Exponential Matrix 2 - complex eigenvalues Euler












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Given a matrix $A=begin{pmatrix} sigma & omega \ -omega & sigma end{pmatrix}$ with two complex eigenvalues $sigmapm iomega$, using the Euler formula $e^{i omega t}=cos(omega t)+i sin(omega t)$ my objective is to show that $e^{At}=e^{sigma t}begin{pmatrix} cos(omega t) & sin(omega t)\ -sin(omega t)& cos(omega t) end{pmatrix}$.

It should be done without using Taylor expansion.






complex-numbers eigenvalues-eigenvectors matrix-exponential







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  • $begingroup$
    Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
    $endgroup$
    – Timothy
    Jan 9 at 18:41










  • $begingroup$
    This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
    $endgroup$
    – amd
    Jan 9 at 20:31
















0












$begingroup$


Given a matrix $A=begin{pmatrix} sigma & omega \ -omega & sigma end{pmatrix}$ with two complex eigenvalues $sigmapm iomega$, using the Euler formula $e^{i omega t}=cos(omega t)+i sin(omega t)$ my objective is to show that $e^{At}=e^{sigma t}begin{pmatrix} cos(omega t) & sin(omega t)\ -sin(omega t)& cos(omega t) end{pmatrix}$.

It should be done without using Taylor expansion.






complex-numbers eigenvalues-eigenvectors matrix-exponential







share|cite|improve this question











$endgroup$












  • $begingroup$
    Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
    $endgroup$
    – Timothy
    Jan 9 at 18:41










  • $begingroup$
    This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
    $endgroup$
    – amd
    Jan 9 at 20:31














0












0








0





$begingroup$


Given a matrix $A=begin{pmatrix} sigma & omega \ -omega & sigma end{pmatrix}$ with two complex eigenvalues $sigmapm iomega$, using the Euler formula $e^{i omega t}=cos(omega t)+i sin(omega t)$ my objective is to show that $e^{At}=e^{sigma t}begin{pmatrix} cos(omega t) & sin(omega t)\ -sin(omega t)& cos(omega t) end{pmatrix}$.

It should be done without using Taylor expansion.






complex-numbers eigenvalues-eigenvectors matrix-exponential







share|cite|improve this question











$endgroup$




Given a matrix $A=begin{pmatrix} sigma & omega \ -omega & sigma end{pmatrix}$ with two complex eigenvalues $sigmapm iomega$, using the Euler formula $e^{i omega t}=cos(omega t)+i sin(omega t)$ my objective is to show that $e^{At}=e^{sigma t}begin{pmatrix} cos(omega t) & sin(omega t)\ -sin(omega t)& cos(omega t) end{pmatrix}$.

It should be done without using Taylor expansion.






complex-numbers eigenvalues-eigenvectors matrix-exponential




complex-numbers eigenvalues-eigenvectors matrix-exponential






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edited Jan 9 at 19:44









user376343

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asked Jan 9 at 18:20









TimothyTimothy

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  • $begingroup$
    Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
    $endgroup$
    – Timothy
    Jan 9 at 18:41










  • $begingroup$
    This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
    $endgroup$
    – amd
    Jan 9 at 20:31


















  • $begingroup$
    Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
    $endgroup$
    – Timothy
    Jan 9 at 18:41










  • $begingroup$
    This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
    $endgroup$
    – amd
    Jan 9 at 20:31
















$begingroup$
Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
$endgroup$
– Timothy
Jan 9 at 18:41




$begingroup$
Yes, sorry I've just added that the proof cannot include the Taylor definition of $e^{At}$.
$endgroup$
– Timothy
Jan 9 at 18:41












$begingroup$
This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
$endgroup$
– amd
Jan 9 at 20:31




$begingroup$
This matrix is diagonalizable over the complex numbers. Find eigenvectors and multiply it out.
$endgroup$
– amd
Jan 9 at 20:31










1 Answer
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$begingroup$

The eigen values of the matrix are :



λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are
$v=begin{bmatrix}1 \ iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} + ibegin{bmatrix}0\ 1end{bmatrix},$ and $v¯= begin{bmatrix}1 \ −iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} − ibegin{bmatrix}0 \ 1end{bmatrix}$



Write,
$A = αI + ωJ, : I = begin{bmatrix}1 & 0\ 0& 1end{bmatrix},J = begin{bmatrix}0 &−1 \ 1& 0end{bmatrix}$



The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.



We have,



$e^{αtI} = begin{bmatrix}e^{αt} & 0 \ 0 & e^{αt}end{bmatrix}, : e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$



which gives the desired result.



Check that this holds true
$e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$
The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.



For any matrix $A in mathbb{R}^{n times n}$ and $t in mathbb{R}$, we have,



$ frac{d}{dt} e^{At} = A e^{At} $



Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{omega tJ} $ above (i.e. for $A = omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.






