$A-A_i-A_j$ is a idempotent matrix

Question

Let $A_1$,$A_2$,...,$A_k$ be symmetric and idempotent matrices and let A=$\sum_{i=1}^k\ A_k$ a idempotent matrix. Can be proved that $A-A_i-A_j$ is a idempotent matrix?

Answer 1

If $i\ne j$, yes, and we can do better:

Proposition. Let $P_1,P_2,\ldots,P_n$ be some $n$ orthogonal projections (by this I mean self-adjoint idempotent linear operators) on a real or complex inner product space $X$ of finite or infinite dimensions. If their sum is also an orthogonal projection, then $P_iP_j=0$ for all $i\ne j$. Consequently, $\sum_{i\in\mathcal I}P_i$ is an orthogonal projection for any subset $\mathcal I$ of $\{1,2,\ldots,n\}$.

Proof 1. The case $n=1$ is trivial. When $n>1$, modify the proof by Jonas Meyer as follows. Let $\Pi=\sum_{i=1}^nP_i$. For any $j\in\{1,2,\ldots,n\}$, we have \begin{aligned} 0&=P_j\left(\sum_iP_i\right)P_j-P_j\Pi P_j\\ &=\sum_{i\ne j}P_jP_iP_j+P_j-P_j\Pi P_j\\ &=\sum_{i\ne j}P_jP_iP_j+P_j(I-\Pi)P_j\\ &=\sum_{i\ne j}P_jP_i^2P_j+P_j(I-\Pi)^2P_j\\ &=\sum_{i\ne j}P_j^\ast P_i^\ast P_iP_j+P_j^\ast(I-\Pi)^\ast(I-\Pi)P_j\\ &=\sum_{i\ne j}(P_iP_j)^\ast(P_iP_j)+\left((I-\Pi)P_j\right)^\ast\left((I-\Pi)P_j\right). \end{aligned} This is a sum of $n$ Gram operators. Hence each Gram operator is zero and $P_iP_j=0$ when $i\ne j$.

Proof 2. We use mathematical induction on $n$. The case $n=1$ is trivial and the case $n=2$ is well-known (see e.g. sec. 42, p.74 of Halmos' Finite-Dimensional Vector Spaces for a proof that works for general projections on infinite-dimensional vector spaces). In the inductive case, recall the following basic properties of any orthogonal projection $P$ on $X$:

1. $X=\ker(P)\oplus\ker(P)^\perp$. It is the sum of the two subspaces because every $x\in X$ can be written as the sum of two components $v=x-Px\in\ker(P)$ and $w=Px\in \ker(P)^\perp$. The sum is direct because it is an orthogonal sum.
2. $\operatorname{range}(P)=\ker(P)^\perp$. For any $Px\in\operatorname{range}(P)$ and $v\in\ker(P)$ we have $\langle Px,v\rangle=\langle x,Pv\rangle=0$ and hence $\operatorname{range}(P)\subseteq\ker(P)^\perp$. For the reverse inclusion, let $w\in\ker(P)^\perp$. Then $w-Pw\in\ker(P)\cap\ker(P)^\perp=0$ and hence $w=Pw\in\operatorname{range}(P)$.
3. $P$ is positive semidefinite, because $ \langle Px,x\rangle=\langle P^2x,x\rangle=\langle Px,P^\ast x\rangle=\langle Px,Px\rangle\ge0$.
4. For any subspace $V$ of $\ker(P)$, both $V$ and $V^\perp$ are invariant subspaces of $P$ because $PV=0\subseteq V$ and $PV^\perp\subseteq\operatorname{range}(P)=\ker(P)^\perp\subseteq V^\perp$.

Now let $\Pi=\sum_{i=1}^nP_i$ and $V=\ker(\Pi)$. By properties 1 and 2, $X=V\oplus V^\perp$ and $V^\perp=\operatorname{range}(\Pi)$. Since each $P_i$ is positive semidefinite (property 3), if $\Pi x=0$, we must have $P_ix=0$ for each $i$. Hence $V\subseteq\ker(P_i)$ for each $i$. Consequently, by property 4, both $V$ and $V^\perp$ are invariant subspaces of each $P_i$. It follows that $\sum_{i=1}^nP_i|_{V^\perp}=\left(\sum_{i=1}^nP_i\right)|_{V^\perp}=\Pi|_{\operatorname{range}(\Pi)}=\operatorname{Id}$, the identity map on $\operatorname{range}(\Pi)=V^\perp$. But then $\sum_{i\ne k}P_i|_{V^\perp}=\operatorname{Id}-P_k|_{V^\perp}$ is an orthogonal projection for every $k\in\{1,2,\ldots,n\}$. So, by induction assumption, we have $P_i|_{V^\perp} P_j|_{V^\perp}=0$ for any two distinct indices $i,j\ne k$. Since $k$ is arbitrary and all $P_i$s are zero on $V$, we conclude that $P_iP_j=0$ on $X$ for any $i\ne j$.