for (ii): from part (i) we know that the eigenvalues for E are all 1 (since none are 0). in particular, the eigenspace E0 contains only the 0-vector

(there are NO non-zero vectors v for which Ev = 0). that is, ker(E) = {0}, or nullity(E) = 0.

thus dim(V) = rank(E), so E is invertible (it is 1-1 and onto). hence from E^2 = E we have:

E^2(E^-1) = E(E^-1)

E(EE^-1) = I........so ?

for question 2: note that

note as well that

can <x,x> be negative?