Largest orthogonal matrix that can be created using only 1 and -1 as elements I am trying to understand what is the largest possible orthogonal matrix that can be created using only 1 and -1 as its elements.
Clearly for a $\ 4 \times 4 $ matrix:
We can construct an orthogonal matrix as :
$\  \begin{bmatrix}1&1&1&-1\\1&1&-1&1\\1&-1&-1&-1\\1&-1&1&1\end{bmatrix} $
But I want to understand whether it is possible to prove that we can or cannot create such a matrix for larger size of the matrix.
Any guidance is appreciated :)
My attempts until now have been restricted to finding bounds on column rank of a matrix with all  possible column vectors constructed from 1 and -1.
But even then, I am unable to link it to orthogonality of vectors.
 A: It is straightforward to see that there is a $(2^r) \times (2^r)$ solution matrix for each integer $r\geq 2$. Indeed, consider a matrix $A=(a_{I,J})$ where the two indices $I$ and $J$ are subsets of $\lbrace 1,2,\ldots ,r \rbrace$, and 
$$a(I,J)=(-1)^{|I\cap J|}$$
Let us show that this matrix is indeed orthogonal. Let $I_1 \neq I_2$, we must show that the sum
$$
S(I_1,I_2)=\sum_{J\subseteq [r]} a(I_1,J)a(I_2,J) \tag{2}
$$
is zero. If $A\cup B$ is any partition of $[r]$, we can decompose $S(I_1,I_2)$ using this partition, and rewrite it as
$$
S(I_1,I_2)=\sum_{J_A\subseteq A}\sum_{J_B\subseteq B} a(I_1,J_A)a(I_1,J_B)a(I_2,J_A)a(I_2,J_B) \tag{2'}
$$
In particular, if we take $A=I_1\cup I_2$ and $B=[r] \setminus A$ in (2'), the terms $a(I_1,J_B)$ and $a(I_2,J_B)$ are always equal to 1, so
$$
S(I_1,I_2)=2^{|B|} \sum_{J_A\subseteq A} \sum_{J_B\subseteq B} a(I_1,J_A)a(I_2,J_B) \tag{3}
$$
In other words, we may assume without loss that $I_1\cup I_2 = [r]$. Similarly, using (2') again with  $A=I_1\cap I_2$, we may also assume without loss that $I_1\cap I_2 = \emptyset$.
At this point $(I_1,I_2)$ is a partition of $[r]$. We may then use (2') a third time, with $A=I_1,B=I_2$ :
$$
S(I_1,I_2)=\sum_{J_A\subseteq I_1}\sum_{J_B\subseteq I_2} a(I_1,J_A)a(I_1,J_B)a(I_2,J_A)a(I_2,J_B)=
\sum_{J_A\subseteq I_1}\sum_{J_B\subseteq I_2}(-1)^{|J_A|+|J_B|}=
\bigg(\sum_{J_A\subseteq I_1}(-1)^{|J_A|}\bigg)\bigg(\sum_{J_B\subseteq I_2}(-1)^{|J_B|}\bigg)
 \tag{4}
$$
Now each factor in this product is zero, using the Newton binomial for $(1-1)^n$.
A: Thanks to @PeterTaylor for guiding me in the right direction.
This problem comes under the genre of construction of Hadamard matrices which are orthogonal matrices built using only 1 and -1 as elements.
https://en.wikipedia.org/wiki/Hadamard_matrix
There are many constructions possible for big matrices of size $2^k $ like Sylvestors construction:
Sylvester's construction(From Wikipedia) :
Examples of Hadamard matrices were actually first constructed by James Joseph Sylvester in 1867. Let H be a Hadamard matrix of order n. Then the partitioned matrix
\begin{bmatrix}H&H\\H&-H\end{bmatrix} 
is a Hadamard matrix of order 2n. This observation can be applied repeatedly and leads to the following sequence of matrices, also called Walsh matrices.
\begin{aligned}H_{1}&={\begin{bmatrix}1\end{bmatrix}},\\H_{2}&={\begin{bmatrix}1&1\\1&-1\end{bmatrix}},\\H_{4}&={\begin{bmatrix}1&1&1&1\\1&-1&1&-1\\1&1&-1&-1\\1&-1&-1&1\end{bmatrix}},\end{aligned}
Also construction of such matrices of arbitrary sizes is still under research with an open conjecture called the Hadamard conjecture which states that a Hadamard matrix of order 4k exists for every positive integer k.
Just a fun fact to add: These matrices are extremely useful especially for error correcting codes to correct messages transmitted over very unreliable channels.
