Editing Hadamard matrix (section)

==Properties==
Let ''H'' be a Hadamard matrix of order ''n''. The [[transpose]] of ''H'' is closely related to its [[inverse matrix|inverse]]. In fact:

: <math>H H^\textsf{T} = n I_n</math>

where ''I<sub>n</sub>'' is the ''n''&thinsp;×&thinsp;''n'' [[identity matrix]] and ''H''<sup>T</sup> is the transpose of ''H''. To see that this is true, notice that the rows of ''H'' are all orthogonal vectors over the [[field (mathematics)|field]] of [[real number]]s and each have length <math>\sqrt{n}\,.</math> Dividing ''H'' through by this length gives an [[orthogonal matrix]] whose transpose is thus its inverse:

: <math>\frac{1}{\sqrt{n}} H^\textsf{T} = \sqrt{n} H^{-1} </math>

Multiplying by the length again gives the equality above. As a result,

: <math>\operatorname{det}(H) = \pm\, n^{n/2},</math>

where det(''H'') is the determinant of ''H''.

Suppose that ''M'' is a [[complex number|complex]] matrix of order ''n'', whose entries are bounded by |''M<sub>ij</sub>''| ≤ 1, for each ''i'', ''j'' between 1 and ''n''. Then [[Hadamard's inequality|Hadamard's determinant bound]] states that

: <math>|\operatorname{det}(M)| \leq n^{n/2}.</math>

Equality in this bound is attained for a real matrix ''M'' [[if and only if]] ''M'' is a Hadamard matrix.

The order of a Hadamard matrix must be 1, 2, or a multiple of 4.<ref>{{cite web|publisher=UC Denver|access-date=11 February 2023
|url=http://math.ucdenver.edu/~wcherowi/courses/m6406/hadamard.pdf|title=Hadamard Matrices and Designs}}</ref>

===Proof===
The [[mathematical proof|proof]] of the nonexistence of Hadamard matrices with dimensions other than 1, 2, or a multiple of 4 follows:

If <math>n>1</math>, then there is at least one scalar product of 2 rows which has to be 0. The scalar product is a sum of ''n'' values each of which is either 1 or −1, therefore the sum is [[parity (mathematics)|odd]] for odd ''n'', so ''n'' must be [[parity (mathematics)|even]].

If <math>n = 4 m + 2</math> with <math>m \geq 1</math>, and there exists an <math>n \times n</math> Hadamard matrix <math>H = (h_{i,j})_{i,j \in \{0,1,...,n-1\}}</math>, then it has the property that for any <math>k \neq l</math>:
:<math>\sum_{i=0}^{n-1} h_{k,i} h_{l,i} = 0</math>

Now we define the matrix <math>A = (a_{i,j})_{i,j \in \{0,1,...,n-1\}}</math> by setting <math>a_{i,j} = h_{0,j}h_{i,j}</math>.
Note that <math>A</math> has all 1s in row 0.
We check that <math>A</math> is also a Hadamard matrix:
:<math>\sum_{i=0}^{n-1} a_{k,i} a_{l,i} = \sum_{i=0}^{n-1} h_{0,j} h_{k,i} h_{0,j} h_{l,i} = \sum_{i=0}^{n-1} h_{0,j}^2 h_{k,i} h_{l,i} = \sum_{i=0}^{n-1} h_{k,i} h_{l,i} = 0.</math>

Row 1 and row 2, like all other rows except row 0, must have <math>n/2</math> entries of 1 and <math>n/2</math> entries of −1 each. (*)

Let <math>\alpha</math> denote the number of 1s of row 2 beneath 1s in row 1.
Let <math>\beta</math> denote the number of −1s of row 2 beneath 1s in row 1.
Let <math>\gamma</math> denote the number of 1s of row 2 beneath −1s in row 1.
Let <math>\delta</math> denote the number of −1s of row 2 beneath −1s in row 1.

Row 2 has to be orthogonal to row 1, so the number of products of entries of the rows resulting in 1, <math>\alpha + \delta</math>, has to match those resulting in −1, <math>\beta + \gamma</math>.
Due to (*), we also have <math>n/2 = \alpha + \gamma = \beta + \delta</math>, from which we can express <math>\gamma = n/2 - \alpha</math> and <math>\delta = n/2 - \beta</math> and substitute:

:<math>\alpha + \delta = \beta + \gamma</math>
:<math>\alpha + \frac{n}{2} - \beta = \beta + \frac{n}{2} - \alpha</math>
:<math>\alpha - \beta = \beta - \alpha</math>
:<math>\alpha = \beta</math>

But we have as the number of 1s in row 1 the odd number <math>n/2 = \alpha + \beta</math>, [[proof by contradiction|contradiction]].