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Math Help - Linear independence of exponential functions

  1. #1
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    Linear independence of exponential functions

    I'm honestly a bit baffled by this problem...

    Let a_1, ... , a_n = distinct numbers that do not equal 0. show that e^(a_1*t), ... , e^(a_n*t) are linearly independent over the numbers.

    It says hint: suppose c_1*e^(a_1*t) + ... + c_n*e^(a_n*t) = 0 for constants c_1, ..., c_n, differentiate n-1 times. The determinant of the coefficients of the system of linear equations should be 0 (Why?)


    I tried the hint, but I don't see how I can prove that the determinant of the coefficients have to equal 0.. Help!
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  2. #2
    Senior Member roninpro's Avatar
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    This is called the Wronskian. You might try reading about it first. Be sure to come back if you have questions about it!

    Wronskian - Wikipedia, the free encyclopedia
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  3. #3
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    Look at the simple case of e^{x} and e^{2x}. If C_1e^x+ C_2e^{2x}= 0 then, differentiating, C_1e^xf+ 2C_2e^{2x}= 0. Set x= 0 in both equations: C_1+ C_2= 0 and C_1+ 2C_2= 0. Those are clearly independent equations and have only C_1= C_2= 0 as solution.

    But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is \left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1 not 0.

    They are saying that if any system of equations has more than one solution, then the determinant of the coefficients must be 0 (if it were not, the matrix of determinants would have an inverse). The point here is to show that the determinant of the coefficients is NOT 0 so there is only the trivial solution.

    In general you have
    \left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|
    and you want to show that, as long as all of the numbers a_i are distinct, that is NOT 0.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by HallsofIvy View Post
    Look at the simple case of e^{x} and e^{2x}. If C_1e^x+ C_2e^{2x}= 0 then, differentiating, C_1e^xf+ 2C_2e^{2x}= 0. Set x= 0 in both equations: C_1+ C_2= 0 and C_1+ 2C_2= 0. Those are clearly independent equations and have only C_1= C_2= 0 as solution.

    But I think you are misunderstanding the hint. In this simple example, the determinant of the coefficients is \left|\begin{array}{cc}1 & 1 \\ 1 & 2\end{array}\right|= 1 not 0.

    They are saying that if any system of equations has more than one solution, then the determinant of the coefficients must be 0 (if it were not, the matrix of determinants would have an inverse). The point here is to show that the determinant of the coefficients is NOT 0 so there is only the trivial solution.

    In general you have
    \left|\begin{array}{ccccc}1 & 1 & 1 & \cdot\cdot\cdot & 1\\a_1 & a_2 & a_3 & \cdot\cdot\cdot & a_n \\ a_1^2 & a_2^2 & a_3^2 & \cdot\cdot\cdot & a_n^2 \\\cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot & \cdot\cdot\cdot a_1^{n-1} & a_2^{n-1} & a_3^{n-1} & \cdot\cdot\cdot & a_n^{n-1}\end{array}\right|
    and you want to show that, as long as all of the numbers a_i are distinct, that is NOT 0.
    This isn't a necessarily simple task. May I suggest you consider A^T. Then, things might look a little more familiar
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  5. #5
    MHF Contributor FernandoRevilla's Avatar
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    Out of context, consider:

    D\in \textrm{End}\;\mathcal{C}^{\infty}(\mathbb{R},\mat  hbb{R}),\;D(x(t))=x'(t)

    Then, e^{a_1t},\ldots,e^{a_nt} are eigenvectors associated respectively to the pairwise distinct eigenvalues a_1,\ldots,a_n.

    So, e^{a_1t},\ldots,e^{a_nt} are linearly independent.

    Regards.
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  6. #6
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    Sweet! Thank you so much. Makes sense now
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  7. #7
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    Re: Linear independence of exponential functions

    Hi and how are you
    I have the following function:
    Let z be a complex number. Is the functions (n^z) are linearly independent for all n natural. If this is true can you indicate a reference for that.
    Thank you in advance.
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  8. #8
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    Re: Linear independence of exponential functions

    This follows from the problem at hand. If n > 1

    n^z = e^{z\log n}

    Here a_n = \log n. For a countable set of functions to be independent, every finite subset of functions must be independent (definition). By the problem above we see that this is true, so the n^z are all independent for natural n>1.
    Thanks from zeraoulia
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  9. #9
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    Re: Linear independence of exponential functions

    Thank you very much for your answer. Can you suggest a reference such as book or a paper about this topic. It is very imporatnt to me to see the refernce cited in my reaserch.
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