Linear dependence of three functions: $f(x) = sin(x)$, $g(x) = cos(x)$ and $h(x)=x$












3












$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40


















3












$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40
















3












3








3





$begingroup$



Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces







share|cite|improve this question











$endgroup$





Let $mathbb{R}^mathbb{R} $ be a vector-space of functions $f:mathbb{R}→mathbb{R}$.
The following functions in the vector space are defined as follows:
$$f(x) = sin(x),qquad g(x) = cos(x),qquad h(x) = x.$$
Is the triple $(f,g,h)$ linearly dependent?




I have a feeling it is, because $h(x)$ is the identity function, so $h(sin(x)) = sin(x)$, and I know the definitions, I can't figure out how to put my explanation (if it is correct) properly.



Thanks in advance.






linear-algebra functions vector-spaces




linear-algebra functions vector-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 1 '18 at 14:53









Martin Sleziak

44.8k10118272




44.8k10118272










asked Dec 1 '18 at 7:32









TegernakoTegernako

908




908












  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40




















  • $begingroup$
    Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
    $endgroup$
    – Fakemistake
    Dec 1 '18 at 7:38












  • $begingroup$
    $h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
    $endgroup$
    – Anthony Ter
    Dec 1 '18 at 7:40


















$begingroup$
Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:38






$begingroup$
Linear dependent means there are coefficients $a,b,c$ (not all equal to zero) such that $asin(x)+bcos(x)+cx=0$ for all $xinmathbb{R}$
$endgroup$
– Fakemistake
Dec 1 '18 at 7:38














$begingroup$
$h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
$endgroup$
– Anthony Ter
Dec 1 '18 at 7:40






$begingroup$
$h(x)$ is not the identity function when the operation is function addition, $f(x) = 0$ is. Function composition does not form a vector space, so linear dependence is as defined in the above comment.
$endgroup$
– Anthony Ter
Dec 1 '18 at 7:40












2 Answers
2






active

oldest

votes


















7












$begingroup$

Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
$$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
Now by letting $x=0,pi/2,pi$, we have three linear equations
$$begin{cases}
Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
end{cases}$$
What may we conclude about $A,B,C$? Can you take it from here?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14












  • $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26










  • $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45












  • $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01










  • $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49





















2












$begingroup$

Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    7












    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49


















    7












    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49
















    7












    7








    7





    $begingroup$

    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?






    share|cite|improve this answer











    $endgroup$



    Your feeling is not correct. Recall that, by definition, if such functions are linearly dependent then there are real numbers $A,B,C$ not all zero such that
    $$Asin(x)+Bcos(x)+Cx=0 quad forall xin mathbb{R}.$$
    Now by letting $x=0,pi/2,pi$, we have three linear equations
    $$begin{cases}
    Acdot 0+Bcdot 1+Ccdot 0=0qquad &(x=0)\
    Acdot 1+Bcdot 0+Ccdot frac{pi}{2}=0qquad &(x=frac{pi}{2})\
    Acdot 0+Bcdot (-1)+Ccdot pi=0qquad &(x=pi)\
    end{cases}$$
    What may we conclude about $A,B,C$? Can you take it from here?







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 1 '18 at 10:44

























    answered Dec 1 '18 at 7:52









    Robert ZRobert Z

    97.4k1066137




    97.4k1066137












    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49




















    • $begingroup$
      @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:14












    • $begingroup$
      You mean "your feeling is incorrect", right?
      $endgroup$
      – Pedro A
      Dec 1 '18 at 10:26










    • $begingroup$
      @PedroA Yes, you are right, thanks for pointing out!
      $endgroup$
      – Robert Z
      Dec 1 '18 at 10:45












    • $begingroup$
      Yes, but I realize issue is with my approach
      $endgroup$
      – caverac
      Dec 1 '18 at 12:01










    • $begingroup$
      @RobertZ that we ultimately get that A=B=C=0 for all cases?
      $endgroup$
      – Tegernako
      Dec 1 '18 at 12:49


















    $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14






    $begingroup$
    @caverac What do you mean? OP is interested in the vector space $mathbb{R}^mathbb{R}$, so the domain for $x$ is the whole real line.
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:14














    $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26




    $begingroup$
    You mean "your feeling is incorrect", right?
    $endgroup$
    – Pedro A
    Dec 1 '18 at 10:26












    $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45






    $begingroup$
    @PedroA Yes, you are right, thanks for pointing out!
    $endgroup$
    – Robert Z
    Dec 1 '18 at 10:45














    $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01




    $begingroup$
    Yes, but I realize issue is with my approach
    $endgroup$
    – caverac
    Dec 1 '18 at 12:01












    $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49






    $begingroup$
    @RobertZ that we ultimately get that A=B=C=0 for all cases?
    $endgroup$
    – Tegernako
    Dec 1 '18 at 12:49













    2












    $begingroup$

    Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.






        share|cite|improve this answer









        $endgroup$



        Asserting that ${f,g,h}$ is linearly independent means that$$(forall a,b,cinmathbb{R}):af+bg+ch=0implies a=b=c=0.$$If $af+bg+ch=0$, then $af(0)+bg(0)+ch(0)=0$, which means that $b=0$. You still have $af+ch=0$. But then $afleft(fracpi2right)+cfracpi2=0$, which means that $a+cfracpi2=0$, and $af(pi)+ch(pi)=0$, which means that $cpi=0$. So, $a=c=0$ too.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 1 '18 at 7:56









        José Carlos SantosJosé Carlos Santos

        161k22127232




        161k22127232






























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