Prove that if $T:U to V$ is an isomorphism and $dim_F (V)=nin mathbb{N}$, show that $dim_F(U)=nin mathbb{N}$
Prove that if $T:U to V$ is an isomorphism and $dim_F (V)=nin mathbb{N}$, show that $dim_F(U)=nin mathbb{N}$
my attempt: [any of these is true? please tell me if there exists any mistake, My prof is so careful - and sorry I don't speak English well. Thanks]
- First proof:
Let ${T(u_1),...,T(u_n)}$ is a basis for $V$.
it is obvious that $span{u_1,...,u_n} subseteq U$.
if $uin U$, then there exists $T(u) in rang(T)=V$ [since $T$ is onto, then $rang(T)=V$]
Hence, $T(u)$ can be written as a linear combination of $T(u_1),...,T(u_n)$.
$T(u)=alpha_1 T(u_1)+...+alpha_n T(u_n)$, where $alpha_1, ..., alpha_1 in F$
by linearity of $T$
$T(u)=T(alpha_1 u_1)+...+T(alpha_n u_n)=T(alpha_1 u_1+...+alpha_n u_n)$
since $T$ is one-to-one, then
$u=alpha_1 u_1+...+alpha_n u_n$
Thus, $u in operatorname{span} {u_1,...,u_n}$.
Therefore, $U subseteq operatorname{span}{u_1,...,u_n}$
As a result, $U=operatorname{span}{u_1,...,u_n}$.
If $beta_1 u_1+...+beta_n u_n=0_U$, where $beta_1, ...,beta_n in F$
Then $T(beta_1 u_1+...+beta_n u_n)=beta_1 T(u_1)+...+beta_n T(u_n)= 0_V$
since $T(u_1),...,T(u_n)$ are linearly independent, so
$beta_1=...=beta_n=0$.
Therefore, $u_1,...,u_n$ are linearly independent.
As a result, ${u_1,...,u_n}$ is a basis for $U$, and $dim_F U =n$.
- Second proof:
Since $T$ is isomorphism, then $T$ invertible, $operatorname{Ker(T)}={0_U}$ and $operatorname {Rang(T)}=V$.
Thus $dim_F operatorname{Ker(T)}=0$ and $dim_F operatorname{Rang(T)}=dim_F V=n$
Therefore, $dim_F U=dim_F operatorname{Ker(T)} + dim_F operatorname{Rang(T)}=0+n=n$.
linear-algebra
asked Nov 30 '18 at 0:00
DimaDima
601416
|
show 3 more comments
Prove that if $T:U to V$ is an isomorphism and $dim_F (V)=nin mathbb{N}$, show that $dim_F(U)=nin mathbb{N}$
my attempt: [any of these is true? please tell me if there exists any mistake, My prof is so careful - and sorry I don't speak English well. Thanks]
- First proof:
Let ${T(u_1),...,T(u_n)}$ is a basis for $V$.
it is obvious that $span{u_1,...,u_n} subseteq U$.
if $uin U$, then there exists $T(u) in rang(T)=V$ [since $T$ is onto, then $rang(T)=V$]
Hence, $T(u)$ can be written as a linear combination of $T(u_1),...,T(u_n)$.
$T(u)=alpha_1 T(u_1)+...+alpha_n T(u_n)$, where $alpha_1, ..., alpha_1 in F$
by linearity of $T$
$T(u)=T(alpha_1 u_1)+...+T(alpha_n u_n)=T(alpha_1 u_1+...+alpha_n u_n)$
since $T$ is one-to-one, then
$u=alpha_1 u_1+...+alpha_n u_n$
Thus, $u in operatorname{span} {u_1,...,u_n}$.
Therefore, $U subseteq operatorname{span}{u_1,...,u_n}$
As a result, $U=operatorname{span}{u_1,...,u_n}$.
If $beta_1 u_1+...+beta_n u_n=0_U$, where $beta_1, ...,beta_n in F$
Then $T(beta_1 u_1+...+beta_n u_n)=beta_1 T(u_1)+...+beta_n T(u_n)= 0_V$
since $T(u_1),...,T(u_n)$ are linearly independent, so
$beta_1=...=beta_n=0$.
Therefore, $u_1,...,u_n$ are linearly independent.
As a result, ${u_1,...,u_n}$ is a basis for $U$, and $dim_F U =n$.
- Second proof:
Since $T$ is isomorphism, then $T$ invertible, $operatorname{Ker(T)}={0_U}$ and $operatorname {Rang(T)}=V$.
