$\lambda^2$ is an eigenvalue of $A^2$. Prove that $\lambda$ or $-\lambda$ is an eigenvalue of the matrix $A$ Let $A$ be a $n\times n$ real matrix. Let $\lambda \in \mathbb{R}$ such that $\lambda^2$ is an eigenvalue of the matrix $A^2$. Prove that $\lambda$ or $-\lambda$ is an eigenvalue of the matrix  $A$.
I know how to prove the converse (and there are multiple threads regarding it), but I'm not sure how to show the other direction
I have:$$$$
$$\begin{gather}
    A^2\bar{x}=AA\bar{x}=\lambda^2\bar{x} \\ 
   \text{since if}\:A \:\text{and}\:B\: \text{share eigenvectors, so does} \: AB,\: \text{we let}\: A\bar{x}=\mu\bar{x}\Rightarrow A^{-1}\bar{x}=\frac{1}{\mu}\bar{x}, \\
A^{-1}AA\bar{x}=A\bar{x}=\frac{1}{\mu}\lambda^2\bar{x} \\
\Rightarrow  \frac{1}{\mu}\lambda^2=\mu \Rightarrow \lambda^2=\mu^2 \\
\therefore\mu=\pm\lambda
\end{gather}$$
Here I'm using the fact that if $A\bar{x}=k\bar{x},\:\text{then}\: A^2\bar{x}=c\bar{x}$ but I feel like I will need the converse of that statement to make my proof valid

Let $x$ be the eigenvector associated to $\lambda^2$, consider $V=Vect(x,A,(x))$ it is stable by $A$. The matrix of $A$ in $\{x,A(x)\}$ is:
$\pmatrix{0&\lambda^2\cr 1&0}$ this implies that the characteristic polynomial of the restriction of $A$ to $V$ is $X^2-\lambda^2$.
