If $\lambda$ is an eigenvalue of $A^2$, then either $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$

Since $\lambda$ is an eigenvalue of $A^2$, we know that $$\det (A^2 - \lambda I) = 0$$ From here we conclude that $$\det (A^2 - \lambda I) = \det((A - \sqrt{\lambda}I)(A + \sqrt{\lambda}I)) = \det(A - \sqrt{\lambda}I) \times\det ( A + \sqrt{\lambda}I)= 0$$ Hence $\sqrt{\lambda}$ or $-\sqrt{\lambda}$ is an eigenvalue of $A$.

First note that: $$A^2 - \lambda I = (A-\sqrt\lambda I)(A+\sqrt\lambda I)$$ Let $v$ be an eigenvector of $A^2$ with eigenvalue $\lambda$. We can use $v$ to find an explicit eigenvector of $A$ with eigenvalue that is either $\sqrt\lambda$ or $-\sqrt\lambda$.

Since $(A^2-\lambda I)v = 0$, we must have either $(A+\sqrt\lambda I)v = 0$, in which case $v$ is also an eigenvector of $A$ with eigenvalue $-\sqrt\lambda$, or $(A+\sqrt\lambda I)v \neq 0$, in which case we set $ u:=(A+\sqrt\lambda I)v$, and note that $u$ is an eigenvector of $A$ with eigenvalue $\sqrt\lambda$, since $(A-\sqrt\lambda I)u=0$.