Thus $dim_F operatorname{Ker(T)}=0$ and $dim_F operatorname{Rang(T)}=dim_F V=n$
Therefore, $dim_F U=dim_F operatorname{Ker(T)} + dim_F operatorname{Rang(T)}=0+n=n$.
linear-algebra
asked Nov 30 '18 at 0:00
DimaDima
601416
1
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
Use$operatorname{text}$for $operatorname{text}$, whenever "text" is, say, "span".
– Shaun
Nov 30 '18 at 0:17
1
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
1
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
1
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04
|
show 3 more comments
Prove that if $T:U to V$ is an isomorphism and $dim_F (V)=nin mathbb{N}$, show that $dim_F(U)=nin mathbb{N}$
my attempt: [any of these is true? please tell me if there exists any mistake, My prof is so careful - and sorry I don't speak English well. Thanks]
- First proof:
Let ${T(u_1),...,T(u_n)}$ is a basis for $V$.
it is obvious that $span{u_1,...,u_n} subseteq U$.
if $uin U$, then there exists $T(u) in rang(T)=V$ [since $T$ is onto, then $rang(T)=V$]
Hence, $T(u)$ can be written as a linear combination of $T(u_1),...,T(u_n)$.
$T(u)=alpha_1 T(u_1)+...+alpha_n T(u_n)$, where $alpha_1, ..., alpha_1 in F$
by linearity of $T$
$T(u)=T(alpha_1 u_1)+...+T(alpha_n u_n)=T(alpha_1 u_1+...+alpha_n u_n)$
since $T$ is one-to-one, then
$u=alpha_1 u_1+...+alpha_n u_n$
Thus, $u in operatorname{span} {u_1,...,u_n}$.
Therefore, $U subseteq operatorname{span}{u_1,...,u_n}$
As a result, $U=operatorname{span}{u_1,...,u_n}$.
If $beta_1 u_1+...+beta_n u_n=0_U$, where $beta_1, ...,beta_n in F$
Then $T(beta_1 u_1+...+beta_n u_n)=beta_1 T(u_1)+...+beta_n T(u_n)= 0_V$
since $T(u_1),...,T(u_n)$ are linearly independent, so
$beta_1=...=beta_n=0$.
Therefore, $u_1,...,u_n$ are linearly independent.
As a result, ${u_1,...,u_n}$ is a basis for $U$, and $dim_F U =n$.
- Second proof:
Since $T$ is isomorphism, then $T$ invertible, $operatorname{Ker(T)}={0_U}$ and $operatorname {Rang(T)}=V$.
Thus $dim_F operatorname{Ker(T)}=0$ and $dim_F operatorname{Rang(T)}=dim_F V=n$
Therefore, $dim_F U=dim_F operatorname{Ker(T)} + dim_F operatorname{Rang(T)}=0+n=n$.
linear-algebra
asked Nov 30 '18 at 0:00
DimaDima
601416
Prove that if $T:U to V$ is an isomorphism and $dim_F (V)=nin mathbb{N}$, show that $dim_F(U)=nin mathbb{N}$
my attempt: [any of these is true? please tell me if there exists any mistake, My prof is so careful - and sorry I don't speak English well. Thanks]
- First proof:
Let ${T(u_1),...,T(u_n)}$ is a basis for $V$.
it is obvious that $span{u_1,...,u_n} subseteq U$.
if $uin U$, then there exists $T(u) in rang(T)=V$ [since $T$ is onto, then $rang(T)=V$]
Hence, $T(u)$ can be written as a linear combination of $T(u_1),...,T(u_n)$.
$T(u)=alpha_1 T(u_1)+...+alpha_n T(u_n)$, where $alpha_1, ..., alpha_1 in F$
by linearity of $T$
$T(u)=T(alpha_1 u_1)+...+T(alpha_n u_n)=T(alpha_1 u_1+...+alpha_n u_n)$
since $T$ is one-to-one, then
$u=alpha_1 u_1+...+alpha_n u_n$
Thus, $u in operatorname{span} {u_1,...,u_n}$.
Therefore, $U subseteq operatorname{span}{u_1,...,u_n}$
As a result, $U=operatorname{span}{u_1,...,u_n}$.
If $beta_1 u_1+...+beta_n u_n=0_U$, where $beta_1, ...,beta_n in F$
Then $T(beta_1 u_1+...+beta_n u_n)=beta_1 T(u_1)+...+beta_n T(u_n)= 0_V$
since $T(u_1),...,T(u_n)$ are linearly independent, so
$beta_1=...=beta_n=0$.
Therefore, $u_1,...,u_n$ are linearly independent.
As a result, ${u_1,...,u_n}$ is a basis for $U$, and $dim_F U =n$.
