# Simple Proof Regarding Eigenvalues of a Square of a Matrix

eigenvalues-eigenvectorslinear algebraminimal-polynomials

Good day! Last week, I was presented with the following short question:

Given a square matrix $$A$$ s.t. $$\lambda^2$$ is an eigenvalue of $$A^2$$, show that $$\lambda$$ or $$-\lambda$$ must be an eigenvalue of $$A$$.

I started simply by looking at what I am able to deduce about the minimal polynomial of $$A^2$$.
$$m_{A^2}(t) = (t – \lambda^2)p(t)$$
Where $$m_{A^2}(t)$$ denotes the minimal polynomial of $$A^2$$ and $$p(t)$$ is some polynomial such that $$(t – \lambda^2)$$ and $$p(t)$$ are coprime (or else $$\lambda^2$$ would have a greater than 1 algebraic multiplicity as an eigenvalue).
This definition of the minimal polynomial of $$A^2$$ can also be seen as:
\begin{align*} & q(t) = m_{A^2}(t) \circ (t^2) = (t^2 – \lambda^2)p(t^2) = (t – \lambda)(t + \lambda)p(t^2) \\ & \text{Where we have } q(A) = 0 \end{align*}
As $$q(A) = 0$$, it must be true that $$m_A(t) | q(t)$$. This is sort of where I got stumped.
It doesn't seem as any of the above statements allows one to claim that $$\lambda$$ or $$-\lambda$$ is a root of $$m_A(t)$$. I am not too sure if my proof is missing a key idea or if I have chosen a wrong path…

Sincerest thanks to any readers! Any hints or ideas would be greatly welcomed.
Have a nice day, evening or night!

$$0 = \text{det}(A^2 - \lambda^2I) = \text{det}( (A - \lambda I) (A + \lambda I)) = \text{det}(A - \lambda I) \text{det}(A + \lambda I).$$
Therefore, either $$\lambda$$ or $$-\lambda$$ is an eigenvalue of $$A$$.