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Consider projection operators $\rho_1,\ldots,\rho_k$ defined on vector space $V$ over field of characteristic $0$, such that $$ \rho_1+\cdots+\rho_k = 1 $$ Projections $\rho, \pi$ are said to be orthogonal, if $\rho\circ\pi=\pi\circ\rho=0$.

Question: Are $\rho_1,\ldots,\rho_k$ necessarily pairwise orthogonal?

If $V$ has finite dimension, the answer is yes. I expect it not to be the case in general, but I can't seem to come up with a counterexample.

Bonus question: What about possibly infinite set of projections, where $$ \sum_{\rho\in P} \rho = 1 $$ is understood as "for each $v\in V$, only finitely many of $\rho v$ are nonzero, and their sum is $v$" ?

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    $\begingroup$ Would the downvoter care to comment what should be improved? $\endgroup$
    – Marcin Łoś
    Jun 26, 2014 at 22:37
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    $\begingroup$ From these assumptions one can prove $\rho_i \circ (\sum_{j\ne i}\rho_j)=0$. Hence the conclusion is valid for $k=2$. $\endgroup$
    – daw
    Jun 27, 2014 at 8:10
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    $\begingroup$ @daw Indeed. It holds for $k=3$ as well - if $\rho_1+\rho_2+\rho_3=1$, then $\rho_1+\rho_2 = 1-\rho_3$ is a projection, and one can prove that sum of projections is a projection iff they're orthogonal, hence $\rho_1\perp \rho_2$. Unfortunately, I don't think it generalizes easily. $\endgroup$
    – Marcin Łoś
    Jun 27, 2014 at 8:38
  • $\begingroup$ Analogous problem on Hilbert space, with orthogonal projections ($p^*=p=p^2$): math.stackexchange.com/q/117702 $\endgroup$
    – Jonas Meyer
    Dec 3, 2014 at 18:25

1 Answer 1

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That isn't true in general, so if we take our space $V=\mathbb{R}^{\mathbb{Z}}_0$ (that mean the space of sequences $(u_n)_{n\in \mathbb{Z}}$ with finitely nonzero sequences ) and we define the canonical algebraic bases $e_n$ and the projections : $$ \left\{\begin{array}{l} P_1(e_n)=\frac{n}{2} (e_n+e_{2-n})\\ P_2(e_n)=\frac{n}{2} (e_n-e_{2-n})\\ P_3(e_n)=\frac{-n}{2} (e_n+e_{-2-n})\\ P_4(e_n)=\frac{-n}{2} (e_n-e_{-2-n})\\ P_5(e_n)=e_n \end{array}\right. $$ So you can easily verify that $P_1+P_2+P_3+P_4+P_5=1$ but there are not orthogonal.

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  • $\begingroup$ And You can find more detail in this paper : H. Bart, T. Ehrhardt and B. Silbermann ''ZERO SUMS OF IDEMPOTENTS IN BANACH ALGEBRAS'' Integr Equat Oper Th Vol. 19 (1994) pp 125-134 $\endgroup$
    – Hamza
    Dec 3, 2014 at 7:12
  • $\begingroup$ no but the space of all sequences $\endgroup$
    – Hamza
    Dec 3, 2014 at 7:18
  • $\begingroup$ oh yes yes you are right i need to change the word basis or taking the subspace of $\mathbb{R}^{\mathbb{Z}}$ of finitely nonzero sequences. $\endgroup$
    – Hamza
    Dec 3, 2014 at 7:27
  • $\begingroup$ Thank you - it's a pleasant surprise to get an answer after few months! $\endgroup$
    – Marcin Łoś
    Dec 3, 2014 at 12:29

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