1\ \begin{matrix}-1\\vdots\-1\end{matrix} &A_n&\begin{matrix}1\\vdots\1\end{matrix}\-1&\begin{matrix}-1&\cdots&-1\end{matrix}&0\end{pmatrix}$$
  It's not hard to check that this recursive step keeps all row and column sums distinct.

Odd $n$: Seems impossible—need proof ideas #

When $n$ is odd, I can't find any matrix that satisfies the condition, and it feels like it's actually impossible. But I need to formalize this: prove that for any odd $n$, every $n \times n$ {-1,0,1}-matrix must have at least two rows or columns with the same sum.

I'm guessing parity arguments or analyzing the possible range of sums might help here, but I'm stuck on how to tie everything together. Does anyone have suggestions on how to approach this proof?

备注：内容来源于stack exchange，提问作者aleph2