# proof that Hadamard matrix has order 1 or 2 or 4n

Let $m$ be the order of a Hadamard matrix^{}. The matrix $[1]$ shows that order 1
is possible, and the entry has a $2\times 2$ Hadamard matrix
, so assume $m>2$.

We can assume that the first row of the matrix is all 1’s by multiplying
selected columns by $-1$. Then permute columns as needed to arrive at a
matrix whose first three rows have the following form, where $P$ denotes a submatrix^{} of one row
and all 1’s and $N$ denotes a submatrix of one row and all $-1$’s.

$$\begin{array}{cc}\hfill \begin{array}{cccc}\hfill x\hfill & \hfill y\hfill & \hfill z\hfill & \hfill w\hfill \end{array}\hfill & \hfill \begin{array}{c}\hfill \hfill \end{array}\hfill \\ \hfill \left[\begin{array}{cccc}\hfill \stackrel{\u23de}{P}\hfill & \hfill \stackrel{\u23de}{P}\hfill & \hfill \stackrel{\u23de}{P}\hfill & \hfill \stackrel{\u23de}{P}\hfill \\ \hfill P\hfill & \hfill P\hfill & \hfill N\hfill & \hfill N\hfill \\ \hfill P\hfill & \hfill N\hfill & \hfill P\hfill & \hfill N\hfill \end{array}\right]\hfill \end{array}$$ |

Since the rows are orthogonal^{} and there are $m$ columns we have

$\{\begin{array}{cc}x+y+z+w\hfill & =m\hfill \\ x+y-z-w\hfill & =0\hfill \\ x-y+z-w\hfill & =0\hfill \\ x-y-z+w\hfill & =0.\hfill \end{array}$

Adding the 4 equations together we get

$$4x=m.$$ |

so that $m$ must be divisible by 4.

Title | proof that Hadamard matrix has order 1 or 2 or 4n |
---|---|

Canonical name | ProofThatHadamardMatrixHasOrder1Or2Or4n |

Date of creation | 2013-03-22 16:50:56 |

Last modified on | 2013-03-22 16:50:56 |

Owner | Mathprof (13753) |

Last modified by | Mathprof (13753) |

Numerical id | 10 |

Author | Mathprof (13753) |

Entry type | Proof |

Classification | msc 05B20 |

Classification | msc 15-00 |