Sufficient condition for symmetric matrices to be an orthogonal projection matrix

Question

I am trying to prove a version of Cochran's theorem(Theorem 5.14) given here

The problem:-

Let $A_1, A_2, \cdots , A_m$ be $n\times n$ symmetric matrices and $A = \sum_i A_i $ with $\text{rank}(A_j)$$=n_j$. Then,
If

1. $A_jA_k=0,\forall j\ne k$
2. $\sum n_j= \text{rank}(A)$ ( should $\text{rank}(A)$ be further constrained to be equal to $n$?? )

Then,

1. $A_j$ is an orthogonal projection for all $j$.

What i have tried:-

$$\text{Since, each } A_i\text{ is symmetric }\implies N(A_i) = (C(A_i))^{\perp} \tag{a}$$
$$\because A_jA_k=0,\forall j\ne k\implies C(A_i)\cap C(A_j)=\{0\} \tag{b}$$
$$\because (A = \sum A_i)\text{ and (b)} \implies Ax = A_i x, \forall x\in C(A_i) \tag{c}$$
$$\because A_jA_k=0,\forall j\ne k\implies A^2 = \sum A_i^2\text{ and } A^2x = A_i^2 x, \forall x\in C(A_i) \tag{d}$$
$$\because (A = \sum A_i) \implies C(A) \subset C(A_1)+C(A_2)+\cdots\cdots+C(A_m) \tag{e}$$
$$\because \sum n_j= \text{rank}(A) \implies \dim C(A) = \dim C(A_1) + \dim C(A_2)+\cdots\cdots+ \dim C(A_m) \tag{f}$$
$$\because \sum n_j= \text{rank}(A) \text{ and (e) and (f) } \implies C(A_i)\cap\big(\bigcup_{j\ne i}C(A_j)\big)={0} \tag{g}$$

$(e)+(f)+(g) \implies C(A) = C(A_1)\oplus C(A_2)\oplus \cdots \cdots \oplus C(A_m) \tag{h}$

$C(A) = $column space of $A$; $N(A) = $null space of $A$

Answer 1

For $n=4$ let $$A_1=\begin{pmatrix} 2 & 1& 0& 0\\ 1 & 1 & 0 & 0\\ 0 & 0& 0 & 0\\ 0 & 0& 0 & 0 \end{pmatrix} \quad A_2=\begin{pmatrix}0 & 0& 0 & 0\\ 0 & 0& 0 & 0 \\ 0 & 0 & 2 & 1\\ 0 & 0 & 1 & 1 \end{pmatrix}$$ and $A=A_1+A_2.$ Then $A_1A_2=A_2A_1=0,$ $\operatorname{rank}(A_1)=\operatorname{rank}(A_2)=2$ and $\operatorname{rank}(A)=4.$ None of the matrices is a projection or a multiple of a projection.