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Assuming that the Hadamard ConjectureHadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by

Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by

Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by

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TheNote that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$. 

We build matrix $\mathrm A$ by normalizing the rows

The $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$. We build $\mathrm A$

Note that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$. 

We build matrix $\mathrm A$ by normalizing the rows

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If $m$ is a power of $2$, then $\mathrm H_m$ can be built recursively using the Sylvester construction

$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_{2} = \begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}$$$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_1 = 1$$

If $m$ is a power of $2$, then $\mathrm H_m$ can be built using the Sylvester construction

$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_{2} = \begin{bmatrix} 1 & 1\\ 1 & -1\end{bmatrix}$$

If $m$ is a power of $2$, then $\mathrm H_m$ can be built recursively using the Sylvester construction

$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_1 = 1$$

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