Properties of the Inclusion-Exclusion Principle

Question

The problem:
Let $A_1,A_2, \cdots A_k$ be orthogonal projections. Let $A = \sum_{i=1}^{k}A_i$. Now,
If $A$ is also an orthogonal projection, then $A_iA_j=0,\forall i\neq j$.

What i have done:

Let $M = C(A) = \text{column space of $A$} $ and $M_i = C(A_i) = \text{column space of $A_i$} $

$$\implies rank(A) = trace(A) = trace(\sum_{i=1}^{k}A_i) = \sum_{i=1}^{k}trace(A_i) = \sum_{i=1}^{k}rank(A_i)$$
$$\implies \dim M = \sum_{i=1}^{k}\dim(M_i) \tag{a}$$

Also,
$\because A = \sum_{i=1}^{k}A_i \implies M\subset M_1+M_2+\cdots +M_k$
$$\implies \dim(M) \le \sum_{i=1}^{k}\dim(M_i)-\sum_{i<j}\dim(M_i\cap M_j)+\sum_{i<j<l}\dim(M_i\cap M_j\cap M_l)+\cdots \cdots+(-1)^{k+1} \dim(M_1\cap\cdots \cap M_k) \tag{b}$$

subtracting equation equation $(a)$ from inequality $(b)$, we get

$\sum_{i<j}\dim(M_i\cap M_j)-\sum_{i<j<l}\dim(M_i\cap M_j\cap M_l) \cdots+(-1)^{k}\dim(M_1\cap\cdots \cap M_k)\le 0$

Now, from here, i need two results which look fairly obvious, but i am unable to prove them.

Let $B_1, B_2, \cdots B_n$ be $n$ sets. Then,

$|B_1\cup B_2 \cup \cdots\cup B_n|$ $= \sum_{i}|B_i|-\sum_{i<j}|B_i\cap B_j| + \sum_{i<j<k}|B_i\cap B_j \cap B_k|+\cdots\cdots +(-1)^{n+1}|B_1\cap B_2 \cap \cdots \cap B_n|$

The results that i need are:-

1. $ \sum_{i<j}|B_i\cap B_j| - \sum_{i<j<k}|B_i\cap B_j \cap B_k|+\cdots +(-1)^{n}|B_1\cap B_2 \cap \cdots \cap B_n| \ge 0$
2. $\sum_{i<j}|B_i\cap B_j| - \sum_{i<j<k}|B_i\cap B_j \cap B_k|+\cdots +(-1)^{n}|B_1\cap B_2 \cap \cdots \cap B_n| = 0$
   if and only if $|B_i\cap B_j|=0, \forall i\ne j$

How do i prove results $1$ and $2$. Are they even true??

Answer 1

The first inequality follows from the formula stated just before the two questions, as $$|B_1\cup B_2 \cup \cdots\cup B_n|\le \sum_{i}|B_i|$$ The second equality is equivalent to $$|B_1\cup B_2 \cup \cdots\cup B_n|= \sum_{i}|B_i|$$ The equality holds if and only if the sets are pairwise disjoint.

I think the following provides a simpler way of solving the problem stated at the very beginnimg of the post.

For any orthogonal projection $P$ we have $$\langle v,v\rangle \ge \langle Pv,v\rangle =\langle Pv,Pv\rangle \ge 0$$ Assume $$A=\sum_{i=1}^kA_i$$ and $v=A_ju.$ Then $$\langle v,v\rangle \ge \langle Av,v\rangle =\sum_{i=1}^k\langle A_iv,v\rangle \ge \langle A_jv,v\rangle =\langle v,v\rangle$$ Therefore $\langle A_iv,v\rangle =0$ for $i\neq j.$ Thus $A_iv=0,$ i.e. $A_iA_ju=0$ for $i\neq j$ and any $u.$ Thus $A_iA_j=0$ for $i\neq j.$