Finding the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$? 
Q. What is the degree of minimal polynomial of a $10 \times 10$ matrix with entries $a_{ij}=1-(-1)^{i+j}$?

My approach : Let the matrix be denoted by $A$. Then $$A=
    \left(\begin{matrix}
    0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
    2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
     0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 \\
2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 & 2 & 0 \\
    \end{matrix}\right)
$$
Some observations follow : $\text {Rank} A=2$, $\det A = 0$ and $\text {tr} A=0$.
Since $\det A=0$, there exists an eigenvalue which is $0$.
Since $\text {Rank} A=2$, number of non-zero eigenvalues can not exceed $2$.
Since $\text {tr} A=0$, number of non-zero eigenvalues are either $2$ or $0$.
Case 1. Number of non-zero eigenvalues is $2$: In this case they must be $\lambda$ and $-\lambda$ for trace should be zero. Thus the matrix is forced to be diagonalizable. This implies that the minimal polynomial is $P(x)=x(x-\lambda)(x+\lambda).$
Case 2. Number of non-zero eigenvalues is  $0$: Here all diagonal entries of the Jordan form are $0$. We have either one $3\times 3$ jordan block or two $2\times 2$ Jordan blocks in order to preserve rank of $A$. Thus the minimal polynomial is $P(x)=x^3$ or $P(x)=x^2$ respectively.
However I checked that for even order matrices with order $\gt 2$ we have eigenvalues of the type we found in Case 1 above. That means for $\text {even $\times$ even}$ matrices, eigenvalues are always $0,\lambda,-\lambda$ for some $\lambda \neq 0$. How do I show this in order to discard the case 2 entirely?

Check that $(1, 1, 1, 1, 1, 1, 1, 1, 1, 1)$ is an eigenvector with eigenvalue $10$.
