Question:
Let $H$ be a Hadamard matrix and $n$ be its order. Can $H$ always be made equivalent (by multiplying row/column by -1 or swapping rows/columns) to a symmetric one?
The answer can be divided into two parts.
Part 1:
For $n<16$, the answer is always YES.
First, we can eazily prove that
$$ n<16 \implies n=1,2,4,8,12 $$
Then from Reference $1$ to Reference $5$, we found that at least one symmetric Hadamard matrix $S$ exists for each order above.
According to Reference $6$, we know that the total number of equivalent classes is always $1$.
So given any Hadamard matrix $H$, we always have $H$ and $S$ are in the same equivalent class.
Part 2:
For $n>15$, the answer is NO.
For $n=16$, from Reference 6, we know that the total number of equivalent classes is $5$. We cannot guarantee that $H$ and $S$ are in the same equivalent class any more.
In fact, by computer bruteforce search, we can see that the following matrix cannot be made equivalent to a symmetric one. (Here, $−$ stands for $−1$ and $+$ for $+1$.)
$++++++++++++++++$
$+-+-+-+-+-+-+-+-$
$++--++--++--++--$
$+--++--++--++--+$
$++++----++++----$
$+-+--+-++-+--+-+$
$++----++++----++$
$+--+-++-+--+-++-$
$++++++++--------$
$+++-+------+-+++$
$++-+---+--+-+++-$
$++---++---+++--+$
$+-++-+---+--+-++$
$+-+---++-+-+++--$
$+--++-+--++--+-+$
$+---++-+-+++--+-$
References:
- https://blog.csdn.net/mcs_liyun/article/details/139578602
- https://blog.csdn.net/mcs_liyun/article/details/139578530
- https://blog.csdn.net/mcs_liyun/article/details/139578492
- https://blog.csdn.net/mcs_liyun/article/details/139578576
- https://blog.csdn.net/mcs_liyun/article/details/139578443
- https://oeis.org/A007299