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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 $n = r^2$ ?

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A regular Hadamard matrix has all row and column sums equal, for instance $$\pmatrix{1&-1&1&1\\1&1&-1&1\\1&1&1&-1\\-1&1&1&1}.$$

ADDED IN EDIT

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 $n=r^2$.

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  • $\begingroup$ Thanks, so it doesn't have to be symmetric. But still $HJ=H^TJ=rJ$,and $HH^TJ=nJ$. Do these imply that $n=r^2$? $\endgroup$ – mdryizk Dec 17 '18 at 22:18

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