= Solution
Suppose first that $E+F$ is a <linear projection>. Part (i) gives $EF=-FE$. Multiplying on the left by $E$ gives $EF=-EFE$, and multiplying on the right by $E$ gives $EFE=-FE=EF$. Combining them yields
$$
EF=-EF,\qquad 2EF=0.
$$
In a <field> of <characteristic> different from two, the nonzero <scalar> $2$ is invertible. Hence $EF=0$, and $FE=-EF=0$ too. Conversely these two zero products make the expansion in part (i) equal zero. Therefore
$$
\boxed{E+F\text{ is a linear projection}\quad\Longleftrightarrow\quad EF=FE=0.}
$$
The argument cancels a nonzero <scalar>, never a possibly singular <linear map>.
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