If $A^4$ has an eigenvalue $\mu$, prove $A$ has an eigenvalue $\lambda$ with $\lambda^4=\mu$

Question

If $A^4$ has an eigenvalue $\mu$, prove $A$ has an eigenvalue $\lambda$ with $\lambda^4=\mu$ ?

Is the following sound:

Let $\chi_{A}(t)$ by the characteristic polynomial of $A$. Then $\chi_{A^2}(t^2) = \chi_{A}(t)\chi_{A}(-t)$.

We know \begin{align} 0 &=\chi_{A^4}(\mu) \\ &=\chi_{A^2}(\sqrt{\mu})\chi_{A^2}(-\sqrt{\mu}) \\ &=\chi_{A}({\mu}^\frac{1}{4})\chi_{A}(-{\mu}^\frac{1}{4})\chi_{A}((-\sqrt{\mu})^\frac{1}{2})\chi_{A}(-(-\sqrt{\mu})^\frac{1}{2})\\ \end{align}

At least one of these must be equal to zero, from which the result follows?

Answer 1

Your argument is correct and is based on the Theorem that $\det(A B)=\det(A)\det(B)$.

You could write it out like this: \begin{equation} \chi_{A^4}(\mu)=\det(A^4-\mu \operatorname{Id}) = \det(A^2-\sqrt\mu \operatorname{Id}) \cdot \det(A^2+\sqrt\mu \operatorname{Id}) = \dots \end{equation}

In the end it indeed follows that at least one of the fourth square roots of $\mu$ is an eigenvalue of $A$.

Answer 2

I dont understand this $\chi_{A^2}(t^2) = \chi_{A}(t)\chi_{A}(-t)$. Here is a proof:

I suppose the base field is algebraically closed, there exists a basis where the matrix of the linear operator represented by $A$ is an upper triangular matrix $B$. $B^4$ is an upper triangular matrix and the elements on its diagonal are $c^4$ where $c$ is on the diagonal of $B$. There exists an invertible matrix $P$ such that $B=PAP^{-1}$.

Answer 3

The $A$ is a square matrix finite size the the argument of Maximilian Janisch works. It infinite dimensions, a slight modification of his argument works. Suppose $A$ is a bounded operator on a topological vector space $X$ and that $(A^4-\mu I)\mathbf{x}=\mathbf{0}$ for some $\mu\in\mathbb{C}$ and $\mathbf{x}\neq\mathbf{0}$. Suppose $\mu=re^{ i \theta}$. As $$(A^4-\mu I)\mathbf{x}=(A-r^{1/4}e^{ i\theta/4}I)(A+r^{1/4}e^{ i\theta/4}I)(A+ir^{1/4}e^{ i\theta/4}I)(A-ir^{1/4}e^{ i\theta/4}I)\mathbf{x}=0$$ It follows that at least of the terms in $\{r^{1/4}e^{ i (\theta+\frac{2\pi}{k})}:k=0,1,2,3\}$ is an eigenvalue of $A$.