# Orthogonal projection of power of a linear operator.

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.
• Very nice, but what about the case of vectors outside $W$? Can we then bound the norm $||P_V(x)||$ by a constant multiple of $||P_W(x)||$? That is, in the case when $V,W$ have the above structure.
• What do you mean by outside $W$ ? If $x\notin W$ the inequality does not hold. We can take $x_0\in W\cap V^\perp$ with $\|x_0\|=1$ and $0\neq x_1\perp W,$ $\|x_1\|=1.$ The for $x=x_0+x_1$ we have $P_Wx=x_0$ and $P_Vx=0.$ Commented Dec 7, 2023 at 14:50