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If $P_1, P_2: V \to V$ are linear projection operators on the vector space $V$ with $R := P_1(V) = P_2(V)$, is it true that any convex combination of $P_1$ and $P_2$ is again a projection operator $P_3$ with $P_3(V) = R$?

I'm trying to figure out whether or not the convex combination of two Ehresmann connections on a fiber bundle is again an Ehresmann connection, as is seemingly implied by the first paragraph of $\S$2 of this paper.

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Yes, this is true.

Let $P_1, P_2:V \to V$ be projections onto $R$. I.e., $P_iP_i = P_i$ and $P_i|_R$ is the identity for $i = 1,2$. Let $c_1, c_2 \geq 0$ be such that $c_1 + c_2 = 1$. Define $P_3:= c_1 P_1 + c_2 P_2$.

It is immediate that $P_3(V) \subseteq R$. We have $$P_3P_3 = (c_1P_1+c_2P_2)(c_1P_1 + c_2 P_2) = c_1^2 P_1^2 + c_2^2 P_2^2 + c_1 c_2 P_1P_2 + c_1 c_2 P_2 P_1 = c_1^2 P_1 + c_2^2 P_2 + c_1c_2(P_1 + P_2) = c_1(c_1+c_2)P_1 + c_2(c_1+c_2)P_2 = c_1 P_1 + c_2 P_2 = P_3,$$

so $P_3$ is indeed a projection. Finally, consider any $v \in R$. Then $$P_3 v = c_1P_1 v + c_2P_2 v = c_1 v + c_2 v = v,$$

which completes the proof. Note also that, since $V = P_3(V) \oplus \ker P_3$ for any linear operator $P_3$, we have $V = R \oplus \ker P_3$.

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  • $\begingroup$ I don't get the equation $$c_1^2 P_1^2 + c_2^2 P_2^2 + c_1 c_2 P_1P_2 + c_1 c_2 P_2 P_1 = c_1^2 P_1 + c_2^2 P_2 + c_1c_2(P_1 + P_2).$$ $\endgroup$ – goblin Jun 22 '16 at 23:44
  • $\begingroup$ @goblin: $P_1^2 = P_1$ and $P_2^2 = P_2$ since $P_1$ and $P_2$ are projections. Since the image of both $P_1$ and $P_2$ is $R$ and $P_1|_R$, $P_2|_R$ are both the identity map on $R$, it follows that $P_1P_2 = P_2$ and $P_2P_1 = P_1$. $\endgroup$ – Matthew Kvalheim Jun 23 '16 at 15:48
  • $\begingroup$ Oh, true. That didn't occur to me! $\endgroup$ – goblin Jun 23 '16 at 17:43

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