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
 A: The rank of your matrix is $2$, which implies that $\lambda=0$ is an eigenvalue with multiplicity at least $2n+1-rank=2n-1$.
Now, if $\lambda_1, \lambda_2$ are the remaining eigenvalues, since $tr(A)$ is the sum of the eigenvalues you get $\lambda_1+\lambda_2=0$. [I assume the row and column have the same index]. This Yields $\lambda_2=-\lambda_1$.
Finally, the eigenvalues of $A^2$ are $0,0,.., 0, \lambda_1^2, \lambda_2^2$, thus 
$$
tr(A^2)=2 \lambda_1^2 \,.
$$
note that 
$$A^2 =\begin{bmatrix}
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
... & ... & ... & ...& ... & ... &... \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..& 1 & .. &0 \\
\end{bmatrix}
\begin{bmatrix}
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
0 & 0 & 0 & ..& 1 & ... &0 \\
... & ... & ... & ...& ... & ... &... \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..& 1 & .. &0 \\
\end{bmatrix}
=\begin{bmatrix}
1 & 1 & 1 & ..& 0 & ... &1 \\
1 & 1 & 1 & ..& 0 & ... &1 \\
1 & 1 & 1 & ..& 0 & ... &1 \\
... & ... & ... & ...& ... & ... &... \\
0 & 0 & 0 & ..&  2n& ... &0 \\
... & ... & ... & ...& ... & ... &... \\
 1& 1 & 1 & ..& 0 & .. &1 \\
\end{bmatrix}$$
Therefore 
$$2 \lambda_1^2=tr(A^2)=4n$$
A: 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.
A: Write down $det(A-xI)$ and you do basic determinant operation then you will find $characteristics~polynomial$ from there you can find eigen values.
Hint: The characteristics polynomial will be
$det(A-xI)=x^{2n+1}-2nx^{2n-1}$
From here you will find that eigen values of this matrix are $ 0~and~\pm \sqrt{2n}$
A: You can fully characterize the eigenvalues as follows. There are $2n-1$ zero eigenvalues; this follows because there are only two linearly independent columns. There is at least one eigenvalue of either $+\sqrt{2n}$ or $-\sqrt{2n}$. You can see this through the characterization of the SVD: the first singular value is the largest possible norm of the image of any unit vector under the matrix. This is $\sqrt{2n}$ in this situation (just look at the norms of the images of $e_1,\dots,e_{2n+1}$). Now for a symmetric matrix the singular values are the absolute values of the eigenvalues. Finally, since there's only one eigenvalue left, to have the trace be zero we must have both $\sqrt{2n}$ and $-\sqrt{2n}$ as eigenvalues.
A: Here is a much simpler way to do the calculation, if you are familiar with quadratic forms:
Consider the quadratic form $Q[x]=x^TAx$ given by this symmetric matrix.
Then 
$$Q[x]=2x_1x_j+2x_2x_j+...+2x_{j-1}x_j+2x_{j+1}x_j+..+2x_{2n+1}x_j=2(x_1+..+x_{j-1}+x_{j+1}+..+x_{2n+1})x_j \\ $$
Now, make the orthogonal change of variable
$$
y_1=\frac{1}{\sqrt{2n}} (x_1+..+x_{j-1}+x_{j+1}+..+x_{2n+1}) \\
y_2=x_j \\
y_{3, 2n+1}=\mbox{anything}$$
then your quadratic form becomes
$$Q[y]=\sqrt{2n} (2y_1y_2)=\sqrt{2n} (\frac{1}{2} (y_1+y_2)^2 -\frac{1}{2} (y_1-y_2)^2  )$$
So after another orthogonal change of variables 
$$z_1=\frac{y_1+y_2}{\sqrt{2}} \\
z_1=\frac{y_1-y_2}{\sqrt{2}} $$
Your quadratic form becomes diagonal:
$$Q[z]=\sqrt{2n} z_1^2-\sqrt{2n} z_2^2$$
