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Hadamard matrix H8

Short answer

Using Sylvester construction, H8 = [H4 H4; H4 -H4] in ±1 form, with H4 = [1 1 1 1; 1 -1 1 -1; 1 1 -1 -1; 1 -1 -1 1]. Rows are mutually orthogonal, so H8 H8^T = 8I.

Construct Hadamard matrix for H8.

A Hadamard matrix of order N is a square matrix whose rows are mutually orthogonal. It is constructed recursively by the Sylvester method:

H₂ₙ = [ Hₙ Hₙ ; Hₙ −Hₙ ] in ±1 form, or H₂ₙ = [ Hₙ Hₙ ; Hₙ Hₙ′ ] in 0/1 form, where Hₙ′ is the complement of Hₙ.

Step 1: H₁ = [+1]

Step 2:

 H2 = | +1 +1 |
      | +1 -1 |

Step 3:

 H4 = | H2  H2 | = | +1 +1 +1 +1 |
      | H2 -H2 |   | +1 -1 +1 -1 |
                   | +1 +1 -1 -1 |
                   | +1 -1 -1 +1 |

Step 4:

 H8 = | H4  H4 |
      | H4 -H4 |

      | +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 |

In binary form (+1 → 0, −1 → 1):

         c1 c2 c3 c4 c5 c6 c7 c8
 W0:      0  0  0  0  0  0  0  0
 W1:      0  1  0  1  0  1  0  1
 W2:      0  0  1  1  0  0  1  1
 W3:      0  1  1  0  0  1  1  0
 W4:      0  0  0  0  1  1  1  1
 W5:      0  1  0  1  1  0  1  0
 W6:      0  0  1  1  1  1  0  0
 W7:      0  1  1  0  1  0  0  1

Check: the inner product of any two different rows is 0 (e.g. row 2 · row 3 = 1 − 1 − 1 + 1 + 1 − 1 − 1 + 1 = 0), and H8·H8ᵀ = 8·I. These 8 rows are the Walsh codes used for channelization in CDMA.

✓ Answer checked by Claude Opus 5.5 (high effort)

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