If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.












4












$begingroup$



Let $V$ be a vector space with $dim V=n$ .



If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.




Now let $mathcal B={v_1,v_2,ldots ,v_n}$ be a basis of $V$,



since $f,g$ are non-zero linear functionals then $exists v_iin mathcal B $ such that $g(v_i)neq 0implies f(v_i)neq 0$



Take $i=1$ without any loss of generality so take $g(v_1)neq 0,f(v_1)neq 0$.



Now take $c=dfrac{f(v_1)}{g(v_1)}$



Then we need to show that $(f-cg)(v_i)=0forall i$



Now $(f-cg)(v_1)=0$



How to show that $(f-cg)(v_i)=0forall ige 2$



Can someone please help?



Note::Another Question Why do we need the dimension of the vector space to be finite?






linear-algebra vector-spaces linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 10:40












  • $begingroup$
    @AniruddhAgarwal,yes u r right
    $endgroup$
    – user596656
    Dec 23 '18 at 10:40










  • $begingroup$
    Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
    $endgroup$
    – Yanko
    Dec 23 '18 at 15:12
















4












$begingroup$



Let $V$ be a vector space with $dim V=n$ .



If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.




Now let $mathcal B={v_1,v_2,ldots ,v_n}$ be a basis of $V$,



since $f,g$ are non-zero linear functionals then $exists v_iin mathcal B $ such that $g(v_i)neq 0implies f(v_i)neq 0$



Take $i=1$ without any loss of generality so take $g(v_1)neq 0,f(v_1)neq 0$.



Now take $c=dfrac{f(v_1)}{g(v_1)}$



Then we need to show that $(f-cg)(v_i)=0forall i$



Now $(f-cg)(v_1)=0$



How to show that $(f-cg)(v_i)=0forall ige 2$



Can someone please help?



Note::Another Question Why do we need the dimension of the vector space to be finite?






linear-algebra vector-spaces linear-transformations







share|cite|improve this question









$endgroup$












  • $begingroup$
    When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 10:40












  • $begingroup$
    @AniruddhAgarwal,yes u r right
    $endgroup$
    – user596656
    Dec 23 '18 at 10:40










  • $begingroup$
    Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
    $endgroup$
    – Yanko
    Dec 23 '18 at 15:12














4












4








4


2



$begingroup$



Let $V$ be a vector space with $dim V=n$ .



If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.




Now let $mathcal B={v_1,v_2,ldots ,v_n}$ be a basis of $V$,



since $f,g$ are non-zero linear functionals then $exists v_iin mathcal B $ such that $g(v_i)neq 0implies f(v_i)neq 0$



Take $i=1$ without any loss of generality so take $g(v_1)neq 0,f(v_1)neq 0$.



Now take $c=dfrac{f(v_1)}{g(v_1)}$



Then we need to show that $(f-cg)(v_i)=0forall i$



Now $(f-cg)(v_1)=0$



How to show that $(f-cg)(v_i)=0forall ige 2$



Can someone please help?



Note::Another Question Why do we need the dimension of the vector space to be finite?






linear-algebra vector-spaces linear-transformations







share|cite|improve this question









$endgroup$





Let $V$ be a vector space with $dim V=n$ .



If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.




Now let $mathcal B={v_1,v_2,ldots ,v_n}$ be a basis of $V$,



since $f,g$ are non-zero linear functionals then $exists v_iin mathcal B $ such that $g(v_i)neq 0implies f(v_i)neq 0$



Take $i=1$ without any loss of generality so take $g(v_1)neq 0,f(v_1)neq 0$.



Now take $c=dfrac{f(v_1)}{g(v_1)}$



Then we need to show that $(f-cg)(v_i)=0forall i$



Now $(f-cg)(v_1)=0$



How to show that $(f-cg)(v_i)=0forall ige 2$



Can someone please help?



Note::Another Question Why do we need the dimension of the vector space to be finite?






linear-algebra vector-spaces linear-transformations




linear-algebra vector-spaces linear-transformations






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 23 '18 at 10:36







user596656



















  • $begingroup$
    When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 10:40












  • $begingroup$
    @AniruddhAgarwal,yes u r right
    $endgroup$
    – user596656
    Dec 23 '18 at 10:40










  • $begingroup$
    Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
    $endgroup$
    – Yanko
    Dec 23 '18 at 15:12


















  • $begingroup$
    When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
    $endgroup$
    – Aniruddh Agarwal
    Dec 23 '18 at 10:40












  • $begingroup$
    @AniruddhAgarwal,yes u r right
    $endgroup$
    – user596656
    Dec 23 '18 at 10:40










  • $begingroup$
    Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
    $endgroup$
    – Yanko
    Dec 23 '18 at 15:12
















$begingroup$
When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 10:40






$begingroup$
When you say linear functionals, you mean that $f, g : V rightarrow k$ (where $k$ is the field $V$ is over) right?
$endgroup$
– Aniruddh Agarwal
Dec 23 '18 at 10:40














$begingroup$
@AniruddhAgarwal,yes u r right
$endgroup$
– user596656
Dec 23 '18 at 10:40




$begingroup$
@AniruddhAgarwal,yes u r right
$endgroup$
– user596656
Dec 23 '18 at 10:40












$begingroup$
Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
$endgroup$
– Yanko
Dec 23 '18 at 15:12




$begingroup$
Just realized that this question is a duplicate of this one math.stackexchange.com/questions/60460/…
$endgroup$
– Yanko
Dec 23 '18 at 15:12










2 Answers
2






active

oldest

votes


















7












$begingroup$

There is no need for finite dimensionality and no need to use bases. Let $f(x) neq 0$, $y$ be arbitrary and consider $y-frac {f(y)} {f(x)} x$. By linearity we get $f(y-frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=frac {g(x)} {f(x)}$. Hypothesis implies that $c neq 0$ so we can write $f=frac 1 c g$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Did you switch up $f$ and $g$?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 15:09










  • $begingroup$
    @ShubhamJohri Thanks for the comment. I have corrected the answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 23 '18 at 23:15





















2












$begingroup$

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).



By definition $f,g:Vrightarrowmathbb{R}$ are non-zero linear functionals (you can replace $mathbb{R}$ with $mathbb{C}$ or any field). By the rank-nullity theorem we have that $dim ker f = dim ker g = n-1$ Since $ker f subseteq ker g$ we conclude that $ker f = ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)



Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)not = 0$ are real numbers.



Take $c=frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $text{span} ({v})$ the rest of the claim is immediate






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice approach !it does not answer for the infinite dimensional case
    $endgroup$
    – user596656
    Dec 23 '18 at 12:03











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









7












$begingroup$

There is no need for finite dimensionality and no need to use bases. Let $f(x) neq 0$, $y$ be arbitrary and consider $y-frac {f(y)} {f(x)} x$. By linearity we get $f(y-frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=frac {g(x)} {f(x)}$. Hypothesis implies that $c neq 0$ so we can write $f=frac 1 c g$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Did you switch up $f$ and $g$?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 15:09










  • $begingroup$
    @ShubhamJohri Thanks for the comment. I have corrected the answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 23 '18 at 23:15


















7












$begingroup$

There is no need for finite dimensionality and no need to use bases. Let $f(x) neq 0$, $y$ be arbitrary and consider $y-frac {f(y)} {f(x)} x$. By linearity we get $f(y-frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=frac {g(x)} {f(x)}$. Hypothesis implies that $c neq 0$ so we can write $f=frac 1 c g$.






share|cite|improve this answer











$endgroup$









  • 1




    $begingroup$
    Did you switch up $f$ and $g$?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 15:09










  • $begingroup$
    @ShubhamJohri Thanks for the comment. I have corrected the answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 23 '18 at 23:15
















7












7








7





$begingroup$

There is no need for finite dimensionality and no need to use bases. Let $f(x) neq 0$, $y$ be arbitrary and consider $y-frac {f(y)} {f(x)} x$. By linearity we get $f(y-frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=frac {g(x)} {f(x)}$. Hypothesis implies that $c neq 0$ so we can write $f=frac 1 c g$.






share|cite|improve this answer











$endgroup$



There is no need for finite dimensionality and no need to use bases. Let $f(x) neq 0$, $y$ be arbitrary and consider $y-frac {f(y)} {f(x)} x$. By linearity we get $f(y-frac {f(y)} {f(x)} x)=0$. By hypothesis this implies $g(y-frac {f(y)} {f(x)} x)=0$. Hence $g(y)=cf(y)$ where $c=frac {g(x)} {f(x)}$. Hypothesis implies that $c neq 0$ so we can write $f=frac 1 c g$.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 24 '18 at 6:08

























answered Dec 23 '18 at 11:49









Kavi Rama MurthyKavi Rama Murthy

64.8k42766




64.8k42766








  • 1




    $begingroup$
    Did you switch up $f$ and $g$?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 15:09










