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Question: Two $k$-dimensional subspaces $W_1,W_2$ with associated orthogonal projections $P_1, P_2$ are isoclinic with parameter $\lambda \ge 0$ if $P_1P_2P_1=\lambda P_1$ and $P_2P_1P_2=\lambda P_2$.

I was able to show that if $W_1\perp W_2$, then $W_1,W_2$ are isoclinic with $\lambda=0$. Now I've read the following definitions in a paper.

  • Two subspaces $W_1,W_2$ are $\epsilon$-nearly orthogonal if for all unit vectors $\phi\in W_1$ and $\psi\in W_2$ we have $|\langle\phi,\psi\rangle|<\epsilon$.
  • Two k-dimensional subspaces $W_1,W_2$ are $\epsilon$-nearly isoclinic if there exists $\lambda\geq0$ such that $(\lambda-\epsilon^2)P_1\le P_1P_2P_1\le(\lambda+\epsilon^2)P_1$ and $(\lambda-\epsilon^2)P_2\le P_2P_1P_2\le(\lambda+\epsilon^2)P_2$.

Then the claim is again, if $W_1,W_2$ have same dimension, then $\epsilon$-nearly orthogonal implies $\epsilon$-nearly isoclinic with parameter $\lambda=0$. In the paper they treat it like it is trivial. But for me, regrettably, it is not. Is there really a short proof for that or does anybody know where i can find a proof for that (if it is not so short)?

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  • $\begingroup$ Phrase there exists $\lambda=0\ $ puzzles me. $\endgroup$ Commented Jan 23, 2015 at 0:07
  • $\begingroup$ How do you define an inequality $\ P\le Q\ $ for two linear operators? (I have a candidate but would like to be sure). Also, does space mean Euclidean space (I am sure that it does but it'd be nice to say so explicitly). $\endgroup$ Commented Jan 23, 2015 at 0:15
  • $\begingroup$ Should there be norms in your second condition--then inequalities would make sense without any additional (exotic :-) definitions.. $\endgroup$ Commented Jan 23, 2015 at 0:19
  • $\begingroup$ Ah sorry, I did not mention that. $Q\geq 0$ is defined by $\langle Q\varphi, \varphi\rangle\geq 0$ for all $\varphi\in\mathcal{H}$ where $\mathcal{H}$ is a Hilbert space. $\endgroup$
    – Roman
    Commented Jan 28, 2015 at 21:59
  • $\begingroup$ So $W_1$ and $W_2$ are subspaces of an arbitrary Hilbert space, not necessarily an Euclidien space. Also other scalar products are allowed. Greetings =) $\endgroup$
    – Roman
    Commented Jan 28, 2015 at 22:12

1 Answer 1

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Hey I just found a solution on my own.
Let $a$ be an arbitrary Element of a Hilbert space.
We have
$\Vert P_2 P_1 a\Vert^2=\langle P_2 P_1a,P_1a\rangle=\Vert P_2 P_1 a\Vert\Vert P_1a\Vert\langle \frac{P_2P_1a}{\Vert P_2 P_1 a\Vert},\frac{P_1a}{\Vert P_1 a\Vert}\rangle<\epsilon\Vert P_2 P_1 a\Vert\Vert P_1a\Vert$.
$\implies \Vert P_2 P_1 a\Vert<\epsilon\Vert P_1a\Vert$
$\implies\Vert P_2 P_1 a\Vert^2<\epsilon^2\Vert P_1a\Vert^2$
$\implies\langle P_1P_2P_1a,a\rangle\le\epsilon^2\langle P_1a,a\rangle$.
Since
$-\epsilon\Vert P_1a\Vert^2<-\Vert P_2 P_1a\Vert^2<\Vert P_2 P_1a\Vert^2$,
we also have
$-\epsilon^2\langle P_1a,a\rangle\le\langle P_1P_2P_1a,a\rangle$.
Changing the roles of $P_1$ and $P_2$ yields the claim.

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