If $ker fsubset ker g$ where $f,g $ are non-zero linear functionals then show that $f=cg$ for some $cin F$.
$begingroup$
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
$endgroup$
add a comment |
$begingroup$
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
$endgroup$
$begingroup$
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
add a comment |
$begingroup$
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
$endgroup$
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
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
add a comment |
$begingroup$
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
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
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 $f=frac 1 c g$.
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
$endgroup$
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
add a comment |
$begingroup$
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
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
$endgroup$
$begingroup$
Nice approach !it does not answer for the infinite dimensional case
$endgroup$
– user596656
Dec 23 '18 at 12:03
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
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 $f=frac 1 c g$.
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
$endgroup$
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
add a comment |
$begingroup$
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 $f=frac 1 c g$.
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
$endgroup$
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
add a comment |
$begingroup$
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 $f=frac 1 c g$.
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
$endgroup$
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 $f=frac 1 c g$.
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
edited Dec 24 '18 at 6:08
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
answered Dec 23 '18 at 11:49
Kavi Rama MurthyKavi Rama Murthy
64.8k42766
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
add a comment |
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
add a comment |
$begingroup$
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
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
$endgroup$
$begingroup$
Nice approach !it does not answer for the infinite dimensional case
$endgroup$
– user596656
Dec 23 '18 at 12:03
add a comment |
$begingroup$
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
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
$endgroup$
$begingroup$
Nice approach !it does not answer for the infinite dimensional case
$endgroup$
– user596656
Dec 23 '18 at 12:03
add a comment |
$begingroup$
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
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
$endgroup$
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
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
answered Dec 23 '18 at 11:16
YankoYanko
7,4801729
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
add a comment |
$begingroup$
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
add a comment |
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$begingroup$
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