Why does the spectral radius satisfy $r(A^{2})=r(A)^{2}$?

Question

I am unsure about one direction of the equality: $r(A^{2})=r(A)^{2}$ where $A$ is $n\times n$ dimensional matrix.

It is clear that for any Eigenvalue $\lambda$ of $A$, and an associated eigenvector $v$, we have that:

$Av=\lambda v\Rightarrow A^{2}v=\lambda^{2}v$ and hence $\sigma(A)^{2}\subseteq \sigma (A^{2})$ where $\sigma (A)$ is the spectrum of $A$ but I am unsure how to show $\sigma(A^{2})\subseteq \sigma (A)^{2}$. Surely this is necessary in order to show that $r(A^{2})=r(A)^{2}$?

Answer 1

Let $A^2v = \lambda v$ for some vector $v \ne 0$. Then $(A-\lambda I)v = 0$.

If $\omega, -\omega$ are the complex square roots of $\lambda$, we have $$0 = (A-\lambda I)v = (A-\omega I)(A+\omega I)v$$ If $(A+\omega I) v = 0$, then $Av = -\omega v$ so $-\omega$ is an eigenvalue of $A$. On the other hand, if $(A+\omega I) v \ne 0$, then $$A\big((A+\omega I)v\big) = \omega \big((A+\omega I)v\big)$$ so $\omega$ is an eigenvalue of $A$.

In either case, $\lambda = (\pm\omega)^2$ so $\lambda$ is a square of an eigenvalue of $A$, or $\lambda \in \sigma(A)^2$.