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Let $T$ be a bounded operator on a Hilbert space with the property that $T^*(T-I)= 0$. I'd like to show that $T$ is an orthogonal projection.

I'm not really sure how to show that an operator is an orthogonal projection. If $A:X\rightarrow U$ is a projection onto a closed subspace of X, then $\langle x- Ax, u_i\rangle = 0 \;\;\forall u_i \in U$ ?

Expanding we get $T'T = T' \Longleftrightarrow TT' = T\ (T'' = T$ in Hilbert spaces?) $$Tx = T'(Tx) \Longleftrightarrow y = T'y$$ Hence $T'$ has $\lambda = 1$ after $Tx$, do $T'$ and $T$ have the same eigenvalues?
There are a lot of question marks here.

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Your equation is $T^*T=T^*$. The key observation here is that $(T^*T)^*=T^*T$. Then $$ T=(T^*)^*=(T^*T)^*=T^*T=T^*. $$ so $T$ is selfadjoint, and moreover now your original equation reads $T^2=T$, so $T$ is an orthogonal projection.

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Here is one way to see this. We have

$T^*T - T^* = T^*(T-I) = 0 = (T^*(T-I))^* = (T^*-I)T = T^* T - T$

hence cancelling the $T^*T$ you get that $T$ is self-adjoint, in particular normal, and so you may appeal to the spectral theorem, which in this case tells you that the fuction $\bar{x}(x-1) = 0$ on the spectrum of $T$.

We want to show that $\sigma(T) \subset \{0,1\}$ for $T$ to be a projection.

Now we have $\bar{x}(x-1) = |x|^2 - \bar{x} = 0$ i.e., since $|x|^2$ is real that for all $x \in \sigma(T)$ you have $x = |x|^2$.

This can happen only if $x = 0$ or $x = 1$ as we wanted.

EDIT: Marc van Leeuwen pointed out nicely that after the first line we actually already have $T^* = T$ and using this $T^2 = T$ as well, hence the claim.

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    $\begingroup$ After the first display, could you not just say that because of $T^*=T$ it becomes $T^2=T$, so $T$ is a projection, and being self-adjoint an orthogonal projection? In any case any linear operator satisfying $T^2=T$ is diagonalizable with $\sigma(T)\subseteq\{0,1\}$. $\endgroup$ – Marc van Leeuwen Dec 8 '12 at 10:33
  • $\begingroup$ for the first part, you are definitely right. In the second statement, why does an operator with $T^2 = T$ have spectrum contained in $\{0,1\}$? $\endgroup$ – mland Dec 8 '12 at 10:52
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    $\begingroup$ If $T^2=T$ then every vector $v$ can be written as a sum $v=\frac12(v-T(v))+\frac12(v+T(v))$ of an eigenvector (if nonzero) with eigenvalue $0$ and an eigenvector (if nonzero) with eigenvalue $1$, so the eigenspaces of $T$ for $\lambda=0$ and for $\lambda=1$ span the whole space, whence my claim. $\endgroup$ – Marc van Leeuwen Dec 8 '12 at 11:05
  • $\begingroup$ Yes. Obviously I think more in terms of the spectral theorem, which is absolutely not necessary here. Thanks for pointing this out. $\endgroup$ – mland Dec 8 '12 at 11:24
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    $\begingroup$ @mland: $\sigma(T^2-T)=\{\lambda^2-\lambda:\ \lambda\in\sigma(T)\}$. $\endgroup$ – Martin Argerami Dec 8 '12 at 11:54

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