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











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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.










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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40

















up vote
3
down vote

favorite













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.










share|cite|improve this question
























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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40















up vote
3
down vote

favorite









up vote
3
down vote

favorite












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.










share|cite|improve this question
















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






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edited Dec 1 at 14:53









Martin Sleziak

44.6k7115269




44.6k7115269










asked Dec 1 at 7:32









Tegernako

596




596












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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40




















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












  • $h(x)$ 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.
    – Anthony Ter
    Dec 1 at 7:40


















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






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














$h(x)$ 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.
– Anthony Ter
Dec 1 at 7:40






$h(x)$ 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.
– Anthony Ter
Dec 1 at 7:40












2 Answers
2






active

oldest

votes

















up vote
7
down vote



accepted










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























  • @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.
    – Robert Z
    Dec 1 at 10:14












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










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












  • Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01










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




















up vote
2
down vote













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.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

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    active

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    active

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    up vote
    7
    down vote



    accepted










    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























    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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

















    up vote
    7
    down vote



    accepted










    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























    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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















    up vote
    7
    down vote



    accepted







    up vote
    7
    down vote



    accepted






    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














    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 at 10:44

























    answered Dec 1 at 7:52









    Robert Z

    92.2k1058129




    92.2k1058129












    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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




















    • @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.
      – Robert Z
      Dec 1 at 10:14












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










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












    • Yes, but I realize issue is with my approach
      – caverac
      Dec 1 at 12:01










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


















    @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.
    – Robert Z
    Dec 1 at 10:14






    @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.
    – Robert Z
    Dec 1 at 10:14














    You mean "your feeling is incorrect", right?
    – Pedro A
    Dec 1 at 10:26




    You mean "your feeling is incorrect", right?
    – Pedro A
    Dec 1 at 10:26












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






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














    Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01




    Yes, but I realize issue is with my approach
    – caverac
    Dec 1 at 12:01












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






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












    up vote
    2
    down vote













    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

























      up vote
      2
      down vote













      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























        up vote
        2
        down vote










        up vote
        2
        down vote









        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












        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 at 7:56









        José Carlos Santos

        146k22117217




        146k22117217






























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