Q: Does $N_{r-1}\subset N_r$ or $N_r\subset N_{r-1}$? Let $r\in\mathbb{N}$ and $0<\lambda_1\dots\lambda_r<\dots$ be a sequence of eigenvalues and let $E_{r}$ the corresponding eigenspaces. We denote
$$N_r =\oplus_{i=1}^r E_i\qquad\mbox{ and }\qquad M_r = (N_r)^{\perp}.$$
My question is: it is possible to determine some relation of inclusions between the spaces $N_r$ and $M_r$?
I mean, does an inequality like $N_{r-1}\subset N_r$ or $N_{r}\subset N_{r-1}$ hold? Similarly, we can say something about the spaces $M_r$.
Intuitively, I would say that $N_{r-1}\subset N_r$ and $M_r\subset M_{r-1}$ hold, but I would like to prove it.
Could someone help me or give me a reference?
Thank you in advance!

A: Obviously $N_{r-1}\subset N_r$ because for $v$ in the direct sum we have $v=e_1+...+e_r$, take $e_r=0$ then we get $N_{r-1}$. The inequality is strict because eigenspaces are nonempty and disjoint.
Let $v\in M_{r}$. Then $v$ is orthogonal to everything in $E_1,...,E_r$. So $v$ is orthogonal to everything in $E_1,...,E_{r-1}$. Therefore $M_r\subseteq M_{r-1}$. The equality case it not clear, it depends on the field because of the Spectral theorem. On a complex vector space, an orthonormal basis of eigenvectors exists if and only if the operator is normal. So if the operator is not normal, we cannot guarantee $M_r\neq M_{r-1}$ (is that clear?)
