I am a bit curious about an exercise.

I was supposed to prove that there is no matrix that has $1,2,3,4$ in each line and is symmetric. I did, by examination of all such matrices.

Now, there is a such a matrix for $1,2,3$, namely

$$\left(\begin{array}{ccc} 1&3&2\\ 3&2&1\\ 2&1&3\\ \end{array}\right)$$

How does the result generalize? Is it a odd vs even thing? Or $n \gt 4$ is always impossible?

Proofs using linear algebra are especially welcome!

  • $\begingroup$ It's unlikely there will be a "linear algebra" proof here, since there are no questions about linear equations, linear independence, vector spaces, linear transformations, etc. For any $n$ just start with any row and do a cyclic shift to the left by 1 for each row. $\endgroup$ – Morgan Rodgers May 20 at 22:44
  • $\begingroup$ @MorganRodgers, fair enougth, but strang has this tendency of linking linear algebra to everything, that makes me wonder $\endgroup$ – josinalvo May 21 at 12:58

On the contrary... it is possible for every $n$.

Let $A_{i,j} = i+j\pmod{n}$

By commutativity of addition it is obvious that the matrix is symmetric. Further, by fixing $i$ or fixing $j$ and letting the other range over all possible values it is clear that there is no repetition in any row or column.


There are 2 other ways for building such a matrix, for most dimensions.

1) Thinking to (symmetric) "latin squares" https://en.wikipedia.org/wiki/Latin_square ; here is a method : https://pdfs.semanticscholar.org/d669/4364b5650f32e84255ab0a76138c55049593.pdf

2) (connected) Thinking to Cayley tables of abelian groups (https://en.wikipedia.org/wiki/Cayley_table).

Indeed, the table of a group is known to be a latin square : (https://en.wikipedia.org/wiki/Latin_square_property). (reciprocal untrue).

If moreover, this group is commutative, it provides an answer to the question.

Let us take an example : the multiplicative group of the finite field $\mathbb{F}_p$ where $p$ is a prime number ; we can build a matrix $(p-1) \times (p-1)$ in the following way

$$A_{i,j}=i*j \ \text{mod} \ p$$

(reminds the solution of @JMoravitz !) that fulfills your condition because multiplication mod $p$ gives the set of nonzero classes $\{\overline{1},\overline{2},\cdots \overline{p-1}\}$ an abelian group structure.


Why can't you say?

\begin{bmatrix} 1&2&3&4\\ 2&1&4&3\\ 3&4&1&2\\ 4&3&2&1 \end{bmatrix}


\begin{bmatrix} 1&2&3&4\\ 2&3&4&1\\ 3&4&1&2\\ 4&1&2&3 \end{bmatrix}

This seems more like abstract algebra though, as these arrays are the Cayley tables for the two group structures of order four.


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