Thread: Eigenvalue Questions

I can do i) but just a bit stuck on ii).

my working for i) is: E^2-E=0 so that means is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then and so \lambda = 0 or 1

cheers

[hr]



I sorta have an idea but dono how to finish this Q

If we let x be a eigenvector of corresponding to , then we have

Now just playing around:

but dono how to get to the result

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?

An alternative for the first question: for all . Consider , if then is an eigenvector associated to , if then is an eigenvector associated to .