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Suppose that $V$ is a $\mathbb C$-vector space. I'm eventually interested in the infinite-dimensional case, but let's say for now that it's finite dimensional. Suppose that $\mathscr S$ is a collection of rank-1 projectors on $V$, not necessarily commuting or respecting any interesting inner product, such that $\sum_{p \in \mathscr S} pV = V$; and that $T$ is an endomorphism of $V$ such that $\operatorname{tr} T \ne 0$. Is there necessarily some $p \in \mathscr S$ so that $p T p \ne 0$? I think of this loosely as: is there necessarily some line in a collection of lines that witnesses a non-0 trace?

(This question arises for me while trying to generalise some old work of Moy and Prasad on minimal K-types. They need only to deduce from the fact that $\operatorname{tr} T \ne 0$ that there is some $p \in \mathscr S$ with $T p \ne 0$, which is clear; but I seem to need the stronger version above.)

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  • $\begingroup$ $V$ is $2$-dimensional with basis $\left(e,f\right)$. The projectors are projecting on $\mathbb C e$ and $\mathbb C f$, each with kernel $e+f$. The map $T$ projects on $\mathbb C \left(e+f\right)$. This is a counterexample, right? $\endgroup$ Mar 11, 2015 at 1:18
  • $\begingroup$ So I think the argument of @darijgrinberg shows that, in finite dimension, your statement is true IFF the projections are mutually orthogonal to each other $\endgroup$ Aug 26, 2019 at 21:10
  • $\begingroup$ @AliTaghavi, what does it mean for projections to be orthogonal? (There is no inner product inherently in sight.) $\endgroup$
    – LSpice
    Aug 26, 2019 at 23:07
  • $\begingroup$ @LSpice I think a projection, by definition s an element (an operator) $A$ with $A=A^*=A^2$ and $*$ requires an inner product. On the other hand if by projection youmerely mean $A^2=A$, I guess that we can equip the space with an inner product such that each $p$ in the set you mentioned would be an orthonormal project.(however it would be probably another uestion?). Now we may prove that your question has affirmative answer IFF the one dimensional spaces(Im(P)'s would be mutually orthogonal. any way my previous comment was based on the following assumption: we have the standard..... $\endgroup$ Aug 26, 2019 at 23:18
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    $\begingroup$ @AliTaghavi, for me a projector $p$ need only satisfy $p^2 = p$; and, as indicated in the post, I very much do not want to assume that the projectors are orthogonal with respect to any common inner product. ‍@‍darijgrinberg's counterexample ruled out the proof technique I had in mind. $\endgroup$
    – LSpice
    Aug 27, 2019 at 1:57

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Counterexample:

$V$ is $2$-dimensional with basis $\left(e,f\right)$. The projectors are projecting on $\mathbb C e$ and $\mathbb C f$, each with kernel $e+f$. The map $T$ projects on $\mathbb C \left(e+f\right)$.

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