Eigenvalues of different symmetric $(2n+1)\times(2n+1)$ matrix 
I ve looked at other similar post but I could not find help with them

I have not seen a solution that also give the eigenvectors, so here it goes:
We use the standard notation of $e_k$ being a vector in $\mathbf{R}^{2n+1}$ with one on position $k$ and zeros elsewhere.
It is straight forward to show that the vectors 
$$e_1-e_k$$
where $k\not\in\{1,n+1\}$ give $2n-1$ eigenvectors corresponding to eigenvalue zero. Let us give two more eigenvectors. A moment of thought suggests vectors of the form
$$
(1,1,\ldots,1,a,1,\ldots,1,1),
$$
where the $a$ is on position $n+1$. The image of such a vector is
$$
(a,a,\ldots,a,2n,a,\ldots,a,a),
$$
so we find that this is an eigenvector precisely if
$$
a/1=2n/a,
$$
that is if and only if $a=\pm\sqrt{2n}$. Also, in that case the eigenvalue is given by $a$.
Sum up:
Zero is an eigenvalue of multiplicity $2n-1$ and $\pm\sqrt{2n}$ are eigenvalues of multiplicity one.
