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.
Best Answer
For a symmetric Hadamard matrix $H$ of order $m$ we have the minimal polynomial $x^2-m$, i.e. eigenvalues $\pm\sqrt{m}$. Indeed, by Cayley-Hamilton theorem $HH^\top=H^2=mI$, as $HH^\top=mI$ for any Hadamard matrix.
In general, eigenvalues are not $\pm\sqrt{m}$. E.g. if I take a skew-symmetric Hadamard matrix of order 4, I get minimal polynomial $x^2-2x+4$.So you get complex eigenvalues. Actually, for any skew-symmetric order $m$ Hadamard matrix $H$ with constant diagonal (i.e. $(H-I)^\top=-H+I$) the minimal polynomial is $p(x)=x^2-2x+m$. Indeed, one gets $H=2I-H^\top$, and plugging this into $p(x)$ gives $H(2I-H^\top)-2H+mI=0$, as required.
The evidence from Sagemath was helpful: