Necessity of linear map conditions












1














I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : mathbb{R}^n rightarrow mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.



I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.






linear-algebra







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  • 4




    It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
    – Eevee Trainer
    Nov 19 '18 at 3:15






  • 3




    It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
    – Lucas Corrêa
    Nov 19 '18 at 3:18


















1














I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : mathbb{R}^n rightarrow mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.



I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.






linear-algebra







share|cite|improve this question


















  • 4




    It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
    – Eevee Trainer
    Nov 19 '18 at 3:15






  • 3




    It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
    – Lucas Corrêa
    Nov 19 '18 at 3:18
















1












1








1


1





I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : mathbb{R}^n rightarrow mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.



I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.






linear-algebra







share|cite|improve this question













I am reading a linear algebra textbook and it mentions that, to be linear, a function $f : mathbb{R}^n rightarrow mathbb{R}^m$ must fulfill $f(x+y) = f(x) + f(y)$ and $f(cx) = c(fx)$.



I'm trying to understand why the second condition is necessary, since taking $y = (c-1)x$ and recursively expanding the sum into $f(x)$ terms seems to be a way to arrive at the same conclusion from only the first condition.






linear-algebra




linear-algebra






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asked Nov 19 '18 at 3:11









zipzapboing

979




979








  • 4




    It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
    – Eevee Trainer
    Nov 19 '18 at 3:15






  • 3




    It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
    – Lucas Corrêa
    Nov 19 '18 at 3:18
















  • 4




    It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
    – Eevee Trainer
    Nov 19 '18 at 3:15






  • 3




    It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
    – Lucas Corrêa
    Nov 19 '18 at 3:18










4




4




It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
– Eevee Trainer
Nov 19 '18 at 3:15




It's probably just a convenience sort of thing, and more helpful for non-integers $c$ (if I'm understanding your argument correctly). Personally I always see the two combined: for constants $alpha,beta$, $f$ is linear if $$f(alpha x + beta y) = alpha f(x) + beta f(y)$$
– Eevee Trainer
Nov 19 '18 at 3:15




3




3




It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
– Lucas Corrêa
Nov 19 '18 at 3:18






It seems to me that you want to write $cx$ as $x + cdots + x$ $c$ times. But if $c$ is not integer? So it's more convenient to check $f(cx) = cf(x)$
– Lucas Corrêa
Nov 19 '18 at 3:18












2 Answers
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1














A problem with the proposed way of defining $f(cx)$,



$f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$



is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take



$f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$



since we have $f(-x) = -f(x)$ from



$f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$



But with $c notin Z$, we will never arrive at a result for $f(cx)$.



If $c in Bbb Q$, one can make some progress in this direction via the observation that with



$c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$



we can write



$pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$



whence



$f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$



We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .



Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have



$c_n f(x) = f(c_n x) to f(cx) tag 7$



by the continuity of $f(x)$.






share|cite|improve this answer





























    1














    You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.



    Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.






    share|cite|improve this answer





















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      2 Answers
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      active

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      2 Answers
      2






      active

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      active

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      active

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      1














      A problem with the proposed way of defining $f(cx)$,



      $f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$



      is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take



      $f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$



      since we have $f(-x) = -f(x)$ from



      $f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$



      But with $c notin Z$, we will never arrive at a result for $f(cx)$.



      If $c in Bbb Q$, one can make some progress in this direction via the observation that with



      $c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$



      we can write



      $pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$



      whence



      $f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$



      We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .



      Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have



      $c_n f(x) = f(c_n x) to f(cx) tag 7$



      by the continuity of $f(x)$.






      share|cite|improve this answer


























        1














        A problem with the proposed way of defining $f(cx)$,



        $f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$



        is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take



        $f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$



        since we have $f(-x) = -f(x)$ from



        $f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$



        But with $c notin Z$, we will never arrive at a result for $f(cx)$.



        If $c in Bbb Q$, one can make some progress in this direction via the observation that with



        $c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$



        we can write



        $pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$



        whence



        $f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$



        We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .



        Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have



        $c_n f(x) = f(c_n x) to f(cx) tag 7$



        by the continuity of $f(x)$.






        share|cite|improve this answer
























          1












          1








          1






          A problem with the proposed way of defining $f(cx)$,



          $f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$



          is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take



          $f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$



          since we have $f(-x) = -f(x)$ from



          $f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$



          But with $c notin Z$, we will never arrive at a result for $f(cx)$.



          If $c in Bbb Q$, one can make some progress in this direction via the observation that with



          $c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$



          we can write



          $pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$



          whence



          $f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$



          We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .



          Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have



          $c_n f(x) = f(c_n x) to f(cx) tag 7$



          by the continuity of $f(x)$.






          share|cite|improve this answer












          A problem with the proposed way of defining $f(cx)$,



          $f(cx) = f(x + (c - 1)x) = f(x) + f((c - 1)x), ; text{and so forth}, tag 1$



          is that, unless $c in Bbb N$, the recursive process won't terminate. What happens is $c = sqrt 2$ or $c = pi$, for example? Or even if $0 > c in Bbb Z$? Of course here one may take



          $f(cx) = f(-x + (c + 1)x) = f(-x) + f((c + 1)x), tag 2$



          since we have $f(-x) = -f(x)$ from



          $f(x) + f(-x) = f(x + (-x)) = f(0) = 0. tag 3$



          But with $c notin Z$, we will never arrive at a result for $f(cx)$.



          If $c in Bbb Q$, one can make some progress in this direction via the observation that with



          $c = dfrac{p}{q}, ; p, q in Bbb Z, tag 4$



          we can write



          $pf(x) = f(px) = f left (q dfrac{p}{q}x right ) = qf left (dfrac{p}{q}x right ), tag 5$



          whence



          $f left (dfrac{p}{q}x right ) = dfrac{p}{q}f(x). tag 6$



          We can handle $f(cx) = cf(x)$, $c in Bbb Z$, via $f(x + y) = f(x) + f(y)$ without axiomatizing $f(cx) = cf(x)$ since the addition axiom essenially posits an abelian group homomporphism between $Bbb R^m$ and $Bbb R^n$, and such homomorphisms extend in a natural way to $Bbb Z$-module homomorphisms; as we have seen, rational $c$ then obey $f(cx) = cf(x)$; but for $c$ irrational we are faced with a non-terminating process . . .



          Typically the assumption that $f(x)$ is continuous may be invoked to address the case of irrational $c$. Then if $c_n to c$ with $c_n in Bbb Q$, we have



          $c_n f(x) = f(c_n x) to f(cx) tag 7$



          by the continuity of $f(x)$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 19 '18 at 3:58









          Robert Lewis

          43.7k22963




          43.7k22963























              1














              You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.



              Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.






              share|cite|improve this answer


























                1














                You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.



                Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.






                share|cite|improve this answer
























                  1












                  1








                  1






                  You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.



                  Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.






                  share|cite|improve this answer












                  You are correct that induction combined with $f(x+y)=f(x)+f(y)$ yields $f(nx)=nf(x)$ for $nin mathbb{N}$. However, for $c=pi$ (say) it isn't obvious how one would show that it follows that $f(pi x)=pi f(x)$.



                  Of course, we want our maps to respect scalar multiplication, so it necessitates making the definition $f(lambda x)=lambda f(x)$ for all $lambda in mathbb{R}$.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 19 '18 at 3:43









                  Antonios-Alexandros Robotis

                  9,15741640




                  9,15741640






























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