It's known that Hadamard matrices can only exist for orders $1$, $2$ and $4k$. It's easy to show that there are no Hadamard matrices of order $2k+1$. But what is the proof that there are no Hadamard matrices of order $4k+2$?
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What you say in your first sentence is correct. But note that it is an open question whether Hadamard matrices exist for all integers of the form $4k$. (You probably know this, but casual readers may not.) According to that link, the smallest value of $4k$ for which no Hadamard matrix is known is $668$. – TonyK Jun 26 '20 at 15:36
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Presumably you mean $4k + 2$ for $k \geq 1$ – Ben Grossmann Jun 26 '20 at 15:36
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@TonyK Yeah, I was reading exactly that, and that's why I got this question :) – emptysamurai Jun 26 '20 at 15:41
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Assume the Hadamard matrix has $\ge3$ rows.
Consider the top row. You may as well assume it is all ones. (otherwise change the signs of various columns). Then row two and row three each consist of $n/2$ ones and $n/2$ minus ones. So $n$ is even. As row $2$ and row $3$ are orthogonal, then they agree in $n/2$ entries. So if row $2$ and row $3$ both have ones in $k$ columns, then in $n/2-k$ columns, row $2$ has a one and row $3$ a $-1$ and so in $k$ columns, row $2$ and row $3$ both have $-1$s. So they agree in $2k$ entries: $2k=n/2$ and $n$ is a multiple of $4$.
Angina Seng
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The following is taken from this wikipedia page:
Any two rows of an $n\times n$ Hadamard matrix are orthogonal. For a $\{1, −1\}$ matrix, it means any two rows differ in exactly half of the entries, which is impossible when $n$ is an odd number. When $n \equiv 2\pmod 4$, two rows that are both orthogonal to a third row cannot be orthogonal to each other. Together, these statements imply that an $n \times n$ Hadamard matrix can exist only if $n = 1, 2$, or a multiple of $4$.
Ben Grossmann
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