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 $T^m$.


1 Answer 1


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.

  • $\begingroup$ 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. $\endgroup$
    – User
    Commented Dec 7, 2023 at 14:33
  • 1
    $\begingroup$ 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 $P_Vx=0.$ $\endgroup$ Commented Dec 7, 2023 at 14:50

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