Does $D$ have real eigenvalues?












1















Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.




  1. Prove that $D$ has a real eigenvalue.


  2. What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?





My attempt: (I claim 1 is false)



Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get



$$a(1-lambda) sin t = (1+ lambda) b cos t$$



This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.



But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.



And for 2, I know every operator over complex field has eigenvalue; is it real in this case?






linear-algebra eigenvalues-eigenvectors linear-transformations







share|cite|improve this question
























  • Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
    – Yadati Kiran
    Nov 24 at 16:18










  • Thanks it completes 2
    – Cloud JR
    Nov 24 at 16:24










  • By the way it is possible with $a sin t + b cos t$ also. Try it.
    – Yadati Kiran
    Nov 24 at 16:25










  • If scalars are real then we dont get eigen value
    – Cloud JR
    Nov 24 at 16:28










  • Yeah but you have to show it.
    – Yadati Kiran
    Nov 24 at 16:28
















1















Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.




  1. Prove that $D$ has a real eigenvalue.


  2. What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?





My attempt: (I claim 1 is false)



Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get



$$a(1-lambda) sin t = (1+ lambda) b cos t$$



This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.



But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.



And for 2, I know every operator over complex field has eigenvalue; is it real in this case?






linear-algebra eigenvalues-eigenvectors linear-transformations







share|cite|improve this question
























  • Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
    – Yadati Kiran
    Nov 24 at 16:18










  • Thanks it completes 2
    – Cloud JR
    Nov 24 at 16:24










  • By the way it is possible with $a sin t + b cos t$ also. Try it.
    – Yadati Kiran
    Nov 24 at 16:25










  • If scalars are real then we dont get eigen value
    – Cloud JR
    Nov 24 at 16:28










  • Yeah but you have to show it.
    – Yadati Kiran
    Nov 24 at 16:28














1












1








1








Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.




  1. Prove that $D$ has a real eigenvalue.


  2. What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?





My attempt: (I claim 1 is false)



Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get



$$a(1-lambda) sin t = (1+ lambda) b cos t$$



This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.



But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.



And for 2, I know every operator over complex field has eigenvalue; is it real in this case?






linear-algebra eigenvalues-eigenvectors linear-transformations







share|cite|improve this question
















Let $V$ be the subspace of the real vector space of real valued functions
on $mathbb{R}$, spanned by $cos t$ and $sin t$. Let $D : V to V$ be the linear map sending $f(t) in V $ to $dfrac{ df(t)}{dt}$.




  1. Prove that $D$ has a real eigenvalue.


  2. What happens if we consider instead the complex vector space of complex valued functions? Does it have an eigenvalue?





My attempt: (I claim 1 is false)



Suppose $D$ has an eigenvalue.
Let $v$ be an eigenvector $v=a sin t + b cos t$
then by using definition I get



$$a(1-lambda) sin t = (1+ lambda) b cos t$$



This holds for all $t$, $tinmathbb{R}$. Put $t=0$, $b=0$ or $lambda=-1$ and both don't work. So it has no eigenvalue.



But I think my answer to 1 is concrete, I want to see a intuitive and rigorous proof.



And for 2, I know every operator over complex field has eigenvalue; is it real in this case?






linear-algebra eigenvalues-eigenvectors linear-transformations




linear-algebra eigenvalues-eigenvectors linear-transformations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 17:33









José Carlos Santos

148k22117218




148k22117218










asked Nov 24 at 15:58









Cloud JR

806417




806417












  • Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
    – Yadati Kiran
    Nov 24 at 16:18










  • Thanks it completes 2
    – Cloud JR
    Nov 24 at 16:24










  • By the way it is possible with $a sin t + b cos t$ also. Try it.
    – Yadati Kiran
    Nov 24 at 16:25










  • If scalars are real then we dont get eigen value
    – Cloud JR
    Nov 24 at 16:28










  • Yeah but you have to show it.
    – Yadati Kiran
    Nov 24 at 16:28


















  • Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
    – Yadati Kiran
    Nov 24 at 16:18










  • Thanks it completes 2
    – Cloud JR
    Nov 24 at 16:24










  • By the way it is possible with $a sin t + b cos t$ also. Try it.
    – Yadati Kiran
    Nov 24 at 16:25










  • If scalars are real then we dont get eigen value
    – Cloud JR
    Nov 24 at 16:28










  • Yeah but you have to show it.
    – Yadati Kiran
    Nov 24 at 16:28
















Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18




Consider the eigenvectors as $e^{it}$ and $e^{-it}$ instead of $a sin t + b cos t$ .
– Yadati Kiran
Nov 24 at 16:18












Thanks it completes 2
– Cloud JR
Nov 24 at 16:24




Thanks it completes 2
– Cloud JR
Nov 24 at 16:24












By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25




By the way it is possible with $a sin t + b cos t$ also. Try it.
– Yadati Kiran
Nov 24 at 16:25












If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28




If scalars are real then we dont get eigen value
– Cloud JR
Nov 24 at 16:28












Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28




Yeah but you have to show it.
– Yadati Kiran
Nov 24 at 16:28










2 Answers
2






active

oldest

votes


















2














Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
begin{bmatrix}
0 & -1 \
1 & 0
end{bmatrix}
Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.






share|cite|improve this answer





















  • This helps a lot ... Thanks
    – Cloud JR
    Nov 24 at 17:30



















1














Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.



For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$






share|cite|improve this answer





















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    2 Answers
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    2 Answers
    2






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    2














    Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
    begin{bmatrix}
    0 & -1 \
    1 & 0
    end{bmatrix}
    Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.






    share|cite|improve this answer





















    • This helps a lot ... Thanks
      – Cloud JR
      Nov 24 at 17:30
















    2














    Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
    begin{bmatrix}
    0 & -1 \
    1 & 0
    end{bmatrix}
    Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.






    share|cite|improve this answer





















    • This helps a lot ... Thanks
      – Cloud JR
      Nov 24 at 17:30














    2












    2








    2






    Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
    begin{bmatrix}
    0 & -1 \
    1 & 0
    end{bmatrix}
    Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.






    share|cite|improve this answer












    Let $mathscr{B}={cos t,sin t}$ be the basis of $V$; then the matrix of $T$ with respect to this basis is
    begin{bmatrix}
    0 & -1 \
    1 & 0
    end{bmatrix}
    Thus the characteristic polynomial is $X^2+1$, which has roots $i$ and $-i$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 24 at 16:50









    egreg

    177k1484198




    177k1484198












    • This helps a lot ... Thanks
      – Cloud JR
      Nov 24 at 17:30


















    • This helps a lot ... Thanks
      – Cloud JR
      Nov 24 at 17:30
















    This helps a lot ... Thanks
    – Cloud JR
    Nov 24 at 17:30




    This helps a lot ... Thanks
    – Cloud JR
    Nov 24 at 17:30











    1














    Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.



    For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$






    share|cite|improve this answer


























      1














      Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.



      For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$






      share|cite|improve this answer
























        1












        1








        1






        Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.



        For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$






        share|cite|improve this answer












        Your answer to 1) is fine. You can also say that if $fcolonmathbb{R}longrightarrowmathbb R$ is such that $f'=af$ (for a real number $a$), then $f(x)=ke^{ax}$, for some $kinmathbb R$. SUch a function doesn't belong to your space.



        For 2), use the fact that, if $f(t)=cos(t)+sin(t)i$, then$$f'(t)=-sin(t)+cos(t)i=if(t).$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 16:47









        José Carlos Santos

        148k22117218




        148k22117218






























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