Let $A$ be an $n \times n$ matrix. Suppose that the matrix $A^2$ has a real eigenvalue $\lambda>0$. Then show that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of the matrix $A$.

Use the following fact: a scalar $\lambda$ is an eigenvalue of a matrix $A$ if and only if
\[\det(A-\lambda I)=0.\]

Proof.

Since $\lambda$ is an eigenvalue of $A^2$, the determinant of the matrix $A^2-\lambda I$ is zero, where $I$ is the $n \times n$ identity matrix:
\[\det(A^2-\lambda I)=0.\]

Now we have the following factorization.
\begin{align*}
A^2-\lambda I=(A-\sqrt{\lambda} I)(A+\sqrt{\lambda} I).
\end{align*}
Taking the determinant of both sides, we obtain
\begin{align*}
0&=\det(A^2-\lambda I)=\det \left((A-\sqrt{\lambda} I)(A+\sqrt{\lambda} I)\right)\\
&=\det(A-\sqrt{\lambda} I)\det(A+\sqrt{\lambda} I)
\end{align*}
by the multiplicative property of the determinant.

Therefore we have either
\[\det(A-\sqrt{\lambda} I)=0 \text{ or } \det(A+\sqrt{\lambda} I)=0\]
Thus we conclude that either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of the matrix $A$.

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