Prove that the matrix
\[A=\begin{bmatrix}
1 & 1.00001 & 1 \\
1.00001 &1 &1.00001 \\
1 & 1.00001 & 1
\end{bmatrix}\]
has one positive eigenvalue and one negative eigenvalue.

(University of California, Berkeley Qualifying Exam Problem)

Let us put $a=1.00001$. We compute the characteristic polynomial $\det(A-tI)$ of the given matrix $A$ as follows.
We have
\begin{align*}
\det(A-tI)&=\begin{vmatrix}
1-t & a & 1 \\
a &1-t &a \\
1 & a & 1-t
\end{vmatrix}\\
&=(1-t)\begin{vmatrix}
1-t & a\\
a& 1-t
\end{vmatrix}-a\begin{vmatrix}
a & a\\
1& 1-t
\end{vmatrix}+\begin{vmatrix}
a & 1-t\\
1& a
\end{vmatrix}
\end{align*}
by the cofactor expansion corresponding to the first row.

Simplifying this, we obtain
\[\det(A-tI)=-t(t^2-3t+2-2a^2).\]

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
\[\det(A-tI)=-t(t-\lambda_1)(t-\lambda_2)=-t(t^2-(\lambda_1+\lambda_2)+\lambda_1 \lambda_2).\]
Therefore we have
\[\lambda_1 \lambda_2=2-2a^2.\]

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 the 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 eigenvalue.

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