How can I see that a simple $3x3$ matrix (latin square) is orthogonal?

Question

In a book I am reading, there is the following example:

$$\begin{bmatrix}A &B& C\\B &C& A\\C& A &B\end{bmatrix}$$

The authors say: "As you can see, all the column vectors are orthogonal."

I wonder, how can I see it from this design (latin square), I cannot see the zero dot product involved.

Answer 1

Since

$$(A,B,C)^T(B,C,A)=AB+BC+CA$$

$$(A,B,C)^T(C,A,B)=AC+BA+CB$$

$$(B,C,A)^T(C,A,B)=BC+CA+AB$$

they are orthogonal if and only if $AB+BC+CA=0$.

The book is referring to another concept explained here: if each entry of an n × n "Latin square" is written as a triple (r,c,s), where r is the row, c is the column, and s is the symbol, we obtain a set of $n^2$ triples called the orthogonal array representation of the square.

Answer 2

I think I have found the solution - as the point of the design is in the pairwise combinations being unique in each pair of columns, let's use vectors with size equal to the number of pairwise combinations for each pair in a vector of size 3: (the numbers denote the columns, 12 means the first and the second column):

12
AA,AB,AC,BB,BA,BC,CC,CA,CB

13
AA,AB,AC,BB,BA,BC,CC,CA,CB

23
AA,AB,AC,BB,BA,BC,CC,CA,CB

when a combination is used, 1 is put to its place, otherwise 0. So for the middle row in the latin square in my question (BCA), the vector would be:

{0,0,0,0,0,1,0,0,0},{0,0,0,0,1,0,0,0,0},{0,0,0,0,0,0,0,1,0}

because of the design the pairwise combinations are unique and therefore 1s never appear in the same spots, i.e. the dot product will be zero.

I found the solution in the accepted answer for a very similar question: What is the relation between orthogonal arrays and orthogonal matrices?