How does one show that the Sylvester construction of Hadamard matrices. i.e. if $A$ is a Hadamard matrix of size $n$ then you can make a matrix of size $2n$ with effectively "$3$ $A$'s and $1$ negative $A$" that is also Hadamard?
I'm sure there is a result I can use either from the definition of Hadamard matrices or the $$\det(A)=nI$$ where A is a matrix of size n and $I$ is the identity matrix of size n
Any help would be appreciated
Suppose $A$ consists of only $1$ and $-1$, then $-A$ also consists of only $1$ and $-1$.
Let $B= \begin{bmatrix} A & A \\ A & - A\end{bmatrix}$, hence $B$ consists of only $1$ and $-1$.
We have $AA^T= nI_n$.
\begin{align} BB^T&= \begin{bmatrix} A & A \\ A & - A\end{bmatrix}\begin{bmatrix} A & A \\ A & - A\end{bmatrix}^T\\ &=\begin{bmatrix} A & A \\ A & - A\end{bmatrix}\begin{bmatrix} A^T & A^T \\ A^T & - A^T\end{bmatrix}\\ &=\begin{bmatrix} 2AA^T &0 \\ 0 & 2AA^T\end{bmatrix}\\ &=2n\begin{bmatrix} I_n &0 \\ 0 & I_n\end{bmatrix} \end{align}