2

in book of conway, functional analysis, section operators on Hilbert space(projection and idempotent) say that a projection is an idempotent such P that $(kerP)=(rangP)^\perp$.

but from the next proposition part c, we can conclude that for every idempotent we have $(kerP)=(rangP)^\perp$. am i right? if this is true then what is the difference between projection and idempotent? please help me. thanks.

joker
  • 363
  • The proposition says that the sum of $kerP$ and $rangP$ span the hilbert space, but the proposition does NOT say that $kerP$ and $rangP$ are orthogonal. – Horstenson Dec 08 '13 at 10:28
  • but we know that because $KerT$ is a closed subspace of H so $H=KerT\oplus KerT^\perp$. now if we can write $H=KerT\oplus rangT$ can we say that $KerT=rangeT^\perp$? – joker Dec 08 '13 at 10:35
  • Why do you think you can write $H = KerT\oplus rangT$? Don't forget that $T:H\rightarrow H$, so $KerT$ is in the domain (the first $H$), and $rangT$ in the codomain (the second $H$) – Horstenson Dec 08 '13 at 13:08
  • @Horstenson There is no need to consider two $H$'s to be distinct. Much of operator theory on Hilbert spaces revolves around invariant subspaces, iteration, etc. – Post No Bulls Dec 09 '13 at 21:24

2 Answers2

3

The notion of idempotent does not require a Hilbert space structure, you simply require an operator, say acting on a Banach space, to satisfy $T^2 = T$. So $T$ restricted to $Ran(T)$ is the identity.

On the other hand, when you write $^{\perp}$, implicitly there is a Hilbert space structure. Then your condition $Ker (P) = Ran (P)^{\perp}$ means $P$ is a self-adjoint idempotent, i.e. (orthogonal) projection on Hilbert space.

Michael
  • 3,068
3

Proposition 3.2c in Conway's book indeed says that for every idempotent $E$ acting on a Hilbert space $\mathscr{H}$ there holds $$\mathscr{H} = \mathscr{M}+\mathscr{N} \tag{1}$$ where $\mathscr{M}=\operatorname{ran}E$ and $\mathscr{N}=\operatorname{ker}E $.

If in (1), "$+$" meant orthogonal sum, your reasoning would be correct. But this is not what Conway means. He uses $\oplus$ for orthogonal sum, see section 1.6, or later in the section on projections. The meaning of $+$ here is the vector sum of sets: $$A+B = \{a+b:a\in A,b\in B\}$$ So, (1) simply says that every element of $\mathscr{H}$ can be written as the sum of two vectors, one in $\mathscr{M}$ and the other in $\mathscr{N}$. This is quite a weak statement; it does not even exclude the possibility that two spaces could overlap, or be the same: $$\mathscr{H} = \mathscr{H}+\mathscr{H} $$ is a true statement. This is why Conway also puts $\mathscr{M} \cap \mathscr{N} = (0)$ as a part of 3.2c.

Just for the sake of completeness, a simple example: $$\begin{pmatrix} 1 & 0 \\ 1 & 0\end{pmatrix}$$ in two-dimensional space, like $\mathbb R^2$. This is an idempotent operator; the kernel is the vertical axis, the range is the diagonal.

Post No Bulls
  • 9,056