Homomorphisms of vectorial spaces, off by a constant
$begingroup$
Knowing that for every vector $v$ , $g(v)=ccdot f(v)$, with $c$ depending on to which vector is applied $g$ , I have to prove that $g = c cdot f$ for one unique constant $c$.
I don't know from where to start solving this problem. Some tips would be helpful.
Thank you.
linear-algebra vector-spaces
edited Dec 24 '18 at 20:04
amWhy
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
$endgroup$
add a comment |
$begingroup$
Knowing that for every vector $v$ , $g(v)=ccdot f(v)$, with $c$ depending on to which vector is applied $g$ , I have to prove that $g = c cdot f$ for one unique constant $c$.
I don't know from where to start solving this problem. Some tips would be helpful.
Thank you.
linear-algebra vector-spaces
edited Dec 24 '18 at 20:04
amWhy
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
$endgroup$
3
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05
add a comment |
$begingroup$
Knowing that for every vector $v$ , $g(v)=ccdot f(v)$, with $c$ depending on to which vector is applied $g$ , I have to prove that $g = c cdot f$ for one unique constant $c$.
I don't know from where to start solving this problem. Some tips would be helpful.
Thank you.
linear-algebra vector-spaces
edited Dec 24 '18 at 20:04
amWhy
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
$endgroup$
Knowing that for every vector $v$ , $g(v)=ccdot f(v)$, with $c$ depending on to which vector is applied $g$ , I have to prove that $g = c cdot f$ for one unique constant $c$.
I don't know from where to start solving this problem. Some tips would be helpful.
Thank you.
linear-algebra vector-spaces
linear-algebra vector-spaces
edited Dec 24 '18 at 20:04
amWhy
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
edited Dec 24 '18 at 20:04
amWhy
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
edited Dec 24 '18 at 20:04
amWhy
1
edited Dec 24 '18 at 20:04
amWhy
1
edited Dec 24 '18 at 20:04
amWhy
1
1
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
asked Dec 24 '18 at 19:58
Adolfo GarciaAdolfo Garcia
61
61
3
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05
add a comment |
3
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05
3
3
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05
add a comment |
1 Answer
1
active
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$begingroup$
Let's try being more specific. I guess that $f$ and $g$ are linear maps $Vto W$, where $V$ and $W$ are vector spaces and the assumption is that, for every $vin V$, there exists a scalar $c_v$ such that $g(v)=c_vf(v)$.
Your task is to prove that there exists a scalar $c$ such that $g(v)=cf(v)$, for every $vin V$.
A hint to start: suppose $f(v_1)$ and $f(v_2)$ are linearly independent; we can write
begin{align}
& g(v_1)=c_{v_1}f(v_1)=pf(v_1) \
& g(v_2)=c_{v_2}f(v_2)=qf(v_2) \
& g(v_1+v_2)=c_{v_1+v_2}f(v_1+v_2)=rf(v_1+v_2)
end{align}
(using $p,q,r$ for simplicity). We have, by linearity,
$$
pf(v_1)+qf(v_2)=rf(v_1)+rf(v_2)
$$
which implies $p=q=r$.
Now the idea is to generalize this by considering ${f(v_1),dots,f(v_n)}$ a basis for the image of $f$.
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
$endgroup$
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Let's try being more specific. I guess that $f$ and $g$ are linear maps $Vto W$, where $V$ and $W$ are vector spaces and the assumption is that, for every $vin V$, there exists a scalar $c_v$ such that $g(v)=c_vf(v)$.
Your task is to prove that there exists a scalar $c$ such that $g(v)=cf(v)$, for every $vin V$.
A hint to start: suppose $f(v_1)$ and $f(v_2)$ are linearly independent; we can write
begin{align}
& g(v_1)=c_{v_1}f(v_1)=pf(v_1) \
& g(v_2)=c_{v_2}f(v_2)=qf(v_2) \
& g(v_1+v_2)=c_{v_1+v_2}f(v_1+v_2)=rf(v_1+v_2)
end{align}
(using $p,q,r$ for simplicity). We have, by linearity,
$$
pf(v_1)+qf(v_2)=rf(v_1)+rf(v_2)
$$
which implies $p=q=r$.
