Recent content by snakebite

I did what you said on another example where H = [{-2,1+i,1-i},{1-i,-1,-2i},{1+i,2i,-1}] I got the eigenvalues which are 2, -3, -3 (I'm sure about this) and the normalized eigenvectors are e1=[{(1-i)/\sqrt{10},-i\sqrt{2/5},\sqrt{2/5}}] e2 = [{(-1-i)/\sqrt{3},1/\sqrt{3}, 0}] e3 =...

Yea I'm sorry there is no A i meant H. And yes I'm sure i have to correct matrix H on. So the correct way would be to just find the eigenvectors and U would just be a matrix with the eigenvectors as the columns?

Homework Statement Hello, the problem is asking me to find a unitary matrix U such that (U bar)^T(H)(U) is diagonal. And we have H = [{7,2,0},{2,4,-2},{0,-2,5}] The Attempt at a Solution I don't know where to start. I tried getting the eigenvalues of the matrix A but that lead to...

Ah yes I didn't think of that so we have (w bar)^T A = (lambda)(w bar)^T replacing w bar by v and then multiplying both sides by (v bar) we get (v^T)(A)(v bar) = (lambda)(v^T)(v bar) but (v^T)(v bar)= ((z1(z1 bar) + ... + zn(zn bar)) which is positive hence (lambda)(v^T)(v bar) is negative...

So, far we have Aw=\lambdaw so by taking the transpose and getting the conjugate we get (w bar)^T(A bar)^T = (w bar)^T (\lambda bar)^T but A is hermitian and lamba is a real number therefore we get (w bar)^T(A) = \lambda (w bar)^T now if I want to arrive at w^TA(w bar) on the right side...

yea if we start off with (w bar)^T(A bar)^T then this is equivalent to the following(I am only looking at the left side of the equation) (w bar)^T(A bar)^T(w bar) [(w bar)^T(A bar)^T(w bar)]^T since it is a 1x1 matrix (w bar)^T(A bar)(w) [(w bar)^T(A bar)(w)]bar (w^T)(A)(w bar) and on...

I'm not to sure what you mean by w_i? Also, I'm still not completely convinced that Aw=\lambdaw is equivalent to A(w bar) = \lambda(w bar). Is it generally true that the conjugate is also an eigenvector? Thanks

Homework Statement Hello, I have the following problem: Suppose A is a hermitian matrix and it has eigenvalue \lambda <=0. Show that A is not positive definite i.e there exists vector v such that (v^T)(A)(v bar) <=0 The Attempt at a Solution Let w be an eigenvetor we have the following...

great THANK YOU very much, i really appreciated your help :D

yea i think I get it now, after doing the change of basis on A with v1 and v2, then doing P(-1)AP is equivalent to the matrix B in your earlier statement since A and B are the same but in different bases correct?

yea in ur case we would have B = \lambda 1 0 \lambda but in our case we don't have the components of v1 and v2 we can just right A in terms of v11,v12,v21,v22? so would P-1AP cancel out and we're sort of left in the basis e1 e2 to get B?

ok so we have v1 is eigenvector hence Av1 = \lambdav1 hence (A-\lambda)v1 = 0 but we also have v1 = (A-\lambdaI)v2 hence Av1 = A(A-\lambdaI)v2 and -\lambdav1 = -\lambda(A-\lambdaI)v2 hence by calculating it out i find (A-\lambdaI)v1 = (A-\lambda)^2 v2 To get v1 =(A-\lambda)v2 which is...

Ok, but what does this tell us about the matrix A, for example in the second part if we take P=[v1 v2] = [v11 v21 then we have v12 v22] P^-1 = v22/det(P) -v21/det(P) -v12/det(P) v11/det(P) So how does P-1AP =...

So in that case can we say that the minimal polynomial is therefore (x-\lambda)^2 and hence we have (A-\lambda)^2 is equal to 0, therefore we get from the original equation that (A-\lambda)\vec{v1}=0 and therefore v1 is an eigenvector. correct?

well if A [ a b the the characteristic polynomial is c d] \lambda^2 + \lambda(-a-d) + (ad -bc) but since we have 1 eigenvalue then the dicriminant must be 0 and hence (a+d)^2-4(ad - bc) = 0 I already tried doing this and calculating (A-\lambdaI)^2 but that leads to nowhere...