  • $begingroup$
    @ShubhamJohri Thanks for the comment. I have corrected the answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 23 '18 at 23:15
















  • 1




    $begingroup$
    Did you switch up $f$ and $g$?
    $endgroup$
    – Shubham Johri
    Dec 23 '18 at 15:09










  • $begingroup$
    @ShubhamJohri Thanks for the comment. I have corrected the answer.
    $endgroup$
    – Kavi Rama Murthy
    Dec 23 '18 at 23:15










1




1




$begingroup$
Did you switch up $f$ and $g$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 15:09




$begingroup$
Did you switch up $f$ and $g$?
$endgroup$
– Shubham Johri
Dec 23 '18 at 15:09












$begingroup$
@ShubhamJohri Thanks for the comment. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:15






$begingroup$
@ShubhamJohri Thanks for the comment. I have corrected the answer.
$endgroup$
– Kavi Rama Murthy
Dec 23 '18 at 23:15













2












$begingroup$

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).



By definition $f,g:Vrightarrowmathbb{R}$ are non-zero linear functionals (you can replace $mathbb{R}$ with $mathbb{C}$ or any field). By the rank-nullity theorem we have that $dim ker f = dim ker g = n-1$ Since $ker f subseteq ker g$ we conclude that $ker f = ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)



Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)not = 0$ are real numbers.



Take $c=frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $text{span} ({v})$ the rest of the claim is immediate






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice approach !it does not answer for the infinite dimensional case
    $endgroup$
    – user596656
    Dec 23 '18 at 12:03
















2












$begingroup$

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).



By definition $f,g:Vrightarrowmathbb{R}$ are non-zero linear functionals (you can replace $mathbb{R}$ with $mathbb{C}$ or any field). By the rank-nullity theorem we have that $dim ker f = dim ker g = n-1$ Since $ker f subseteq ker g$ we conclude that $ker f = ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)



Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)not = 0$ are real numbers.



Take $c=frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $text{span} ({v})$ the rest of the claim is immediate






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Nice approach !it does not answer for the infinite dimensional case
    $endgroup$
    – user596656
    Dec 23 '18 at 12:03














2












2








2





$begingroup$

Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).



By definition $f,g:Vrightarrowmathbb{R}$ are non-zero linear functionals (you can replace $mathbb{R}$ with $mathbb{C}$ or any field). By the rank-nullity theorem we have that $dim ker f = dim ker g = n-1$ Since $ker f subseteq ker g$ we conclude that $ker f = ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)



Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)not = 0$ are real numbers.



Take $c=frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $text{span} ({v})$ the rest of the claim is immediate






share|cite|improve this answer









$endgroup$



Let me provide another approach: (it might be used to solve the problem with your last line but it is completely independent to your approach).



By definition $f,g:Vrightarrowmathbb{R}$ are non-zero linear functionals (you can replace $mathbb{R}$ with $mathbb{C}$ or any field). By the rank-nullity theorem we have that $dim ker f = dim ker g = n-1$ Since $ker f subseteq ker g$ we conclude that $ker f = ker g$. (In some sense this shows that $f-cg(v_i)=0$ in your solution because $f(v_i)=g(v_i)=0$.)



Now take a basis $v_1,...,v_{n-1}$ for the kernel and take $v$ which is linearily independent of those. Then $f(v),g(v)not = 0$ are real numbers.



Take $c=frac{f(v)}{g(v)}$. Since $f,g$ are non-zero only on $text{span} ({v})$ the rest of the claim is immediate







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 23 '18 at 11:16









YankoYanko

7,4801729




7,4801729












  • $begingroup$
    Nice approach !it does not answer for the infinite dimensional case
    $endgroup$
    – user596656
    Dec 23 '18 at 12:03


















  • $begingroup$
    Nice approach !it does not answer for the infinite dimensional case
    $endgroup$
    – user596656
    Dec 23 '18 at 12:03
















$begingroup$
Nice approach !it does not answer for the infinite dimensional case
$endgroup$
– user596656
Dec 23 '18 at 12:03




$begingroup$
Nice approach !it does not answer for the infinite dimensional case
$endgroup$
– user596656
Dec 23 '18 at 12:03


















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