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!

## Best Answer

It's not so difficult,

$$ 0 = \text{det}(A^2 - \lambda^2I) = \text{det}( (A - \lambda I) (A + \lambda I)) = \text{det}(A - \lambda I) \text{det}(A + \lambda I). $$

Therefore, either $\lambda$ or $-\lambda$ is an eigenvalue of $A$.