Let $T$ be a linear operator on a vector space $V$, and let $W$ be an invariant subspace. If $V(\lambda)$ denotes the $\lambda$-eigenspace of $V$ and $W(\lambda)$ the eigenspace of $T$ on $W$, then does $[V(\lambda)/W(\lambda)] = [V/W](\lambda)$.
If I can show this, then it will help with something I am working on. Is there a nice way to show this (or is it even true)?
It is true, in the form Omnomnomnom suggested in the comment:
If $v \in [V(\lambda)/W(\lambda)]$, then it is the image of some $v_0 \in V(\lambda)$ orthogonal to (all vectors in) $W(\lambda)$. Since it is an eigenvector corresponding to eigenvalue $\lambda$, it is also orthogonal to $W/W(\lambda)$ (which is spanned by eigenvectors of different value), so in fact it's orthogonal to all of $W$, and thus $v_0$ also projects nicely to $[V/W](\lambda)$.
On the other hand, if $v \in [V/W](\lambda)$, then it's the image of some $v_0 \in V$ orthogonal to all of $W$ and having eigenvalue $\lambda$. Therefore it is an element of $V(\lambda)$ and is orthogonal to $W(\lambda) \subset W$, so $v_0$ also projects in the other direction. Therefore, the two sets are equal.