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Prove that if $H$ is a (normalized) Hadamard matrix, then so is the matrix $\pmatrix{ H& H\\\ H& -H}$.

I have been working on this and I know this statement is true. My book just simply says that this is true. Does it have to do with the order of the Hadamard matrix?

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    http://math.stackexchange.com/questions/1326770/hadamard-matrices-and-sub-matrices2015-06-16

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Assume that the order of $H$ is $n$. First, it's clear that all the entries of this new matrix, say $\mathcal H$ are $-1$ of $1$. Now, select two row of this matrix:

  • First case: the indexes of these two row are between $1$ and $n$. Then computing their inner product, we can see that it can we written as two sums of $n$ terms, which are $0$ since $H$ is supposed to ba a Hadamard matrix.
  • Second case: one index is between $1$ and $n$ and the other between $n+1$ and $2n$. We can write the first row $R_1=[r_1 ,r_1]$ and the second $R_2=[r_2,-r_2]$ where $r_1$ and $r_2$ are the corresponding row in $H$. Then the inner product is $r_1\cdot r_2-r_1\cdot r_2=0$.
  • Third case: the indexes of these two row are between $n+1$ and $2n$. It's the same as the first case.
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    So why is the inner product of a row with itself = n?2012-01-28
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Let $A = \pmatrix{ H& H\\\ H& -H}$. We know that all the entries of $A$ are 1's and -1's, so that's half of the problem there. Now just calculate $AA^T$. What's wrong with this approach?