Say $H$ is a Hilbert space with norm $|| \cdot ||$ and $T$ is a bounded linear operator $T: H \to H$. Consider the linear operator $T^2: H \to H$, where $T^2=T \circ T$, or more generally, $T^m$, for any $m\in \mathbb{N}$.
Write $P_{T^m}$ for the orthogonal projection onto $Ker(T^m-I)$, where $Ker$ denotes the Kernel and $I$ is the identity.
For $x\in H$, I would like to know what is the relationship between $||P_{T}(x)||$ and $||P_{T^m}(x)||$. After the counterexample, I want to place the extra restriction of $x \notin Ker(T^m-I)$.
Intuitively, I would expect something like $||P_{T^m}(x)||^2 \leq ||P_{T}(x)||^2 \leq m||P_{T^m}(x)||^2$ to hold. Is either of these inequalities true, and if so, why?
At least for the first inequality, I had in mind the fact that the space of $T$-invariant vectors is contained in the space of $T^m$-invariant vectors, and so the orthogonal projection should have bigger norm for $T$ than for $T^m$.
In general, for two closed subspaces if $V\subset W$ then $\|P_Wx\|\le \|P_Vx\|.$ Moreover if $V\subsetneq W$ there exists $0\neq x\in W$ such that $x\perp V.$ Then $P_Vx=0$ and $P_Wx= x.$
Since $\ker(I-T)\subset \ker(I-T^m)$ then $\|P_{T^m}x\|\le \|P_Tx\|.$ However if $\ker(I-T)\subsetneq \ker(I-T^m),$ the converse inequality with any positive constant does not hold.