Let $F$ be a subfield of the complex numbers (or, a field of characteristic zero). Let $V$ be a finite-dimensional vector space over $F$. Suppose that $E_1,\dots,E_k$ are projections of $V$ and that $E_1+\dots+E_k=I$. Prove that $E_iE_j=0$ for $i\ne j$. (Hint: Use the trace function and ask yourself what the trace of a projection is.
This is Exercise 10 from Section 6.6 (Direct-Sum Decompositions) in Hoffman, Kunze: Linear algebra.
I have following solution
Please help in understanding: If $W_{ij}$ is the range of $E_iE_j$. Then (a) how $tr(E_iE_j)=\dim(W_{ij})$ and (b) if $\dim(W_{ij})=0$ then how $E_iE_j=0$ for $i\ne j$?
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Yes, we do require simultaneous diagonalizability, which then takes some time to prove, so as Berci says, this method is more time taking than another proof which I will show.
For this, we need to show that the projections commute i.e. $E_iE_k = E_kE_i$. For this, write $\sum E_j = I$. Therefore, $\sum_{j \neq i} E_j = I-E_i$. From here, we get $\sum_{j \neq i} E_iE_jE_i =E_i(1-E_i)E_i=0$. Now, note that $E_iE_jE_i = (E_iE_j)(E_iE_j)^H$ since $E_i,E_j$ are projections, so we get $\sum_{j \neq i} (E_iE_j)(E_iE_j)^H = 0$.
(Where $\cdot^H$ denotes the conjugate transpose, or the Hermitian transpose).
From this, clearly $E_iE_j = 0$ for each $i,j$, see this yourself.
This is the shortest method to show that $E_i,E_j$ commute, but it also shows the products are zero, as desired. I think you should go with this proof itself.