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  "body": "\n## Definition\n\nLet $m \\ge 0$ and $N = 2^m$.\nLet $H_0 = (1)$, and for $m \\ge 1$ let $H_m = H_1 \\otimes H_{m-1}$, where $\\otimes$ is the [[Kronecker product|0N7ENQRCXN]] and\n$$H_1 = \\begin{pmatrix} 1 \u0026 1 \\\\ 1 \u0026 -1 \\end{pmatrix}.$$\nThe *Walsh–Hadamard transform* of a vector $x \\in \\C^N$ is the vector $H_m x \\in \\C^N$.\nIndex rows and columns by $0, \\dots, N-1$.\nThen $(H_m)_{jk} = (-1)^{j \\cdot k}$, where $j \\cdot k$ counts the positions where the binary expansions of $j$ and $k$ both have the digit $1$.\n\n## Remarks\n\n$H_m$ is a symmetric [[Hadamard matrix|J0N4J606X0]] of order $N$, so $H_m^2 = N I$ and the inverse transform is $x = N^{-1} H_m (H_m x)$.\nMany authors use the normalized matrix $N^{-1/2} H_m$, which is orthogonal and its own inverse.\nWriting $x$ as two halves $x = (x', x'')$, one has $H_m x = (H_{m-1}(x' + x''), H_{m-1}(x' - x''))$; iterating this gives the fast transform with $m N$ additions and subtractions.\n\nExample: for $m = 2$,\n$$H_2 = \\begin{pmatrix} 1 \u0026 1 \u0026 1 \u0026 1 \\\\ 1 \u0026 -1 \u0026 1 \u0026 -1 \\\\ 1 \u0026 1 \u0026 -1 \u0026 -1 \\\\ 1 \u0026 -1 \u0026 -1 \u0026 1 \\end{pmatrix},$$\nand $x = (1, 0, 0, 0)$ goes to $(1, 1, 1, 1)$, while $x = (1, 1, 1, 1)$ goes to $(4, 0, 0, 0)$.\nCheck of one entry: $(H_2)_{3,3} = (-1)^{2} = 1$, since $3 = 11_2$ shares two digits $1$ with itself.\n\nNon-example: the discrete Fourier transform matrix $(e^{-2\\pi i jk/4})_{j,k}$ of order $4$ has entries $\\pm i$, so it is not $H_2$, although its rows are also orthogonal.\n",
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