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.

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?

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:

 sage: from sage.combinat.matrices.hadamard_matrix import skew_hadamard_matrix
x^2 - 2*x + 4
x^2 - 2*x + 8
x^2 - 2*x + 12
x^2 - 2*x + 16

• sorry. In symmetric case I have this same argument with me. actually my question is about the general Hadamard matrix. – GA316 Mar 10 '16 at 16:57
• well, see my edit... – Dima Pasechnik Mar 10 '16 at 17:11
• in general, one cannot hope for a very nice answer, as one can multiply $H$ by an arbitrary $\pm 1$-permutation matrix, this does not break Hadamard property, but does change eigenvalues. – Dima Pasechnik Mar 10 '16 at 17:43
• oh. so in general the result is not true. I was expecting it to be for my own purposes. any way thanks a lot. – GA316 Mar 10 '16 at 17:48

The question has already been answered, but I thought a bit of illustration were good.

Here is a short list of minimal polynomials of Hadamard-matrices of order $$8$$ (without symmetry, skew-symmetry-properties or any special structure). Between $$100$$ random initializations of my searching-procedure I found only one time the minimal polynomial $$x^8+4096$$ . The exemplar that I'd found had also eigenvalues as $$[1,3,5,7,9,11,13,15]$$ 'th powers of $$\omega_{16}=\sqrt[8]{-1}$$

Here a short (sorted) list of minimal polynomials, factored using Pari/GP. This looks like highly arbitrary/combinatorical:

(x^8 - 8*x^6 + 16*x^4 - 512*x^2 + 4096)
(x^8 - 6*x^7 + 16*x^6 - 48*x^5 + 160*x^4 - 384*x^3 + 1024*x^2 - 3072*x + 4096)
(x^8 + 4096)
(x^8 + 4*x^7 + 8*x^6 + 16*x^5 + 16*x^4 + 128*x^3 + 512*x^2 + 2048*x + 4096)
(x^8 + 4*x^7 + 4*x^6 - 32*x^5 - 160*x^4 - 256*x^3 + 256*x^2 + 2048*x + 4096)
(x^8 + 4*x^7 + 4*x^6 - 24*x^5 - 144*x^4 - 192*x^3 + 256*x^2 + 2048*x + 4096)
(x^8 - 4*x^7 + 4*x^6 + 16*x^5 - 96*x^4 + 128*x^3 + 256*x^2 - 2048*x + 4096)
(x^8 + 4*x^7 + 4*x^6 - 16*x^5 - 96*x^4 - 128*x^3 + 256*x^2 + 2048*x + 4096)
(x^8 + 4*x^7 - 32*x^5 - 112*x^4 - 256*x^3 + 2048*x + 4096)
(x^8 - 4*x^7 + 16*x^5 - 32*x^4 + 128*x^3 - 2048*x + 4096)
(x^8 - 4*x^6 + 8*x^5 - 16*x^4 + 64*x^3 - 256*x^2 + 4096)
(x^8 + 4*x^6 + 64*x^4 + 256*x^2 + 4096)
(x^8 - 4*x^6 - 256*x^2 + 4096)
(x^8 - 4*x^6 - 16*x^5 + 32*x^4 - 128*x^3 - 256*x^2 + 4096)
...
(x^4 - 4*x^3 + 8*x^2 - 32*x + 64).(x^4 - 8*x^2 + 64)
(x^4 - 4*x^3 + 8*x^2 - 32*x + 64).(x^4 + 4*x^3 + 8*x^2 + 32*x + 64)
(x^4 - 4*x^3 + 8*x^2 - 32*x + 64).(x^4 + 2*x^3 - 4*x^2 + 16*x + 64)
(x^4 + 2*x^3 + 16*x + 64)^2
(x^4 - 2*x^3 - 16*x + 64).(x^4 + 2*x^3 + 16*x + 64)
...
(x^2 - 8)^2.(x^4 - 6*x^3 + 20*x^2 - 48*x + 64)
(x^2 - 8)^2.(x^4 - 4*x^3 + 12*x^2 - 32*x + 64)
(x^2 - 8)^2.(x^2 + 8).(x^2 + 4*x + 8)
(x^2 - 8)^2.(x^2 + 2*x + 8)^2
...
(x^2 - 2*x + 8).(x^2 - 8)^2.(x^2 + 4*x + 8)
(x^2 - 2*x + 8).(x^2 - 8)^2.(x^2 + 2*x + 8)
...