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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Does it have to do with the fact that the first row and first column but have positive 1's? – Jackson Hart Jan 28 '12 at 19:54
en.wikipedia.org/wiki/… This could also enlighten you. :) – 000 Jan 28 '12 at 20:14
You've asked a lot of questions in a very brief span. That's usually a sign of someone who needs more help than what m.se can provide. Better to seek out someone locally who can explain the math to you. – Gerry Myerson Jan 28 '12 at 22:22

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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What exactly is meant by indexes? – Jackson Hart Jan 28 '12 at 20:07
By "index $k$", I meant the $k$-th row. – Davide Giraudo Jan 28 '12 at 20:08
The inner product is always just r1*r2 - r1*r2? And we just do matrix multiplication like normal? – Jackson Hart Jan 28 '12 at 20:09
You also said that "we can see that it can be written as two sums of n terms, which are 0". I am not quite sure how to show this is 0. – Jackson Hart Jan 28 '12 at 20:11
The row are $[r_1,r_1]$ and $[r_2,r_2]$ so the inner product is $r_1r_2+r_1r_2$. $r_1r_2$ is a sum of $n$ terms and is equal to $0$ since $H$ is a Hadamard matrix. – Davide Giraudo Jan 28 '12 at 20:21

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?

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