a. How do I prove that if $\lambda^2$ is an eigenvalues of $A^2$ then at least $\lambda$ or $-\lambda$ is an eigenvalue of $A$.
b. I want to show that an invertible linear transoformation $T$ and its inverse $T^{-1}$ have the same eigenvectors.
thank you!
Let $v$ be the eigenvector of $A^2$ corresponding to the eigenvalue ${\lambda}^2.$ Then we have $$(A^2 - \lambda^2 I) v = 0 \implies (A- \lambda I) (A + \lambda I) v = 0.$$ So either $(A+\lambda I) v = 0.$ In which case $-\lambda$ is an eigenvalue of $A$ corresponding to the eigenvector $v.$ Otherwise $$(A+ \lambda I) v \neq 0.$$ Then $\lambda$ will be an eigenvalue of $A$ corresponding to the eigenvector $(A + \lambda I)v$.
For the second one let us take an eigenvector $v$ of $T$ corresponding to the eigen value $\lambda$. Then first show that $\lambda \neq 0$ and then try to show that $v$ is also an eigen vector of $T^{-1}$ corresponding to the eigenvalue $\frac {1} {\lambda}$. The converse can also be similarly proved.