# $\lambda^2$ 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

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 $$A$$ .
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 $$X^2-\lambda^2$$.