If $u,vin V$ then there is a linear transformation $T:Vto V$ such that $T(u)=v$.
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Suppose $V$ is a vector space, not necessarily finite dimensional. If $u,vin V$, is there a linear transformation $T:Vto V$ such that $T(u)=v$.
I proved the result for the finite dimensional case where I used a basis for $V$. What should one do for the infinite dimensional case?
linear-algebra
asked Nov 22 at 18:08
Adel
460216
|
show 4 more comments
up vote
2
down vote
favorite
Suppose $V$ is a vector space, not necessarily finite dimensional. If $u,vin V$, is there a linear transformation $T:Vto V$ such that $T(u)=v$.
I proved the result for the finite dimensional case where I used a basis for $V$. What should one do for the infinite dimensional case?
linear-algebra
asked Nov 22 at 18:08
Adel
460216
1
You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
1
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33
|
show 4 more comments
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Suppose $V$ is a vector space, not necessarily finite dimensional. If $u,vin V$, is there a linear transformation $T:Vto V$ such that $T(u)=v$.
I proved the result for the finite dimensional case where I used a basis for $V$. What should one do for the infinite dimensional case?
linear-algebra
asked Nov 22 at 18:08
Adel
460216
Suppose $V$ is a vector space, not necessarily finite dimensional. If $u,vin V$, is there a linear transformation $T:Vto V$ such that $T(u)=v$.
I proved the result for the finite dimensional case where I used a basis for $V$. What should one do for the infinite dimensional case?
linear-algebra
linear-algebra
asked Nov 22 at 18:08
Adel
460216
asked Nov 22 at 18:08
Adel
460216
asked Nov 22 at 18:08
Adel
460216
asked Nov 22 at 18:08
Adel
460216
asked Nov 22 at 18:08
Adel
460216
460216
1
You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
1
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33
|
show 4 more comments
1
You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
1
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33
1
1
You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
1
1
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33
|
show 4 more comments
1 Answer
1
active
oldest
votes
up vote
-1
down vote
Not always. If $u=0$, then $v$ must be $0$.
If $une0$ and $V$ has an inner product, then $displaystyle T(x)=dfrac{langle x,urangle}{langle u,urangle}v$ works.
Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.
answered Nov 22 at 18:18
lhf
161k9165384
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
-1
down vote
Not always. If $u=0$, then $v$ must be $0$.
If $une0$ and $V$ has an inner product, then $displaystyle T(x)=dfrac{langle x,urangle}{langle u,urangle}v$ works.
Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.
answered Nov 22 at 18:18
lhf
161k9165384
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
add a comment |
up vote
-1
down vote
Not always. If $u=0$, then $v$ must be $0$.
If $une0$ and $V$ has an inner product, then $displaystyle T(x)=dfrac{langle x,urangle}{langle u,urangle}v$ works.
Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.
answered Nov 22 at 18:18
lhf
161k9165384
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
add a comment |
up vote
-1
down vote
up vote
-1
down vote
Not always. If $u=0$, then $v$ must be $0$.
If $une0$ and $V$ has an inner product, then $displaystyle T(x)=dfrac{langle x,urangle}{langle u,urangle}v$ works.
Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.
answered Nov 22 at 18:18
lhf
161k9165384
Not always. If $u=0$, then $v$ must be $0$.
If $une0$ and $V$ has an inner product, then $displaystyle T(x)=dfrac{langle x,urangle}{langle u,urangle}v$ works.
Geometrically, $T$ projects $V$ onto the subspace generated by $u$, which is canonically isomorphic to the subspace generated by $v$.
answered Nov 22 at 18:18
lhf
161k9165384
answered Nov 22 at 18:18
lhf
161k9165384
answered Nov 22 at 18:18
lhf
161k9165384
answered Nov 22 at 18:18
lhf
161k9165384
161k9165384
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
add a comment |
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
I like that you used inner products. But for a general vector space of infinite dimension, is there a way to prove the result without resorting to Zorn's Lemma?
– Adel
Nov 22 at 18:24
1
1
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
This construction fails even in non-Hilbertian Banach spaces that we do know have a basis even without assuming AC, for instance $ell^1$.
– Federico
Nov 22 at 18:35
add a comment |
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You also have a basis in the infinite dimensional case if you accept AC
– Federico
Nov 22 at 18:10
Is there a proof of this result without using a basis?
– Adel
Nov 22 at 18:11
1
This is equivalent to saying that there is a linear functional $fin L(V,mathbb R)$ such that $f(u)=1$. Indeed, in this case you can take $T(x)=f(x)v$.
– Federico
Nov 22 at 18:31
So we want to prove that for every $uin Vsetminus{0}$ there is $fin L(V,mathbb R)$ such that $f(u)neq 0$.
– Federico
Nov 22 at 18:32
Well, assuming the base field is $mathbb R$, otherwise take the appropriate one.
– Federico
Nov 22 at 18:33