Follows from the two laws

(A o B) (C o D) = AC o BD

and

(A o B)^T = A^T o B^T.

(The proofs of these laws are straightforward, and left to the reader.) Indeed,

(A o B) (A o B)^T = (A o B) (A^T o B^T) = AA^T o BB^T = mI_m o nI_n = mnI_mn.

Let F be a field of order q with primitive element g.

Put n = q + 1. For the construction of the Hadamard matrix we use the map r defined by:

r : F -> C,     0 -> 0,   gⁱ -> (-1)ⁱ.

We construct a square matrix C of dimension n as follows:

C =

0 1 1 ·  ·  · 1 r(-1)   ·   · Q ·   r(-1)

where Q is the square matrix with Q_y,z = r(y - z). The rows and columns are indexed by the elements of F.

We claim that the matrix H = I_n + C is a Hadamard matrix. The proof is divided into a two steps.

Some properties of r.

Check that H is indeed a Hadamard matrix.

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We use the same matrix C as in Part 2. Recall

CC^T = qI_n    and    C^T = r(-1)C.

Now the matrix H of dimension 2n = 2(q + 1), given by

H =

I + C   -I + C -I + C   -I - C

is a Hadamard matrix. The verification is easy, and uses r(-1) = 1.

Let F be a field of order q with primitive element g.

Put n = q + 1. For the construction of the Hadamard matrix we use the map r defined by:

r : F -> C,     0 -> 0,   gⁱ -> (-1)ⁱ.

We construct a square matrix C of dimension n as follows:

C =

0 1 1 ·  ·  · 1 r(-1)   ·   · Q ·   r(-1)

where Q is the square matrix with Q_y,z = r(y - z). The rows and columns are indexed by the elements of F.

We claim that the matrix H = I_n + C is a Hadamard matrix. The proof is divided into a two steps.

Some properties of r.

Check that H is indeed a Hadamard matrix.

---

The map r has the following properties.

1. r(-1) = (-1)^(q-1)/2.
2. r(xy) = r(x)r(y) for all x, y F.
3. _xF r(x) = 0.
4. If c0, then _xF r(x) r(x + c) = -1.

Let g be a primitive element of F. Then -1 = g^(q-1)/2, so -1 is a square if (q-1)/2 is even, that is, q = 1 mod 4, and a nonsquare if (q-1)/2 is odd, that is, q = 3 mod 4. Hence r satisfies 1.

The second property of r easily follows from the facts that

the product of square and a nonsquare is a nonsquare,

the product of nonsquare and a nonsquare is a square,

and the product of square and a square is a square, etc.

The third statement follows from the fact that the nonzero squares form a subgroup of the multiplicative group of F of index 2 (and order (q-1)/2).

For, this implies that there are just as many nonzero squares as there are nonsquares.

Finally the last statement follows from

xF r(x) r(x+c) =	xF\{0} r(x) r(x+c)
=	xF\{0} r(x-1) r(x+c)
=	xF\{0} r(1+x-1c)
=	yF\{1} r(y)
=	-1 + yF r(y)
=	-1,

by Part 3.

Let F be a field of order q with primitive element g.

Put n = q + 1. For the construction of the Hadamard matrix we use the map r defined by:

r : F -> C,     0 -> 0,   gⁱ -> (-1)ⁱ.

We construct a square matrix C of dimension n as follows:

C =

0 1 1 ·  ·  · 1 r(-1)   ·   · Q ·   r(-1)

where Q is the square matrix with Q_y,z = r(y - z). The rows and columns are indexed by the elements of F.

We claim that the matrix H = I_n + C is a Hadamard matrix. The proof is divided into a two steps.

Some properties of r.

Check that H is indeed a Hadamard matrix.

---

We claim

CC^T = qI_n    and    C^T = r(-1)C.

For, the inner product of the rows corresponding to y and z equals

1+ _vF r(y - v) r(z - v) = 1 + _vF r(v) r(z - y + v) ,

which equals 0 if y z in view of Property 3 of r, and equals q if y = z. This establishes the first claim.

The second claim follows from r(y - z) = r(-1)r(z - y).

Now, since r(-1) = -1 , H = I_n + C is a Hadamard matrix:

HH^T = I_n + C + C^T + CC^T = I_n + 0 + qI_n = nI_n.