Given a diagonalizable matrix $M$ (that is, a normal matrix), can we determine whether the matrix has degenerate eigenvalues without explicitly calculating all the eigenvalues and eigenvectors?
An example that came to my mind is that $M$ is square of a skew-Hermitian matrix since a skew-symmetric matrix always has pairs of pure imaginary eigenvalues $\pm i \lambda_i$. Similarly, a matrix that is square of a matrix that has pairs of eigenvalues with different signs such as $\lambda_1,-\lambda_1,\dots$ is such a case. However, these things require sorts of decomposition of the matrix $M$, which is another problem!
Another way is calculating the characteristic polynomial $\det(M-\lambda I)=0$ and factorize it, then check the degrees of each terms. But this amounts to calculating all the eigenvalues already.
Do we have other (simple) criteria or ways to determine the degeneracy of eigenvalues of a matrix $M$?