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







share|cite|improve this question


















  • 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?
    – UserA
    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















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







share|cite|improve this question


















  • 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?
    – UserA
    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













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







share|cite|improve this question













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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 22 at 18:08









UserA

496216




496216








  • 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?
    – UserA
    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




    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?
    – UserA
    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?
– UserA
Nov 22 at 18:11




Is there a proof of this result without using a basis?
– UserA
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










1 Answer
1






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-1
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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 $v$.






share|cite|improve this answer





















  • 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?
    – UserA
    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











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-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 $v$.






share|cite|improve this answer





















  • 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?
    – UserA
    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















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 $v$.






share|cite|improve this answer





















  • 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?
    – UserA
    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













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 $v$.






share|cite|improve this answer












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 $v$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 22 at 18:18









lhf

162k9166385




162k9166385












  • 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?
    – UserA
    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?
    – UserA
    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?
– UserA
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?
– UserA
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


















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