What are the relations between the eigenvalues and eigenvectors of a Hermitian matrix $A$ and its square $A^2$?
I do know that if $(\lambda, v)$ is an eigenpair of $A$, then ($\lambda^2$, $v$) is an eigenpair of $A^2$, but what of the converse? For instance, are all eigenvectors of $A^2$ eigenvectors of $A$? And, if $(\xi, v)$ is an eigenpair of $A^2$, then must one of $(\sqrt \xi, v)$ or $(-\sqrt \xi, v)$ be an eigenpair of $A$?
Clearly what you propose cannot be true. A Hermitian operator $T$ is always diagonalisable, and as you remarked the eigenspace for $\lambda\in\Bbb R$ of $T$ is contained in the eigenspace for $\lambda^2$ of $T^2$. The sum of all the eigenspaces of $T$ is the whole space, so one easily sees that for every $\mu\in\Bbb R_{\geq 0}$ the eigenspace of $T^2$ for $\mu$ is the sum of the eigenspaces of $T$ for all eigenvalues $\lambda$ with $\lambda^2=\mu$. But if for some $\mu>0$ there are two (opposite) such values $\lambda$, then each eigen$\,$vector of $T^2$ for $\mu$ does not have to be either an eigenvector of $T$ for $\sqrt\mu$ or an eigenvector for $-\sqrt\mu$, as it can also be a sum of two eigenvectors.
An easy counterexample is $A=\pmatrix{1&0\\0&-1}$, where $(1,\pmatrix{2\\3})$ is an eigenpair for $A^2=I$, but $\pmatrix{2\\3}$ is not an eigenvector of $A$.
