I did see the related question here but my question is more related to the actual explanation of the orthogonality itself.

So the following design is orthogonal (this is a latin square to be precise):


As noted in one of the afore mentioned answers:

"An experimental design is orthogonal if the effects of any factor balance out (sum to zero) across the effects of the other factors."

Can that be demonstrated on the design above? I still cannot see how that is orthogonal (and from a comment it should be immediately obvious when you look at it).

  • 1
    $\begingroup$ That was stated in a few books showing a latin square. The authors stated that it is immediately obvious that rows and columns are orthogonal to treatments. And that is exactly point of my question - I do not see that obvious! $\endgroup$
    – John V
    Commented Jun 22, 2017 at 7:19
  • 1
    $\begingroup$ Why would you assume it should be "immediately obvious"? Latin square designs are actually quite complicated to formulate mathematically. The statistical concept of "orthogonality" is closely related to orthogonality of vectors and to statistical independence. If you want an explanation of "orthogonality", there are much easier ways to explain it than by way of a Latin square experiment! It would be helpful if you clarified whether you are after an explanation of orthogonality in general or of Latin squares in particular. $\endgroup$ Commented Jun 22, 2017 at 7:20
  • $\begingroup$ Well, it may become obvious after a few pages of mathematical derivations. You are perhaps taking the book authors' comment somewhat out of context. $\endgroup$ Commented Jun 22, 2017 at 7:22
  • $\begingroup$ No, surely not. It is a book dealing with entirely different topic and there is just one page mentioning latin squares. Then there is a sample picture of a square and the comment that "you can immediately see that rows and columns are ortogonal to treatments". In another book, there is an explanation "because each treatment appears once in each row" - but to me that is not an explanation of why it is orthogonal. $\endgroup$
    – John V
    Commented Jun 22, 2017 at 7:27

1 Answer 1


I understand you have four row levels, four column levels and the third factor has four levels (1, 2, 3, 4). In total, there are 16 observations.

The model is:

$${\vec Y}_{ijk} = \alpha + \sum_{i=1}^4\alpha_i^{A}{\vec v}_i + \sum_{j=1}^4\alpha_i^{B}{\vec w}_j + \sum_{k=1}^4\alpha_k^{C}{\vec z}_k $$

where $A$, $B$ and $C$ are respectively row, column and the third ("latin letter") factor. Vectors ${\vec v}_i$, ${\vec w}_j$ and ${\vec z}_k$ are such that contain 1 at positions corresponding to combinations of factors present in the design. For instance, you have an experiment in which $A=2$ (second row), $B=2$ (second column) and $C=3$, so in the same position occupied by $Y_{223}$ in vector ${\vec Y}_{ijk}$, vectors ${\vec v}_2$, ${\vec w}_2$ and ${\vec z}_3$ would have 1 and all the others 0.

Now if the usual constraints $\sum_{i=1}^4\alpha_i^{A} = \sum_{j=1}^4\alpha_j^{B} = \sum_{k=1}^4\alpha_k^{C} = 0$ are forced upon the coefficients, it is easy to see that any inner product such as

$$< {\vec v}_i, \sum_{k=1}^4\alpha_k^{C}{\vec z}_k>$$

will be necessarily zero. The "ones" in ${\vec v}_i$ multiply the "ones" in ${\vec z}_k$ only once for each $k$, so:

$$< {\vec v}_i, \sum_{k=1}^4\alpha_k^{C}{\vec z}_k> = \sum_{k=1}^4\alpha_k^{C}<{\vec v}_i,{\vec z}_k> = \sum_{k=1}^4\alpha_k^{C} = 0$$

the last equality because of the zero sum constraint on the coefficients.

EDIT (in response of comment):

It would require much writing to answer using your example (and would perhaps not fit across the page) so let us imagine a simpler case:


that is, two row, column, and "latin letter" levels. The response vector and the design matrix would be as follows: $$ \pmatrix{Y_{111} \\ Y_{122} \\ Y_{212} \\ Y_{221} } = \pmatrix{ 1 & 0 & 1 & 0 & 1 & 0 \cr 1 & 0 & 0 & 1 & 0 & 1 \cr 0 & 1 & 1 & 0 & 0 & 1 \cr 0 & 1 & 0 & 1 & 1 & 0 } $$

The columns of the matrix are respectively $v_1$, $v_2$. $w_1$, $w_2$, $z_1$ and $z_2$. So you see for instance that facing $Y_{212}$ (=in the same row) you have "ones" in vectors $v_2$, $w_1$ and $z_2$. That's what I meant.

Let me add that this is something which doesn't lend itself to lengthy explanations online. I suggest you look at some of the (many) good books on this topic. My favourite is Seber's book (I used much the first edition, but there is now a second one). HTH.

Further EDIT:

Consider for instance $\alpha^B_1 = - \alpha^B_2 = 0.4$ so they add up to zero as required by the constraint. Then,

$$ <{\vec v}_1, \sum_{k=1}^4\alpha_k^{B}{\vec w}_k> = <\pmatrix{1 & 1 & 0 & 0}, 0.4\pmatrix{1 \\ 0 \\ 1 \\0} - 0.4\pmatrix{0 \\ 1 \\ 0\\ 1}> = 0. $$ HTH.

  • $\begingroup$ Thanks for the edit. And in the design matrix, how the orthogonality is now shown? Because when I multiply two column vectors, only the neighoubing ones have product=0. $\endgroup$
    – John V
    Commented Jun 28, 2017 at 9:28
  • $\begingroup$ Vectors in each block are directly orthogonal. Any vector in one block is orthogonal to feasible (=with coefficients satisfying the zero-sum constraint) linear combinations of vectors in any other block. "Block orthogonality" would perhaps be a better description of what is going on. $\endgroup$
    – F. Tusell
    Commented Jun 28, 2017 at 9:37
  • $\begingroup$ Could you please show on the sample design matrix in your example? What you mean by block and what vectors you combine with to show the zeru sum. $\endgroup$
    – John V
    Commented Jun 28, 2017 at 9:42
  • $\begingroup$ I wil further edit answer. $\endgroup$
    – F. Tusell
    Commented Jun 28, 2017 at 9:44
  • $\begingroup$ I see I am missing some basics, I have no idea how you arrived to 0,4 (what is meant by the alpha) $\endgroup$
    – John V
    Commented Jun 28, 2017 at 10:19

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