Now the idea is to generalize this by considering ${f(v_1),dots,f(v_n)}$ a basis for the image of $f$.
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
$endgroup$
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
add a comment |
$begingroup$
Let's try being more specific. I guess that $f$ and $g$ are linear maps $Vto W$, where $V$ and $W$ are vector spaces and the assumption is that, for every $vin V$, there exists a scalar $c_v$ such that $g(v)=c_vf(v)$.
Your task is to prove that there exists a scalar $c$ such that $g(v)=cf(v)$, for every $vin V$.
A hint to start: suppose $f(v_1)$ and $f(v_2)$ are linearly independent; we can write
begin{align}
& g(v_1)=c_{v_1}f(v_1)=pf(v_1) \
& g(v_2)=c_{v_2}f(v_2)=qf(v_2) \
& g(v_1+v_2)=c_{v_1+v_2}f(v_1+v_2)=rf(v_1+v_2)
end{align}
(using $p,q,r$ for simplicity). We have, by linearity,
$$
pf(v_1)+qf(v_2)=rf(v_1)+rf(v_2)
$$
which implies $p=q=r$.
Now the idea is to generalize this by considering ${f(v_1),dots,f(v_n)}$ a basis for the image of $f$.
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
$endgroup$
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
add a comment |
$begingroup$
Let's try being more specific. I guess that $f$ and $g$ are linear maps $Vto W$, where $V$ and $W$ are vector spaces and the assumption is that, for every $vin V$, there exists a scalar $c_v$ such that $g(v)=c_vf(v)$.
Your task is to prove that there exists a scalar $c$ such that $g(v)=cf(v)$, for every $vin V$.
A hint to start: suppose $f(v_1)$ and $f(v_2)$ are linearly independent; we can write
begin{align}
& g(v_1)=c_{v_1}f(v_1)=pf(v_1) \
& g(v_2)=c_{v_2}f(v_2)=qf(v_2) \
& g(v_1+v_2)=c_{v_1+v_2}f(v_1+v_2)=rf(v_1+v_2)
end{align}
(using $p,q,r$ for simplicity). We have, by linearity,
$$
pf(v_1)+qf(v_2)=rf(v_1)+rf(v_2)
$$
which implies $p=q=r$.
Now the idea is to generalize this by considering ${f(v_1),dots,f(v_n)}$ a basis for the image of $f$.
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
$endgroup$
Let's try being more specific. I guess that $f$ and $g$ are linear maps $Vto W$, where $V$ and $W$ are vector spaces and the assumption is that, for every $vin V$, there exists a scalar $c_v$ such that $g(v)=c_vf(v)$.
Your task is to prove that there exists a scalar $c$ such that $g(v)=cf(v)$, for every $vin V$.
A hint to start: suppose $f(v_1)$ and $f(v_2)$ are linearly independent; we can write
begin{align}
& g(v_1)=c_{v_1}f(v_1)=pf(v_1) \
& g(v_2)=c_{v_2}f(v_2)=qf(v_2) \
& g(v_1+v_2)=c_{v_1+v_2}f(v_1+v_2)=rf(v_1+v_2)
end{align}
(using $p,q,r$ for simplicity). We have, by linearity,
$$
pf(v_1)+qf(v_2)=rf(v_1)+rf(v_2)
$$
which implies $p=q=r$.
Now the idea is to generalize this by considering ${f(v_1),dots,f(v_n)}$ a basis for the image of $f$.
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
answered Dec 24 '18 at 21:58
egregegreg
182k1486204
182k1486204
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
add a comment |
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
$begingroup$
Thank you very much , I have understood the idea .
$endgroup$
– Adolfo Garcia
Dec 24 '18 at 22:36
add a comment |
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3
$begingroup$
What exactly are $f$ and $g$?
$endgroup$
– angryavian
Dec 24 '18 at 20:05