Hadamard Matrix

Question

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?

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
http://en.wikipedia.org/wiki/Hadamard_matrix#Sylvester.27s_construction
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
http://math.stackexchange.com/questions/1326770/hadamard-matrices-and-sub-matrices — James Smithson, Jun 16 '15 at 11:33

score 3 · Accepted Answer · answered Jan 28 '12 at 20:03

3

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.

answered Jan 28 '12 at 20:03

Davide Giraudo

181,608

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 r1r2 - r1r2? 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
You said the inner product was 2 different things. Earlier you said it was r1⋅r2−r1⋅r2=0. Now you say it is r1r2+r1r2. r1r2 – Jackson Hart Jan 28 '12 at 20:23
$r_1r_2-r_1r_2$ is for the second case, whereas $r_1r_2+r_1r_2$ is for the first and third one. – Davide Giraudo Jan 28 '12 at 20:26
How do we know r1r2+r1r2 = 0? How do you do this multiplication? – Jackson Hart Jan 28 '12 at 20:29
It's not really a multiplication, but an inner product. $r_1\cdot r_2=0$ since we assume that two different lines of $H$ are orthogonal. – Davide Giraudo Jan 28 '12 at 20:32
So why is the inner product of a row with itself = n? – Jackson Hart Jan 28 '12 at 20:35

score 2 · Answer 2 · answered Jun 23 '14 at 23:52

2

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?

answered Jun 23 '14 at 23:52

Paul Hurst

943

Hadamard Matrix

2 Answers2