Projection matrix that sum to identity are orthogonal

Here is an argument that works for any number of projections. It uses only the algebraic properties of the projections (that they are selfadjoint and positive, and that they equal their own square), without using them as operators. Also, equality to the identity is not required, just that the sum is below the identity.

So, assume$$\sum_{k=1}^NZ_k\leq I.$$ Take indices $i$ and $j$ with $i\ne j$. Then $$ 0\leq Z_iZ_jZ_i\leq Z_i\left(\sum_{k\ne i} Z_k\right)Z_i\leq Z_i(I-Z_i)Z_i=0. $$ It follows that $Z_iZ_jZ_i=0$. But then $$ 0=Z_iZ_jZ_i=Z_iZ_j^2Z_i=(Z_jZ_i)^*Z_jZ_i, $$ and so $Z_jZ_i=0$.

In fact if $P+Q$ is a projection, then $PQ = 0$, because $(P+Q)^2 = (P+Q)$, and so $$ PQ(x) + QP(x) = 0 \Rightarrow PQ = -QP $$ and hence for any $x \in ran(PQ)$, $$ PQx = x \Rightarrow Px = x $$ and similarly, $$ -QPx = x \Rightarrow Qx = x $$ and hence $x = -QPx = -Qx = -x$, and so $x = 0$, which means that $PQ = 0$

Now apply this to $P = Z_i, Q=Z_j$ for $i\neq j$