A few examples [check the definition for each] of Hadamard matrices are;
1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1These matrices were first considered as Hadamard determinants. They were so named because the determinant of an Hadamard matrix satisfies equality in Hadamard's determinant theorem, which states that if X = x_{ij} is a matrix of order n where  x_{ij}  <= 1 for all i and j, then
 det X  n^{n/2}
It is apparent that if the rows and columns of an Hadamard matrix are permuted, the matrix remains Hadamard. It is also true that if any row or column is multiplied by 1, the Hadamard property is retained.[Prove this] Thus, it is always possible to arrange to have the first row and first column of an Hadamard matrix contain only +1 entries. An Hadamard matrix in this form is said to be normalized.
Theorem V.1.1  The order of an Hadamard matrix is 1,2 or 4n, n an integer.
Proof: [1] is an Hadamard matrix of order 1 and the first example above is an Hadamard matrix of order 2. Suppose now that H is an Hadamard matrix of order h > 2. Normalize H and rearrange the first three rows to look like:
1 ..... 1 1 ..... 1 1 ..... 1 1 ..... 1 1 ..... 1 1 ..... 1 1 .....1 1 .....1 1 ..... 1 1 .....1 1 ..... 1 1 .....1 x y z wWhere x,y,z,w are the numbers of columns of each type. Then since the order is h,
x + y + z + w = h
and taking the inner products of rows 1 and 2, 1 and 3, and, 2 and 3 we get
x + y  z  w = 0
x  y + z  w = 0
x  y  z + w = 0.
Solving this system of equations gives, x = y = z = w = h/4. Thus, the integer h must be divisible by 4.
Corollary V.1.2  If H is a normalized Hadamard matrix of order 4n, then every row (column) except the first has 2n minus ones and 2n plus ones, further n minus ones in any row (column) overlap with n minus ones in each other row (column).
Proof: This is a direct result of the above proof since any two rows other than the first can take the place of the second and third rows in the proof. The same arguement can be applied to the columns.
Hadamard matrices are known for many of the possible orders, the smallest order for which the existence of an Hadamard matrix is in doubt is currently 668 (A solution for the previous unknown case of 428 was announced by Kharaghani and TayfehRezaie in June 2004).
Construction V.2.1  Given Hadamard matrices H_{1} of order n and H_{2} of order m the direct product of these two matrices, represented by:
h_{11}H_{2}  h_{12}H_{2}  ...  h_{1n}H_{2} 
h_{21}H_{2}  h_{22}H_{2}  ...  h_{2n}H_{2} 
...  ...  ...  ... 
h_{n1}H_{2}  h_{n2}H_{2}  ...  h_{nn}H_{2} 
Proof: [Left as an exercise].
Example:
Let




H*  H* 
H*  H* 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
1  1  1  1  1  1  1  1 
Homework: Starting with the Hadamard matrix of order 2, repeatedly use the direct product construction to construct an Hadamard matrix of order 32.
Wallis,Street and Wallis, Combinatorics: Room Squares, SumFree Sets, Hadamard Matrices, (Lecture Notes in Mathematics # 292) SpringerVerlag, New York, 1972.
However, this treatment is very detailed and concerned primarily with the problems of construction. A chapter of Hall, M., Combinatorial Theory, Blaisdell, Waltham, Mass., 1967, is devoted to Hadamard Matrices.