# 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.

 $\begin{matrix}\begin{matrix}x&\quad y&\quad z&\quad w\end{matrix}&\begin{% matrix}\end{matrix}\\ \left[\begin{matrix}\overbrace{P}&\overbrace{P}&\overbrace{P}&\overbrace{P}\\ P&P&N&N\\ P&N&P&N\\ \end{matrix}\right]\end{matrix}$

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

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

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 ProofThatHadamardMatrixHasOrder1Or2Or4n 2013-03-22 16:50:56 2013-03-22 16:50:56 Mathprof (13753) Mathprof (13753) 10 Mathprof (13753) Proof msc 05B20 msc 15-00