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    $begingroup$

    The eigen values of the matrix are :



    λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are
    $v=begin{bmatrix}1 \ iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} + ibegin{bmatrix}0\ 1end{bmatrix},$ and $v¯= begin{bmatrix}1 \ −iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} − ibegin{bmatrix}0 \ 1end{bmatrix}$



    Write,
    $A = αI + ωJ, : I = begin{bmatrix}1 & 0\ 0& 1end{bmatrix},J = begin{bmatrix}0 &−1 \ 1& 0end{bmatrix}$



    The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.



    We have,



    $e^{αtI} = begin{bmatrix}e^{αt} & 0 \ 0 & e^{αt}end{bmatrix}, : e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$



    which gives the desired result.



    Check that this holds true
    $e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$
    The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.



    For any matrix $A in mathbb{R}^{n times n}$ and $t in mathbb{R}$, we have,



    $ frac{d}{dt} e^{At} = A e^{At} $



    Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{omega tJ} $ above (i.e. for $A = omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The eigen values of the matrix are :



      λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are
      $v=begin{bmatrix}1 \ iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} + ibegin{bmatrix}0\ 1end{bmatrix},$ and $v¯= begin{bmatrix}1 \ −iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} − ibegin{bmatrix}0 \ 1end{bmatrix}$



      Write,
      $A = αI + ωJ, : I = begin{bmatrix}1 & 0\ 0& 1end{bmatrix},J = begin{bmatrix}0 &−1 \ 1& 0end{bmatrix}$



      The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.



      We have,



      $e^{αtI} = begin{bmatrix}e^{αt} & 0 \ 0 & e^{αt}end{bmatrix}, : e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$



      which gives the desired result.



      Check that this holds true
      $e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$
      The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.



      For any matrix $A in mathbb{R}^{n times n}$ and $t in mathbb{R}$, we have,



      $ frac{d}{dt} e^{At} = A e^{At} $



      Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{omega tJ} $ above (i.e. for $A = omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The eigen values of the matrix are :



        λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are
        $v=begin{bmatrix}1 \ iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} + ibegin{bmatrix}0\ 1end{bmatrix},$ and $v¯= begin{bmatrix}1 \ −iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} − ibegin{bmatrix}0 \ 1end{bmatrix}$



        Write,
        $A = αI + ωJ, : I = begin{bmatrix}1 & 0\ 0& 1end{bmatrix},J = begin{bmatrix}0 &−1 \ 1& 0end{bmatrix}$



        The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.



        We have,



        $e^{αtI} = begin{bmatrix}e^{αt} & 0 \ 0 & e^{αt}end{bmatrix}, : e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$



        which gives the desired result.



        Check that this holds true
        $e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$
        The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.



        For any matrix $A in mathbb{R}^{n times n}$ and $t in mathbb{R}$, we have,



        $ frac{d}{dt} e^{At} = A e^{At} $



        Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{omega tJ} $ above (i.e. for $A = omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.






        share|cite|improve this answer









        $endgroup$



        The eigen values of the matrix are :



        λ=α+iω and λ¯=α−iω. The corresponding eigenvectors are
        $v=begin{bmatrix}1 \ iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} + ibegin{bmatrix}0\ 1end{bmatrix},$ and $v¯= begin{bmatrix}1 \ −iend{bmatrix} = begin{bmatrix}1 \ 0end{bmatrix} − ibegin{bmatrix}0 \ 1end{bmatrix}$



        Write,
        $A = αI + ωJ, : I = begin{bmatrix}1 & 0\ 0& 1end{bmatrix},J = begin{bmatrix}0 &−1 \ 1& 0end{bmatrix}$



        The matrices I and J commute (check), so $e^{tA}=e^{αtI + ωtJ} = e^{αtI} e^{ωtJ}$.



        We have,



        $e^{αtI} = begin{bmatrix}e^{αt} & 0 \ 0 & e^{αt}end{bmatrix}, : e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$



        which gives the desired result.



        Check that this holds true
        $e^{ωtJ} = begin{bmatrix}cosωt & −sinωt \ sinωt & cosωtend{bmatrix}$
        The definition of matrix exponential (refer to any standard book) is based on the converging Taylor series, but since you are not allowed to use that definition to conclude the above, you may instead use the Lemma for the derivative of a matrix exponential to prove that the above expression holds true.



        For any matrix $A in mathbb{R}^{n times n}$ and $t in mathbb{R}$, we have,



        $ frac{d}{dt} e^{At} = A e^{At} $



        Also, note that the above only holds for scalar matrix $A$ and not when A is also a function of time. Using this lemma, you can substitute the expression for $e^{omega tJ} $ above (i.e. for $A = omega J$, in this case) , differentiate with respect to $t$ and show that it holds to complete the proof.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 17 at 1:18









        wootmanwootman

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