A (±1)-matrix is a matrix whose entries are 1 and −1. An $n \times n$ (±1)-matrix is called an Hadamard matrix if the rows are orthogonal.
Equivalently, An $n \times n$ (±1)-matrix $H$ is Hadamard ⇔ $H H^t = nI_n$, where $I_n$ denotes the $n \times n$ identity matrix.
In this paper : http://www.sciencedirect.com/science/article/pii/0024379582902105
A note on the eigenvectors of Hadamard matrices of order $2^n$ R. Yarlagadda J. Hershey
we have the result that, If the order of $H$, is $2^n$, then its $2^n$ eigenvalues as follows:
$2^{n - 1}$ eigenvalues are $2^{\frac{n}{2}}$
$2^{n - 1}$ eigenvalues are $- 2^{\frac{n}{2}}$
This result says that $H_n$ has $\frac{n}{2}$ eigen values equal to $n^{\frac{1}{2}}$ and $\frac{n}{2}$ eigenvalues equalt to $- n^{\frac{1} {2}}$ if $n = 2^n$.
My questions are,
1) Is this result true for any Hadamard matrix of order $n$ (this $n$ is any multiple of 4 for which we know there is a Hadamard matrix of that order) ?
2) What is best known result regarding their eigen values?
Thanks for your valuable timing.
In general, if $H$ is an $n\times n$ Hadamard matrix, then $A:=\frac{1}{\sqrt{n}}H$ is an orthogonal matrix, which implies that if $$ Av=\lambda v, $$ then $$ v^*A^*=\lambda^* v^*, $$ where the superscript "$*$" denotes Hermitian adjoint, and hence that $$ v^*v=v^*A^*Av=\lvert\lambda\rvert^2v^*v. $$ From this it follows that the eigenvalues of $A$ have modulus $1$ and therefore that the eigenvalues of $H$ have modulus $\sqrt{n}$.
It is not the case, however, that the eigenvalues of any Hadamard matrix are $\pm\sqrt{n}$ in equal number. If you take the matrix $H_n$ of the article you reference, constructed by taking repeated Kronecker products of the matrix $$ H_1=\begin{bmatrix} 1 & 1\\ 1 & -1 \end{bmatrix}, $$ and, for example, permute some of the rows of the matrix, the resulting matrix may have complex eigenvalues.
The Hadamard matrix $$ H=\begin{bmatrix} 1 & -1 & 1 & -1\\ 1 & -1 & -1 & 1\\ 1 & 1 & -1 & -1\\ 1 & 1 & 1 & 1 \end{bmatrix}, $$ for instance, has four distinct eigenvalues of the form $\pm\sqrt{2}\pm\sqrt{2}i$.