Recent content by guroten

Okay, so I can use compactness to find a finite open covering that is disjoint from the other neighborhood. How can I turn these open sets in X into disjoint open sets in Y? Since the mapping is closed, I have tried to use complements to find the proper disjoint sets, but it is not working.

Homework Statement Let f : X--> Y be a closed continuous surjective map such that f^(-1)(y) is compact. Show that if X is regular, so is Y .The Attempt at a Solution I'm not sure which piece of info I need to use to start each of these. Any help with the proof would be really appreciated. I...

Homework Statement Show that the map f : R--> S1 given by f(t) =[(t^2-1)/(t^2+1), 2t/(t^2+1)] is a homeomorphism onto S1-{(1, 0)}, where S1 is the unit circle in the plane. I know this is a stereographic projection, but I do not know how to show that it has a continuous inverse. I am...

ok, so I think what I said was (hopefully) a proof that there does exist a B st BAB^-1=R for some x, but it does not show that B has real entries. I just decomposed A and R into diagonal and invertible matrices, and since we can always find an R such that, given A, they have the same...

Let R be the rotation matrix you mentioned for some x. then R=BAB^-1, where A can be diagonalized as PDP^-1 and R can be diagonalized as LDL^-1. Then D=P^-1AP and R=LP^-1APL^-1. So B=LP^-1, since, given A, we can find an R with the same eigenvalues as A. the eigenvalues of R are cosx+-isinx and...

ok, I see that I can turn A into the rotation matrix form using B=LP, where A=PDP^-1 and the rotation matrix=LDL^-1. Is that correct? however, I still would need to show that B is real. Without knowing the eigenvectors of A, but knowing the eigenvectors of the rotation matrix, how do I do that?

ok, so we want a basis st A acts as a rotation. I'm having trouble conceptualizing what basis would do that.

Ok, thanks for the simple example. I think I get the concept now. Any tips on how to manipulate the B's? I need to find a P st PBAB^-1P^-1 is the rotation matrix, right? Does the restriction that det(PB)=1 matter?

Ok, but A is the original matrix, not the one with eigenvalues on the diagonal. Is that what you mean?

Thanks for helping me; sorry I'm a little slow on the uptake. I'll just repeat the parts that I understand. Ok, so the fact that det(A)=1 means that the eigenvalues are the two conjugates you mentioned, I get that. Then we can associate those with 2 eigenvectors that form a basis. How do we know...

I guess I'm not getting it. The eigenvector depends on the eigenvalue. If A=[a,b,c,d] then the eigenvalue is [(a+d)+-sqrt((a+d)^2-4)]/2. If I plug this in and try to find an eigenvector, it gets messy fast. I end up with an equation where an eigenvector is [0,1] and...

yes, that made the computation much easier.

I understand the part about the eigenvalues being complex conjugates (I can use the quadratic formula to find the eigenvalues of an arbitrary 2x2 matrix, and then I get that their product is 1), but I don't understand how to get B.

Homework Statement Let A be 2x2 and det(A)=1 and entries in R. Suppose A does not have any real eigenvalues. Then prove there exists a B st B is 2x2, det(B)=1 and BAB^-1=[cos(x),sin(x),-sin(x),cos(x)] for some x. The Attempt at a Solution I'm not sure how to start this proof. Any...

Let W1 and W2 be subspaces of a finite dimension. Prove that A(W1 intersection with W2) = A(W1) + A(W2) Where A(W1) is the annihilator of W1. I can prove one way, but not the other. How do I prove the rhs is contained in the lhs?