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Linear algebra question

  1. Apr 6, 2010 #1
    1. The problem statement, all variables and given/known data
    Let A be a real n x n matrix. Prove that we can find a subspace V in R^N such that 1 <= dim V < = 2 and A(V) is a subset of V.


    2. Relevant equations
    None I don't think.


    3. The attempt at a solution
    I know that the eigenspace of a matrix satisfies the condition that A(E) is a subset of E since for any vectors v in the eigenspace, Av = λv. Since we know v is in the eigenspace, any multiple of v by λ is also in the eigenspace. Now how am I supposed to incorporate the dimension into the argument?
     
  2. jcsd
  3. Apr 6, 2010 #2

    Mark44

    Staff: Mentor

    Another possibility is the nullspace of A. If null(A) is a line through the origin, then dim(null(A)) = 1. If null(A) is a plane containing the origin, then dim(null(A)) = 2.
     
  4. Apr 6, 2010 #3
    Yes I knew that, but how does it relate to A - λI?
     
  5. Apr 6, 2010 #4

    Mark44

    Staff: Mentor

    It doesn't. Your problem statement doesn't mention anything about eigenvalues or eigenspaces, so why is that a concern?
     
  6. Apr 6, 2010 #5
    I guess I was just trying to stick with the eigenspace argument. So Av = λv and the set of all solutions v that satisfies this is in the eigenspace. And (A-λI)v = 0. So how do I prove that the dimension of the eigenspace is between or equal to 1 and 2?
     
  7. Apr 6, 2010 #6

    Mark44

    Staff: Mentor

    The dimension in your proof has to be either 1 or 2; it can't be some number in between.

    To get to eigenvalues, eigenvectors, etc. you first have to get the characteristic polynomial, and then you have to factor it. The Fund. Thm. of Algebra states that every polynomial of degree n with complex coefficients has at least one complex root.
     
  8. Apr 8, 2010 #7
    Ok, I'm still confused how to do this problem...

    So you have a matrix A. I solve det(A-λI) = 0 and find the eigenvalues. Suppose I take an eigenvalue and plug it into A-λI and solve the equation (A-λI)x = 0. The the set of all solutions to this equation is the eigenspace. I know that the eigenspace satisfies A(V) subset of V because Ax = λx and if x is in the eigenspace, then any scalar multiple of x is also in the eigenspace by the axioms of a vector space. If the dimension is 1, then I have a line, and if the dimension is 2, I have a plane. Is that what they're asking for? That doesn't seem right or complete. What is meant by a space where the dimension is 1<= dimV <= 2?
     
  9. Apr 8, 2010 #8

    Mark44

    Staff: Mentor

    For your last question, 1 <= dim(V) <= 2 means that dim(V) = 1 or dim(V) = 2.

    What this problem boils down to is that you need to show that for any nxn matrix A that there is an eigenspace of dimension 1 or there is an eigenspace of dimension 2. Think about what you need to do to get the eigenvalues.
     
  10. Apr 8, 2010 #9
    Ok, so I know that if A has distinct eigenvalues λ1, ..., λp, then for 1 <= k <= p the dimension of the eigenspace for λk is less than or equal to the multiplicity of the eigenvalue λi. So for this problem, that would imply that there has to be an eigenvalue with multiplicity of 1 or an eigenvalue with multiplicity of 2. But how am I supposed to show that this is the case for any real n x n matrix?
     
  11. Apr 8, 2010 #10
    What if A is real, but it does not have any real eigenvalue (i.e. only complex eigenvalues) like A = [0 -1],[1, 0] a 2x2 matrix?

    I don't think that your line of argument would apply here since there is no eigenspace in R^n with complex eigenvalues. :(
     
  12. Apr 8, 2010 #11
    Gah! I still don't get it. I don't even think I have any sort of argument going... I'm just that confused. I just don't see how we can say that the dimension of V is either 1 or 2 when we don't have much information about A.
     
  13. Apr 9, 2010 #12
    Remember V can be almost anything that you want as long as you define it properly and it satisfies the given conditions. What I did was separating the problem into two cases, the first one is pretty easy, the second one, not so much.

    Case 1: A has some real eigenvalue
    Then there is some eigenvector v corresponding to that particular eigenvalue.
    You can DEFINE V = span (v) and you can check that V satisfy all the given conditions.

    Case 2: Uber hard for me :(, but I got it, and now you going to have to think a bit
    A has ONLY complex eigenvalue a + bi
    How would you define V? And then check it? Remember that there still exist some eigenvector (only that its complex)

    Now I got to do 3 and 4, want to give me some hints on those? :d
     
  14. Apr 9, 2010 #13

    Mark44

    Staff: Mentor

    But where do the eigenvalues come from? Look at what I said in post #6.
     
  15. Apr 9, 2010 #14
    Ok, so for your case 1, we have an eigenvalue that gave us some eigenvector v. I define V to be the span of v. Because it is just the one vector, it is linearly independent (which we know is true since an eigenvector is not the zero vector). Thus, the dimension is equal to 1. And if v is in the eigenspace, then every multiple of v is also in the eigenspace by some axiom somewhere. So the A(V) is a subset of V because this property holds for an eigenspace.

    For case 2, we have a complex eigenvector. If this was of the form [a + bi], couldn't we say that the eigenvector is given by v = [a] + i? And then do the same thing as before, where we define the span of v to be V? Except that would still give me a dimension of one... hmm.



    Not sure when you posted your response, but here's my hint for 3b: Try to prove first that Bx is also an eigenvector of A.
     
  16. Apr 9, 2010 #15
    You can't define V to be span of v for case 2 because V is a subspace in R^n and v is a complex eigenvector (i.e. the span of a complex vector will create complex vectors outside of R^n). Try separate v into Re(v) and Im(v) and make V to be a span of those vectors. (This is where dim = 2 come from).

    Also you will need A Re(x) = Re (Ax) and likewise for the other component to verify the other property of V. You can find this fact on page 340.

    And I'm already done with 3, working on 4 now.
     
  17. Apr 9, 2010 #16
    Ah! Makes so much more sense! Thank you SO much for your help :)

    I just finished 4a. Find a basis for nul((C-lambda I)^2) and relate this to nul(C - lambda I).. if you haven't already.
     
  18. Apr 9, 2010 #17
    Eh not sure what you just said, but I just make P to be the identity matrix, then C is the same as the matrix between P and P^1. After all, it asks for some matrix P.
     
  19. Apr 9, 2010 #18
    I'm stuck on 4b, you got any plan on how to do it?
     
  20. Apr 9, 2010 #19
    Hmm, not entirely sure. If we consider the equation XA = BX, we must be able to show that it has a complex solution, which would account for the A = PBP^-1 and that it also has a real solution, which would account for the A = QBQ^-1. Maybe if we start with one and then work to the other?
     
  21. Apr 9, 2010 #20
    How do you know that the equation xA = Bx has a complex solution? And what does that have to do with PBP^-1? I don't see the reasoning, care to elaborate?
     
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