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Im looking for an explanation of the following: a standard way to prove that, if there exists Hadamard matrix of dimension $n > 2$, then $4|n$, is to suppose that without loss of generality every column of the matrix starts with +. (Otherwise, one can multiply the column by -1, which doesn't change the Hadamard property).

Then there are only 4 possibilities of how the first three entries of each column look like: +++, ++-, +-+ and +--. Let's say there's $a$ columns of the first type, $b$ of the second, $c$ and $d$ of third and fourth respectively.

Obviously, since the matrix is $n\times n$, we obtain $a+b+c+d=n$. But here comes the point of confusion: the final step says that because of the orthogonality relations we also obtain 3 more relations: $$ a + b - c - d = 0 $$ $$ a + c - b - d = 0 $$ $$ a - b - c + d = 0 $$
which all put together yields $4a=n$. Im not sure, how can "number of columns of some type" be mixed with the fact that every two columns are orthogonal? How do we obtain these 3 relations? This might be a stupid question, but I can't really see something that might be obvious.

(How come the orthogonal property of the columns can yield something like "number of type 1 columns + number of type 2 columns - number of type 3 columns - number of type 4 columns" = 0 ??)

NumberFour
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    See http://math.ucdenver.edu/~wcherowi/courses/m6406/csln9.html, or Theorem $1.5$ here: http://faculty.fiu.edu/~draghici/pastcourses/appla-fa11/Hadam_handout.pdf. – Dietrich Burde Apr 14 '14 at 20:59

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Hint: A Hadamard matrix $H$ has the property $HH^T=nI_n$. This implies that $H^T$ is Hadamard whenever $H$ is. Therefore if the columns all are orthogonal to each other, then so are the rows. Guess what you get with the inner products of the first three rows.

Jyrki Lahtonen
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  • It is clear for me that if columns are orthogonal, so are the rows. But I can't see how this relates to the number of the columns of those types. – NumberFour Apr 14 '14 at 20:42
  • First and second row have the same sign on columns of types $a$ and $b$, and a different sign on columns of types $c$ and $d$. Therefore the inner product of the first and the second rows is $a+b-c-d$. Calculate the inner products of 1st & 3rd as well as 2nd and 3rd in the same way... – Jyrki Lahtonen Apr 14 '14 at 20:44
  • Ok, even though I knew this and you didn't actually answer my question, you forced me to write it down again and now I see it really works! Thank you :) – NumberFour Apr 14 '14 at 20:59
  • I get it. First row all $+1,$ second row demanded $(1,1,1,-1,-1,-1).$ Trial third row $(-1,-1,-1,1,1,1),$ inner product $-6 \equiv 2 \pmod 4.$ If I switch one $+1$ entry in the third row to $-1,$ and one $-1$ entry to $+1,$ I have changed the dot product by $\pm 4$ or by $0,$ so it is still $2 \pmod 4.$ – Will Jagy Apr 14 '14 at 21:01
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    Glad to hear that you figured it out, @NumberFour. Well done! I have this habit of dropping hints rather than spelling it all out if at all feasible. I believe that this approach has some pedagogical merit :-) – Jyrki Lahtonen Apr 14 '14 at 21:13
  • Jyrki, superficially similar oddity. True for $n \leq 8,$ counterexamples for any $n \geq 9, ; n \equiv 1 \pmod 8.$ No idea about $n = 10,11,12,13,14,15,16. $ http://math.stackexchange.com/questions/261204/every-integer-vector-in-mathbb-rn-with-integer-length-is-part-of-an-orthogon – Will Jagy Apr 14 '14 at 21:26
  • Curios, indeed, @Willy. Reminds me of orthogonal designs, of which it is known that the only full rate ones correspond to the complex numbers, quaternions and octonions respectively. But there (unlike in orthogonal designs) we are not constrained to just make use permutations of the entries... – Jyrki Lahtonen Apr 14 '14 at 21:35