$W$ = $\begin{bmatrix} 1 & 1 & 1 & 1 \\ 1 & -1 & 1 & -1 \\ 1 & 1 & -1 & -1 \\ 1 & -1 & -1 & 1 \end{bmatrix}$. Use W to build an 8x8 matrix encoding an orthonormal basis in $R^8$ by scaling A = $\begin{bmatrix} W & W \\ W & -W \end{bmatrix}$ in the right way.
Am I wrong here, but is it not just this matrix beside each other 4 times? Doesn't that work?
On
Note that $$W^T\cdot W=4I$$ then
$$\begin{bmatrix}W&0\\0&W\end{bmatrix}^T\begin{bmatrix}W&0\\0&W\end{bmatrix}=4\begin{bmatrix}I&0\\0&I\end{bmatrix}$$
then assume
$$U=\begin{bmatrix} \frac1{\sqrt4} & \frac1{\sqrt4} & \frac1{\sqrt4} & \frac1{\sqrt4} \\ \frac1{\sqrt4} & -\frac1{\sqrt4} & \frac1{\sqrt4} & -\frac1{\sqrt4} \\ \frac1{\sqrt4} & \frac1{\sqrt4} & -\frac1{\sqrt4} & -\frac1{\sqrt4} \\ \frac1{\sqrt4} & -\frac1{\sqrt4} & -\frac1{\sqrt4} & \frac1{\sqrt4} \end{bmatrix}$$
such that $$U^T\cdot U=I$$ then
On
You have $$ A^TA=\begin{bmatrix} W^T & W^T \\ W^T & -W^T \end{bmatrix}\begin{bmatrix} W & W \\ W & -W \end{bmatrix}=\begin{bmatrix}2W^TW&0\\ 0&2W^TW \end{bmatrix}. $$ Since $W^TW=4I$, we got $A^TA=8I$. Then $$ \frac1{2\sqrt2}\,\begin{bmatrix} W & W \\ W & -W \end{bmatrix} $$ is an orthogonal matrix.
On
The matrix $W$ above is a Hadamard matrix of order four and therefore satisfies $WW^\top = W^\top W = 4I_4$.
If $$ A = \frac{1}{\sqrt{8}}\begin{bmatrix} W & W \\ W & -W \end{bmatrix}, $$ then $A$ is an orthogonal matrix.
No. First of all, $W$ is not orthonormal: its columns don't have norm $1$. Normalize it first. Call it $U$. Then consider $\left[\begin{smallmatrix}U&0\\0&U\end{smallmatrix}\right]$.