# Eigenvalue Questions

• Jun 20th 2011, 04:33 AM
usagi_killer
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^2-E=0 so that means $\displaystyle \lambda^2-\lambda$ is an eigenvalue of E^2-E=0, since the zero vector only has an eigenvalue of 0, then $\displaystyle \lambda^2-\lambda=0$ 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 $\displaystyle B^TB$ 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 $\displaystyle \lambda \ge 0$
• Jun 20th 2011, 06:59 AM
Deveno
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 $\displaystyle x^T(B^TB)x = <Bx,Bx>\ \ge 0$

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?
• Jun 20th 2011, 08:28 AM
FernandoRevilla
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 $\displaystyle \lambda=1$ .