
Eigenvalue Questions
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I can do i) but just a bit stuck on ii).
my working for i) is: E^2E=0 so that means is an eigenvalue of E^2E=0, since the zero vector only has an eigenvalue of 0, then and so \lambda = 0 or 1
cheers :)
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http://img854.imageshack.us/img854/1...sqfrom2008.jpg
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

Re: Eigenvalue Questions
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 0vector
(there are NO nonzero 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 11 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?

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