Decompose the identity matrix into several projection matrices. Prove the range of $P_j$ is the subset of the kernel of $P_i$

Question

Decompose a $N \times N$ identity matrix into $N$ projection matrices.

$$ I = \sum_{1 \le i \le N} P_i \\ \forall P_i,\ P_i^2 = P_i $$

Prove

$$\operatorname{Range}(P_j) \subset \operatorname{Kernel}(P_i)\ \text{when } i \ne j$$

Researches that I have done

There are some resources showing if the ranges of a set of projection matrices form a basis then the summation of those projection matrices is the identity matrix.

• Prove that the sum of (symmetric) projection matrices is the identity matrix
• Taboga, Marco (2021). "Projection matrix"

I want to prove the inversed statement.

Answer 1

Let $F$ be the field of scalars. We must suppose that its characteristic $c$ is $0.$ Assuming $$ I = \sum_{1 \le i \le N} P_i\quad\text{where}\quad \forall i\quad P_i^2 = P_i, $$ let $R_i:=\operatorname{im}(P_i)$ and $d_i:=\dim(R_i).$ Then, $$\sum R_i=F^N\quad\text{and}\quad\sum d_i=\operatorname{tr}\left(\sum P_i\right)=N$$ hence $$\bigoplus R_i=F^N.$$ Therefore, for every $x\in R_j,$ since $\sum_{i\ne j}P_i(x)=0,$ all $P_i(x)$'s for $i\ne j$ are $0.$

Note that if $c\ne0,$ the statement is false: a counter-example is $P_1=\dots=P_{c+1}=I,P_{c+2}=\dots=P_N=0.$