What is the proof for existence of B in this scenario?

[guroten]

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 help would be appreciated.

 

[Dick]

If A is real and doesn't have real eigenvalues then it has two complex conjugate eigenvalues and their product is 1. That means they can be written in the form cos(x)+/-i*sin(x). Can you say why all these things are true? Isn't that starting to sound like a rotation matrix? Can you pick B to change the standard basis into a linear combination of those eigenvectors so that you get that form for BAB^(-1)?

 

[guroten]

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.

 

[Dick]

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, but then I get that their product is not 1 for some reason), but I don't understand how to get B.

Did you use the fact that det(A)=1?

 

[guroten]

yes, that made the computation much easier.

 

[guroten]

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 d-a-sqrt((a+d)^2-4)-2cb/[a-d-sqrt((a+d)^2-4)] must equal 0 for the matrix to be singular. What am I supposed to derive from this?

 

[Dick]

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 d-a-sqrt((a+d)^2-4)-2cb/[a-d-sqrt((a+d)^2-4)] must equal 0 for the matrix to be singular. What am I supposed to derive from this?

You are doing this WAY TOO explicitly. You want to show the matrix exists, not explicitly calculate it. You can think of B as just representing a choice of basis. You know (I hope) that you can write the two eigenvalues as cos(x)+isin(x) and cos(x)-isin(x) for some x. Can you explain why? So let v1 and v2 be the corresponding two eigenvectors. That means if you pick the B corresponding to that choice of basis, then BAB^(-1) is diagonal with the two eigenvalues along the diagonal. Can you think of a way to make linear combinations of v1 and v2 to create another basis where A takes the given form?

 

[guroten]

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 that BAB^-1 is a diagonal matrix? By "corresponding", do you mean using the eigenvectors as the columns of B?

 

[Dick]

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 that BAB^-1 is a diagonal matrix? By "corresponding", do you mean using the eigenvectors as the columns of B?

Yes, that's it. In this case I would use the eigenvectors as the columns of B^(-1), Then B^(-1) maps (1,0) and (0,1) into v1 and v2. A multiplies each one by it's eigenvalue and B transforms the eigenvectors back to (1,0) and (0,1) times the eigenvalues. Hence the product is diagonal. Now do some more messing around with the basis to get the form you want.

 

[guroten]

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

 

[Dick]

Look. Here's an example. The matrix A=[[2,1],[1,2]] has eigenvalues 1 and 3. The eigenvectors are simple. Construct a matrix B such that BAB^(-1)=[[1,0],[0,3]]. Just do it for practice, ok? No, A is not diagonal. Now suppose I told only that the matrix A has eigenvalues 1 and 3 without telling what the matrix is. You can still say if I set the matrix B^(-1) to have columns v1 and v2 where v1 has the eigenvalue 1 and v2 has the eigenvalue 3 then BAB^(-1) is [[1,0],[0,3]]. Do you see why? Once you've done that with the eigenvalues cos(x)+/-i*sin(x) you still have to rearrange the basis a bit to get the indicated form for BAB^(-1).

 

[guroten]

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?

 

[Dick]

Ok, thanks for the simple example. I think I get the concept now. Any tips on how to manipulate the B's? The restriction that det(B)=1 seems to make it even more difficult.

det(B)=1 doesn't make it any more difficult. If C=BAB^(-1) then C=(B*k)*A*(B*k)^(-1) for any constant k. I've been trying to explain that you shouldn't worry about the form of B. What you should be trying to do is find a basis where A takes the form you want. Then describe the B. It just does the basis change.

 

[guroten]

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

 

[Dick]

You know a basis that diagonalizes A, right? Try this. Diagonalize [cos(x),sin(x),-sin(x),cos(x)]. You'll get the same diagonal matrix. Now work backwards.

 

[Dick]

Try this. Take the matrix [cos(x),sin(x),-sin(x),cos(x)] and diagonalize it. You'll get the same eigenvalues as A. Now figure out the relation between the basis where A looks like a rotation matrix and the basis of eigenvectors.

 

[guroten]

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?

 

[Dick]

You ask a lot of questions without showing a lot of work, no offence, but what does B=LP mean? And why do you think think B has to be real? I'm going to ask you once more. What are the eigenvalues and eigenvectors of [[cos(x),sin(x)],[-sin(x),cos(x)]]? If you can actually work through this that might give you some insight into how to handle the case were the real matrix A has det(A)=1 and has no real eigenvalues.

 

[guroten]

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 the respective eigenvectors are [+-i,1]. I apologize if I seem like I'm not trying to do any work on this; I did work out the example you gave before, I just did not mention it. Anyway, I hope I'm being clear on what I've got so far. The last thing is to show B is real, since that was part of the problem.

 

[Dick]

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 the respective eigenvectors are [+-i,1]. I apologize if I seem like I'm not trying to do any work on this; I did work out the example you gave before, I just did not mention it. Anyway, I hope I'm being clear on what I've got so far. The last thing is to show B is real, since that was part of the problem.

Ok. The trouble is that I'm not getting enough feedback to tell where you are in the proof? Can you summarize your progress so far? What I really wanted out of that is that the eigenvectors are v1=[-i,1] and v2=[i,1]. In that basis A is diagonal. Notice that the original basis where A looks like a rotation can now be written (0,1)=(v1+v2)/2 and (1,0)=(v2-v1)/(2i). Can you apply that to your problem?

 

[guroten]

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 eigenvalues, then they have the same diagonal matrix with the eigenvalues on the diagonal. Then I just manipulated it to find a B.

 

[Dick]

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 eigenvalues, then they have the same diagonal matrix with the eigenvalues on the diagonal. Then I just manipulated it to find a B.

That's true of any two matrices that have the same two distinct eigenvalues. To say anything about what B is you are going to have to be more specific about the basis. I'll get you started. Here's a big hint. Suppose v1 and v2 are the two eigenvectors of A. What does A look like in the basis u1=(v1+v2)/2 and u2=(v2-v1)/(2i)? Does that get you started?