Let us put $a=1.00001$. We compute the characteristic polynomial $\det(A-tI)$ of the given matrix $A$ as follows.
1-t & a & 1 \\
a &1-t &a \\
1 & a & 1-t
1-t & a\\
a & a\\
a & 1-t\\
by the cofactor expansion corresponding to the first row.
Simplifying this, we obtain
The eigenvalues of $A$ are roots of this characteristic polynomial.
Hence $0$ is an eigenvalue of $A$. Let $\lambda_1$ and $\lambda_2$ be other two eigenvalues of $A$.
Then we have
Therefore we have
Since $2-2a^2=2(1-a^2)<0$ as $a=1.0001>1$, the product $\lambda_1 \lambda_2$ is negative, and we conclude that one of them is positive and the other is negative.
(Note that if the constant $c$ term of a quadratic polynomial $x^2+bx+c$ is negative, then roots of the polynomial are real and one is negative and the other is positive.)
In summary, the eigenvalues of the matrix $A$ are $0$ and one positive eigenvalue and one negative eigenvalues.
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