Are convex combinations of projection operators still projection operators? 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.
 A: 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 $V = R \oplus \ker P_3$.
A: This is essentially the same as Matthew's answer, but instead of directly showing that $P_3$ is idempotent by algebraic manipulations, I'll consider the images of vectors under $P_3$ and $P_3^2$.
By assumption, $P_1$ and $P_2$ share a common range $U=P_1V=P_2V$. Let $t\in\mathbb R$ and $P_3=(1-t)P_1+tP_2$ (note: we don't need $0\le t\le1$). For any $u\in U$, we have $P_1u=u=P_2u$. It follows that $P_3u=u$ and $P_3U=U$.
Now, for any $v\in V$, let $u=P_3v$. Then $u\in P_1V+P_2V\subseteq U$. Therefore $P_3^2v=P_3u=u=P_3v$. That is, $P_3$ is a projection and $P_3V\subseteq U$. However, as $U=P_3U\subseteq P_3V$, we conclude that $P_3V=U$.
If $V$ is finite-dimensional, we can translate the above into matrix language to visualise what all these mean. Let $\mathcal B=\{u_1,\ldots,u_k,v_{k+1},\ldots,v_n\}$ be a basis of $V$ such that the linear span of $\{u_1,\ldots,u_k\}$ is $U$. Since $P_1$ and $P_2$ are projections onto $U$, the matrices of $P_1$ and $P_2$ With respect to the basis $\mathcal B$ are in the form of
$$
\pmatrix{I&\ast\\ 0&0}.
$$
It follows that $P_3=(1-t)P_1+tP_2$ has the same form too, for all real numbers $t$. Hence $P_3$ is a projection whose range is $U$.
