Relationship between eigenvalues of a matrix and its square

Are there any general relations between the eigenvalues of a matrix $M$ and those of $M^2$?

Of course if $\lambda$ is an eigenvalue of $M$, then $\lambda^2$ is an eigenvalue of $M^2$: $$M v =\lambda v\quad\Rightarrow\quad M^2v=M\lambda v=\lambda M v =\lambda^2 v.$$

If the eigenvalues are distinct, then the square matrix $A$ is diagonalizable, namely $$A = Q^{-1}DQ.$$

Then, $$A^2 = (Q^{-1}DQ)^2 = Q^{-1}DQQ^{-1}DQ = Q^{-1}D^2Q.$$

The diagonal entries of $D^2$ are the diagonal entries of $D$, squared.

A useful way to view an eigenspace is that the matrix $M$ just becomes multiplication on the eigenspace.

So, suppose $Mv = \lambda v$. Applying $M$ again just results in another multiplication by $\lambda$, as in $M(M v) = M (\lambda v) = \lambda M v = \lambda^2 v$. Repeating gives $M^kv = \lambda^k v$.

If $p$ is a polynomial, say $p(x) = \sum p_k x^k$, and we let $p(M) = \sum p_k M^k$, then we have $p(M)v = \sum p_k M^k v = \sum p_k \lambda^k v = p(\lambda) v$. So, if $\lambda$ is an eigenvalue of $M$, then $p(\lambda)$ is an eigenvalue of $p(M)$.

The result is true more generally, but that is out of scope here.

If $M$ happened to be invertible, then multiplying an eigenvector by $M^{-1}$ becomes dividing by $\lambda$ (which must be non-zero since I'm assuming invertibility). This follows from $M^{-1} Mv = v = M^{-1} (\lambda v)$, which gives $M^{-1} v = \frac{1}{\lambda} v$. Repeating gives $M^{-k} v = \frac{1}{\lambda^k} v$, so the formula $M^kv = \lambda^k v$ holds for all integers (assuming invertibility).

Clearly, if $\underline M(u) = \lambda u$, then $\underline M^2(u) = \lambda^2 u$.

However, $\underline M^2$ may have eigenvectors that weren't eigenvectors of the original linear operator. Rotation operators come to mind. A 90 degree rotation has no eigenvectors*, but chaining two of these together makes all vectors in the plane of rotation eigenvectors with eigenvalue $-1$.

*Edit: in the plane of rotation.