If I have a Hadamard design $(4t-1, 2t-1, t-1)$ is it possible that 2 subsets share $t$ or more elements?
The text I am working through seems to imply it is obvious that this isnt possible but I seem to be missing something.
I have tried the specific case of $t=3$ and have the below proof that 2 subsets cannot share $t$ or more elements, but I can't seem to generalize it.
In the case of $t = 3$, then suppose we had a 2 subsets, $A$, $B$ with $ |A \cap B| > 2$.
Let $z \in (A \cup B)^{c}$
Then for each $y \in A \cup B $. There are 2 subsets containing both z and y.
These subsets are unique per $y$, since $y$ cannot be in another subset with the elements it has already shared 2 subsets with.
But this means $z$ is in 6 subsets contradicting our assumption of being a $(11, 5, 2)$ design.
Help gratefully recieved :)
Thanks, Dan
In general in a $2$-$(v,k,\lambda)$ design, it is possible for two blocks to intersect in more than $\lambda$ elements. Examples can be found, for example see information about quasi-symmetric designs.
However, the Hadamard designs are special: they are symmetric designs, meaning there are the same number of blocks as there are elements. In a symmetric design, any pair of blocks must intersect in exactly $\lambda$ elements.
To see this, recall that if $A$ is the incidence matrix of a design, then $$AA^{T} = (r-\lambda)I+\lambda J.$$ We can take the determinant to see that $$\det(AA^{T}) = rk(r-\lambda)^{v-1}.$$ The right side is clearly nonzero, and since $A$ is a square matrix in the case of a symmetric design, $$\det(AA^{T}) = \det(A)^{2} = rk(r-\lambda)^{v-1} \neq 0.$$ This means $A$ is invertible, so if we consider the equation $$AA^{T}A = ((r-\lambda)I+\lambda J)A = A(r-\lambda)I+\lambda J = AAA^{T},$$ we can multiply both sides on the left by $A^{-1}$ to see that $$A^{T}A = AA^{T}.$$ Considering the $ij$ entry when $i\neq j$, on the left this gives the intersection size of block $i$ and block $j$, while on the right, the number of blocks containing both element $i$ and element $j$. These values are the same: $\lambda$.