Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Is there any linear-algebraic link with the use of the word "orthogonal" in orthogonal latin squares? I thought about it a little bit and the closest I got to linear algebra was this definition : if $V = \langle v_1,\dots,v_n \rangle_{\mathbb R}$ is an $\mathbb R$-vector space of dimension $n$, a Latin square of order $n$ is a linear map $f : V^2 \to V$ which maps $f(v_i,v_j)$ to $v_{f^*(i,j)}$ where $f^* : [n] \times [n] \to [n]$ is a Latin square in the classical sense (i.e. assigns to a coordinate of an $n \times n$ grid an integer between $1$ and $n$). Then, if we translate the classical definition, two Latin squares $f,g : V^2 \to V$ are orthogonal if $f \times g : V^2 \to V^2$ defined by $(v_i,v_j) \mapsto (f(v_i,v_j),g(v_i,v_j))$ is an isomorphism. This was a quite natural attempt, but it does not seem to involve orthogonality.

(In other words, all I did was replace the occurences of $[n]$ by the $\mathbb R$-vector space it generates, and the rest is just translation.)

Anyone knows the origin of this choice of words, and if yes, is there a link with linear algebra?

share|cite|improve this question
The idea is that you can do a decomposition into orthogonal squares by splitting into separate Greek and Latin squares. – oldrinb Sep 8 '13 at 15:34
I do't know what's true, but I have thought that in the context of Latin squares "orthogonal" means roughly the same as "strongly independent". A traditional use of Latin squares has been in testing, and if you base tests on two orthogonal Latin squares you make the tests as independent as they can be. – Jyrki Lahtonen Sep 8 '13 at 15:37
@oldrinb : You have not understood my question. I totally understand the definition of orthogonal latin squares, I'm asking if there is a link of the choice of words with linear algebra. – Patrick Da Silva Sep 8 '13 at 15:37
@Jyrki Lahtonen : So you're saying the choice of words has nothing to do with linear algebra, but rather this feeling of orthogonality that says "no relations between two orthogonal things" as in linear independence? – Patrick Da Silva Sep 8 '13 at 15:39
That's how I feel, but I know next to nothing about the history of MOLS, so I'm not sure. – Jyrki Lahtonen Sep 8 '13 at 15:47

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.