Past exam of the algebra 1 and differential equations 1 course of the University of Oxford / 2015 / 1 / b

Expand the square using $E^2=E$ and $F^2=F$:ThusThis sum of two linear projections criterion holds in every characteristic; it uses an equality of linear maps, not cancellation of factors in the noncommutative associative algebra of linear endomorphisms.

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 yieldsIn 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. ThereforeThe argument cancels a nonzero scalar, never a possibly singular linear map.

No. On any nonzero vector space over a field of characteristic two, take $E=F=I$. Both are linear projections, andThe correct condition from part (i) in this characteristic is $EF=FE$, because minus and plus coincide. Vanishing of both products remains sufficient, but is no longer necessary.