If $\lambda$ is in spectrum of bounded linear operator $A$, then $\lambda^n$ is in spectrum of $A^n$.

Question

If $\lambda$ is in spectrum of $A$ then for all $n \in N$, $\lambda^n$ is in spectrum of $A^n$.

I try to solve this but I could only solve in the case point spectrum assuming $\lambda$ is an eigenvalue. But I know that it is incomplete.

Any hint is appreciated.

Answer 1

Simply applying the definitions, you want to prove that if $\lambda I - A$ is not invertible, then $\lambda^n I - A^n$ is not invertible. Contrapositively: if $\lambda^n I - A^n$ has an inverse, then so does $\lambda I - A$.

Progress can be made just by generalizing the usual factorization $$ x^n - y^n = (x-y)(x^{n-1} + x^{n-2}y + \dots + x y^{n-2} + y^{n-1}) $$ to a factorization of linear operators $$ \lambda^n I - A^n = (\lambda I - A)(\lambda^{n-1}I + \lambda^{n-2}A + \dots + \lambda A^{n-2} + A^{n-1}) $$ which is easy to check just by distributing and cancelling. (Note that $\lambda I$ always commutes with $A$.)

From here, it's easy to take an inverse of $\lambda^n I - A^n$ and use it to construct an inverse of $\lambda I - A$.