TITLE: Almost Hadamard matrices
QUESTION [11 upvotes]: As well-known, a Hadamard matrix is a square matrix with all coefficients $\pm 1$
and pairwise orthogonal rows or columns. Such matrices exist conjecturally 
in every dimension divisible by $4$. Call a matrix with an odd number $n$ of columns an "almost Hadamard matrix" if all its coefficients are $\pm 1$ and if all scalar products between distinct rows are $\pm 1$. We want to maximize the
number of rows in almost Hadamard matrices: If $n\equiv 3\pmod 4$ this is easy:
erase the last column of a Hadamard matrix of size $n+1$ (provided such a matrix exists). This yields a matrix with $n+1$ rows which is best possible. If $n\equiv 1\pmod 4$ the number of rows cannot exceed $n$ (argument: up to replacing rows by their opposites, we can suppose that all rows have an even 
number of coefficients $-1$. All scalar products between rows are now 
equivalent to $n$ modulo $4$ and a rank computation of the corresponding Gram matrix gives the result). 
A solution with $n-1$ rows in the case $n\equiv 1\pmod 4$
is obtained by adding an arbitrary last column 
(with coefficients $\pm 1$) to a Hadamard matrix of size $n-1$.
For $n=5$ (and of course for $n=1$) there is a solution with $n$ rows but I am
unaware of the existence of solutions with $n$ rows for any $n\equiv 1\pmod 4$
greater than $5$. Are there any?
Is there any literature on such almost Hadamard matrices (perhaps under a different terminology, they are for example related to systems of equiangular lines)?

REPLY [5 votes]: You can find most useful information on this page, which however has not been updated since 2012. The initial question is different, but the partial solution they have found is relevant here.   
Quote:
The Hadamard maximal determinant problem asks when a matrix of a given order with entries -1 and +1 has the largest possible determinant. 
If the determinant of such an extremal matrix $M=M_n$ for $n\equiv1\pmod4$ attains Barba's bound $$\sqrt{ (2n-1)  (n-1)^{n-1}},$$ which obviously needs $(2n-1)$ to be a perfect square, then $$M^TM=MM^T=J_n+(n-1)I_n$$ where $J_n$ is the all-$1$-matrix. So this matrix is an "almost Hadamard" (a.H.) matrix. (Moreover, all scalar products are $+1$.)
Examples are given for $n=5,13,25,41,61,113$.
If $2n-1=(2q+1)^2$ and $q$ is a prime power (note that the examples correspond to $q=1,2,3,4,5,7$), there is an explicit construction according to the paper
A. E. Brouwer, An infinite series of symmetric designs, Math. Centrum
   Amsterdam Report ZW 202/83 (1983).
If $(2n-1)$ is not a perfect square at all, we cannot conclude inversely that no square a.H. matrix exists, though I have a (kind of) heuristic argument against its existence:
What we can conclude is that $M^TM$ must have a determinant smaller than Barba's bound because it must be a square. This means that the set of column vectors cannot be split in two sets $V$ and $W$ such that $v^T_iv_j=w_i^Tw_j=+1$ and $v^T_iw_j=-1$. (Nor, similarly, for the rows.)
Before explaining why, here is an illustration of what I mean with those two sets. Take for instance the matrix $M$ given for $n=13$ in Gottfried Helms' comment. After some row and column permutations we have 
 $$MM^T=-J_{13}+2J_5\oplus J_8+12I_{13}$$ and $$M^TM=-J_{13}+2J_3\oplus J_{10}+12I_{13},$$ written in compact form: $MM^T=$
n 1 1 1 1 - - - - - - - -
1 n 1 1 1 - - - - - - - -
1 1 n 1 1 - - - - - - - -
1 1 1 n 1 - - - - - - - -
1 1 1 1 n - - - - - - - -
- - - - - n 1 1 1 1 1 1 1
- - - - - 1 n 1 1 1 1 1 1
- - - - - 1 1 n 1 1 1 1 1
- - - - - 1 1 1 n 1 1 1 1
- - - - - 1 1 1 1 n 1 1 1
- - - - - 1 1 1 1 1 n 1 1
- - - - - 1 1 1 1 1 1 n 1
- - - - - 1 1 1 1 1 1 1 n

and similarly for $M^TM$. That is, the rows of $M$ come in two sets of sizes $5$ and $8$, while the columns of $M$ form two sets of sizes $3$ and $10$.
Now the determinant of such a block matrix is always $(2n-1)  (n-1)^{n-1}$, independently of the block sizes. So if $(2n-1)$ is not a square, the rows and columns cannot be partitioned in such a way, which means that an $n\times n$ a.H. matrix cannot have this kind of "symmetries". And heuristically, the existence of such a matrix seems to me about as improbable as the existence of a projective plane of non prime-power order...
BUT based on the data of the above site, e.g. the $n=93$ example, I would conjecture that $n-1$ almost orthogonal vectors always exist.