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Math Help - Linear map eigenvalues

  1. #1
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    Linear map eigenvalues

    Could someone please explain to me the following?

    Suppose you have a vector space, and a linear map T. need to find eigenvalues. when T is a matrix (2 by 2 or 3 by 3 or whichever square matrix with actual VALUES) this is easy, just work out the characteristic polynomial det(I*x - T)=0 and then the eigenvalues come out.

    HOWEVER,
    what does one do when T is a function? say, T(f) = f + f ' where f ' denotes the derivative of f?
    so we don't have T as a matrix but as a function...
    say, T is applied along the basis (1, x, x^2, x^3, ... , x^n)
    do we need to apply T to every element of the basis and then get a matrix out of that to work with? or how??
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  2. #2
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    Re: Linear map eigenvalues

    Hey cassius2020.

    You will have to either calculate the value for a specific value given your input (like a particular value of x) or do a symbolic calculation using the construction of your function and derivative if it has an explicit form.

    There is an area of mathematics known as operator algebra's and it allows you to calculate a function of an operator so you can use this to get the value of your operator given a function and if this can be evaluated, then you can get a specific value of your operator under some transformation and apply the usual technique.
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  3. #3
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    Re: Linear map eigenvalues

    in the example you gave, you are looking for eigenfunctions. in this case, T is the linear operator 1+D, where D(f) = f'.

    solving (1+D)(f) = λf leads to the differential equation:

    (1-λ)f + f' = 0. what the solution set is going to be is going to depend a LOT on what kind of function "f" can be (real-valued, complex-valued, polynomial, etc.).

    for example, if f is a polynomial, then f must be the 0-polynomial. if f is a real-valued function on the interval [a,b], then f is of the form:

    f(t) = Ke^{-\left(\frac{t}{1-\lambda}\right)}

    in this example, 1+D might have every real eigenvalue except 1 (the spectrum is not a finite set). if we are given more information about f (often as "boundary conditions"), say f is defined on [0,1], with f(0) = 1, f(1) = 2:

    we get K = 1 (from f(0) = 1), and λ = log(2) + 1 (from f(1) = 2).
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