$\lambda^2$ is an eigenvalue of $A^2$. Prove that $\lambda$ or $-\lambda$ is an eigenvalue of the matrix $A$

Question

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

Answer 1

There is an $x \neq 0$ such that $A^2 x = \lambda^2 x$. This means $ (A-\lambda I)(A+\lambda I)x = 0$. So $(A-\lambda I)(A+\lambda I)$ is a singular matrix, and so one of $(A-\lambda I)$ or $(A+\lambda I)$ must be singular, and hence one of $\lambda$ or $-\lambda$ must be an eigenvalue of $A$ .

Answer 2

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 $X^2-\lambda^2$.