
Eigenvalue Questions
http://img94.imageshack.us/img94/3196/idempotent.jpg
I can do i) but just a bit stuck on ii).
my working for i) is: E^2E=0 so that means $\displaystyle \lambda^2\lambda$ is an eigenvalue of E^2E=0, since the zero vector only has an eigenvalue of 0, then $\displaystyle \lambda^2\lambda=0$ and so \lambda = 0 or 1
cheers :)
[hr]
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 $\displaystyle B^TB$ corresponding to $\displaystyle \lambda$, then we have $\displaystyle B^TBx = \lambda x$
Now just playing around: $\displaystyle B^TBx.x = x. ((B^TB)^Tx)=x.(B^TB)x= x.(\lambda x)$
but dono how to get to the result $\displaystyle \lambda \ge 0$

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 $\displaystyle x^T(B^TB)x = <Bx,Bx>\ \ge 0$
note as well that $\displaystyle x^T(B^TB)x = x^T(\lambda x) = \lambda(x^Tx) = \lambda<x,x>$
can <x,x> be negative?

Re: Eigenvalue Questions
An alternative for the first question: $\displaystyle E^2=E\Rightarrow E(Ex)=Ex$ for all $\displaystyle x\in\mathbb{R}^n$. Consider $\displaystyle x\neq 0$ , if $\displaystyle Ex=0$ then $\displaystyle x$ is an eigenvector associated to $\displaystyle \lambda=0$, if $\displaystyle Ex\neq 0$ then $\displaystyle Ex$ is an eigenvector associated to $\displaystyle \lambda=1$ .