# Are regular Hadamard matrices symmetric?

I am trying to show that a regular Hadamard matrix must have order $$m^2$$ for some integer $$m$$.

So far I have found that if $$H$$ is an $$r$$-regular Hadamard matrix of order $$n$$, then $$HJ = rJ$$ and $$HH^TJ = nJ$$.

Does that mean if I can prove that $$H=H^T$$, then it follows that $$n = r^2$$ ?

A regular Hadamard matrix has all row and column sums equal, for instance $$\pmatrix{1&-1&1&1\\1&1&-1&1\\1&1&1&-1\\-1&1&1&1}.$$
In a regular Hadamard matrix, $$Hu=ru$$ and $$H^Tu=ru$$ where $$u$$ is the all-ones column vector and $$r$$ is the row/column sum. Therefore $$HH^Tu=H(ru)=r^2u.$$ But $$HH^T=nI$$ by Hadamard-ness, so $$HH^Tu=nu.$$ Therfore $$n=r^2$$.
• Thanks, so it doesn't have to be symmetric. But still $HJ=H^TJ=rJ$,and $HH^TJ=nJ$. Do these imply that $n=r^2$? – mdryizk Dec 17 '18 at 22:18