1. ## 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 $\lambda^2-\lambda$ is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then $\lambda^2-\lambda=0$ 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 $B^TB$ corresponding to $\lambda$, then we have $B^TBx = \lambda x$

Now just playing around: $B^TBx.x = x. ((B^TB)^Tx)=x.(B^TB)x= x.(\lambda x)$

but dono how to get to the result $\lambda \ge 0$

2. ## 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 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 $x^T(B^TB)x = \ \ge 0$

note as well that $x^T(B^TB)x = x^T(\lambda x) = \lambda(x^Tx) = \lambda$

can <x,x> be negative?

3. ## Re: Eigenvalue Questions

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