Let \(E_1, \ldots, E_k\) be linear operators on the space \(V\) such that \(E_1 + \cdots + E_k = I\).
Prove that if \(E_iE_j = 0\) for \(i \neq j\), then \(E_i^2 = E_i\) for each \(i\).
In the case \(k = 2\), prove the converse of (a). That is, if \(E_1 + E_2 = I\) and \(E_1^2 = E_1\), \(E_2^2 = E_2\), then \(E_1E_2 = 0\).
Solution
1. Pairwise annihilation implies each \(E_i\) is idempotent.
Fix \(i\). From \[ E_1+\cdots+E_k=I \] we get, by multiplying on the left by \(E_i\), \[ E_i=E_iI=E_i(E_1+\cdots+E_k) =E_i^2+\sum_{j\ne i}E_iE_j. \] The hypothesis gives \(E_iE_j=0\) for \(j\ne i\), so \[ E_i=E_i^2. \] Thus every \(E_i\) is a projection [@HK71].
2. For two summands, idempotence implies the cross-products vanish.
Assume \(k=2\), \(E_1+E_2=I\), and \(E_1^2=E_1\), \(E_2^2=E_2\). Then \[ E_2=I-E_1, \] so \[ E_1E_2=E_1(I-E_1)=E_1-E_1^2=0. \] Similarly, \[ E_2E_1=(I-E_1)E_1=E_1-E_1^2=0. \] Hence the two projections annihilate one another, proving the converse in the case \(k=2\).