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Homework Help: Eigenvalues and eigenvectors for linear transformation

  1. Sep 29, 2011 #1
    1. The problem statement, all variables and given/known data
    V is a vector space consisting all functions f:R->R that is differentiable many times

    (a) Let T:V->V be the transformation T(f)=f'
    Find the (real) eigenvectors and eigenvalues of T

    (b) Let T be transformation T(f)=f"
    Prove that all real number, m is the eigenvalue of T

    2. Relevant equations
    For part a, it means to describe the eigenspace of T for each eigenvalue m

    3. The attempt at a solution
    (a) This means that to find all functions such that T(f)=f'=mf
    I seperate it into 2 cases, 1st is when m=0
    when m=0, all constant function will form the eigenspace of T.
    2nd, m is all real number except 0
    I can only think of 1 type of function that satisfy f'=mf
    which is when f(x)=ae^(bx+c) +d for all real a,b,c,d where a,b=/0 which corresponds to eigenvalue m=ab
    However, it seems very not convincing and there seems to be a better way of writing this.

    (b) This means that we need to show there exist a eigenspace for all m.
    I seperated into 3 cases
    1st, m=0. All functions of the form f(x)=ax+b for all real a,b where a=/0 is the eigenvector.
    2nd, m=-ve. All functions of the form f(x)=a sin(x)+b cos(x) for all non zero a,b.
    3rd, m=+ve. I'm clueless .... T-T
  2. jcsd
  3. Sep 29, 2011 #2


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    Ad b): The second derivative of f(x) = a sin(x)+b cos(x) is -f(x), so that's just an eigenfunction for eigenvalue -1.

    Also, you might want to have a look at ex.

    Just out of curiosity, have you learned about complex numbers yet?
  4. Sep 29, 2011 #3
    So should I put f(x) = ca sin(x)+cb cos(x) instead of f(x) = a sin(x)+b cos(x)?

    For f(x)=ex, isn't it f(x)=f'(x)=f"(x) so it'll work for eigenvalue 1? So if I let f(x)=aex, can I conclude that all non zero real number is the eigenvalue for T?

    Ya, I've learned complex numbers but the question actually state only get the real eigenvectors and eigenvalues.
  5. Sep 29, 2011 #4


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    No, you don't want that restriction on a. a= 0, that is f(x)= b, has the property that f'(x)= 0 which is an eigenvector with m= 0.

    Where did you get this? The second derivative of f(x)= a sin(x)+ b cos(x) is f''(x)= -a sin(x)- b cos(x) which is of the form f''= mf only if m= -1.
    More generally, the solutions to f''= mf, for m negative, are of the form [itex]f(x)= a sin(\sqrt{m} x)+ b cos(\sqrt{m} x)[/itex].

    Try [itex]f(x)= ae^{\sqrt{m}x}+ be^{-\sqrt{m}x}[/itex]
    equivalently, [itex]f(x)= a sinh(\sqrt{m}x)+ b cosh(\sqrt{m}x)[/itex].
  6. Sep 29, 2011 #5


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    What is the second derivative of f(x) = a ex? Can you get it equal to m f(x) for any number m?

    True, but there is an elegant way to solve the problem using complex numbers. You just need to slightly modify the example I gave you. Think about how you can get any number in front when you differentiate ex (the position where HallsOfIvy placed the [itex]\sqrt{m}[/itex] should give you a pretty good hint).
  7. Sep 30, 2011 #6
    Hmm, I know it works but how did you get this 2 equations just by looking at the question? I understand the part where it relates to sin, cos and sinh, cosh.... But just the term inside the bracket.... How did you derive it?
  8. Sep 30, 2011 #7


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    Have you not taken elementary differential equations? The "linear differential equation with constant coefficients", f''= mf, has "characteristic equation" r^2= m which has roots [itex]r= \pm\sqrt{m}[/itex] if m is positive and [itex]r= \pm i\sqrt{m}[/itex] if m is negative. The negative roots give [itex]f= Ce^{i\sqrt{-m}x}+ De^{i\sqrt{-m}x}= C'cos(\sqrt{-m}x)+ D'sin(\sqrt{-m}x)[/itex] while positive roots [itex]f= Ce^{\sqrt{m}x}+ De^{-\sqrt{m}x}[/itex].

    If you did not know that, how did you get the sine and cosine solution?
  9. Sep 30, 2011 #8
    So if we let f(x)=aemx, then we will get f"(x)=am2emx=m2f(x); which is what we want.
    So, is it true that we can conclude f(x)=emx will do coz a is not important?

    Erm, I don't really understand but I'd tried the following but I can't reason....

    let f(x)=a sin(nx) + b cos(nx)
    so f"(x) = -n2f(x)
    Can we derive from here as n can be all real number and n2 is always positive, m will always be a negative value when f(x)=a sin(nx) + b cos(nx). But how does the term [itex]\sqrt{m}[/itex] comes into place?
  10. Sep 30, 2011 #9
    I got the sine and cosine solution by thinking that that is the way a function will get back itself whenever we differentiate it twice.

    I have learned difference equation for discrete calculus. But does it work the same way as I can understand the part of the characteristic equation to get roots [itex]r= \pm\sqrt{m}[/itex].... But, still a bit lost in the later part on getting the function related to ex....
  11. Oct 2, 2011 #10


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    Yep. If you think about eigenvectors for matrices, then it is also true that if v is an eigenvector, then so is r v for any real number r (not equal to zero). The same holds here: if em x is an "eigenvector" (although we usually call them eigenfunctions in the continuous case) then so is r em x for any real number r (not equal to zero).

    Note that actually there are two solutions, and the eigenspace is spanned by these: just like with the sines and cosines the general eigenvector looks like f(x) = a em x + ...

    As for the "elegant" solution I was hinting at: you have shown that f''(x) = m2 f(x). So if you only consider m2 > 0 you have shown that all the positive numbers are eigenvalues. But m2 can also be negative. In that case m must be imaginary and you will get a solutions of the form f(x) = a ei n x (if m = i n where n is a real number). If you again write down the general form and use Euler's identity, you can rewrite this to the solution with the sines and cosines you already had.
  12. Oct 2, 2011 #11


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    Yes, the coefficent "a" is not important because if [itex]\vec{v}[/itex] is an eigenvector of a given linear transformation, then so is any multiple of [itex]\vec{v}[/itex].

    The original equation was f''= -m f,. not f''= -n2. [itex]m= n^2[/itex].

    And, if [itex]f(x)= ae^{nx}+ be^{-nx}[/itex] then [itex]f'(x)= ane^{nx}- bne^{-nx}[/itex] and [itex]f''(x)= an^2e^{nx}+ bn^2e^{-nx}= n^2(ae^{nx}+ be^{-nx})= n^2f(x)[/itex].
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