- Second proof:
Since $T$ is isomorphism, then $T$ invertible, $operatorname{Ker(T)}={0_U}$ and $operatorname {Rang(T)}=V$.
Thus $dim_F operatorname{Ker(T)}=0$ and $dim_F operatorname{Rang(T)}=dim_F V=n$
Therefore, $dim_F U=dim_F operatorname{Ker(T)} + dim_F operatorname{Rang(T)}=0+n=n$.
linear-algebra
linear-algebra
asked Nov 30 '18 at 0:00
DimaDima
601416
asked Nov 30 '18 at 0:00
DimaDima
601416
edited Dec 3 '18 at 3:09
Dima
asked Nov 30 '18 at 0:00
DimaDima
601416
asked Nov 30 '18 at 0:00
DimaDima
601416
asked Nov 30 '18 at 0:00
DimaDima
601416
601416
1
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
Use$operatorname{text}$for $operatorname{text}$, whenever "text" is, say, "span".
– Shaun
Nov 30 '18 at 0:17
1
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
1
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
1
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04
|
show 3 more comments
1
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
Use$operatorname{text}$for $operatorname{text}$, whenever "text" is, say, "span".
– Shaun
Nov 30 '18 at 0:17
1
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
1
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
1
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04
1
1
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
Use
$operatorname{text}$ for $operatorname{text}$, whenever "text" is, say, "span".– Shaun
Nov 30 '18 at 0:17
Use
$operatorname{text}$ for $operatorname{text}$, whenever "text" is, say, "span".– Shaun
Nov 30 '18 at 0:17
1
1
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
1
1
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
1
1
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04
|
show 3 more comments
1 Answer
1
active
oldest
votes
Your proof is basically right. You could just make it simpler and clearer.
Let ${T(u_1),T(u_2),dots,T(u_n)}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. ${u_1,u_2,dots,u_n}$ is linearly independent.
Indeed, if $alpha_1u_1+alpha_2u_2+dots+alpha_nu_n=0$, then also
$$
0=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)=
alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)
$$
forcing $alpha_1=alpha_2=dots=alpha_n=0$.
Second fact. ${u_1,u_2,dots,u_n}$ spans $V$.
Let $vin V$; then $T(v)=alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)$ for some scalars $alpha_1,alpha_2,dots,alpha_n$. This can be rewritten as
$$
T(v)=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=alpha_1u_1+alpha_2u_2+dots+alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $dimoperatorname{range}(T)=dim V=n$; since $T$ is injective, $dimker(T)=0$. The rank-nullity theorem says
$$
dim U=dimker(T)+dimoperatorname{range}(T)=0+n=n
$$
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
Thank you so much.
– Dima
Dec 6 '18 at 22:15
add a comment |
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Your proof is basically right. You could just make it simpler and clearer.
Let ${T(u_1),T(u_2),dots,T(u_n)}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. ${u_1,u_2,dots,u_n}$ is linearly independent.
Indeed, if $alpha_1u_1+alpha_2u_2+dots+alpha_nu_n=0$, then also
$$
0=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)=
alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)
$$
forcing $alpha_1=alpha_2=dots=alpha_n=0$.
Second fact. ${u_1,u_2,dots,u_n}$ spans $V$.
Let $vin V$; then $T(v)=alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)$ for some scalars $alpha_1,alpha_2,dots,alpha_n$. This can be rewritten as
$$
T(v)=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=alpha_1u_1+alpha_2u_2+dots+alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $dimoperatorname{range}(T)=dim V=n$; since $T$ is injective, $dimker(T)=0$. The rank-nullity theorem says
$$
dim U=dimker(T)+dimoperatorname{range}(T)=0+n=n
$$
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
Thank you so much.
– Dima
Dec 6 '18 at 22:15
add a comment |
Your proof is basically right. You could just make it simpler and clearer.
Let ${T(u_1),T(u_2),dots,T(u_n)}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. ${u_1,u_2,dots,u_n}$ is linearly independent.
Indeed, if $alpha_1u_1+alpha_2u_2+dots+alpha_nu_n=0$, then also
$$
0=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)=
alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)
$$
forcing $alpha_1=alpha_2=dots=alpha_n=0$.
Second fact. ${u_1,u_2,dots,u_n}$ spans $V$.
Let $vin V$; then $T(v)=alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)$ for some scalars $alpha_1,alpha_2,dots,alpha_n$. This can be rewritten as
$$
T(v)=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=alpha_1u_1+alpha_2u_2+dots+alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $dimoperatorname{range}(T)=dim V=n$; since $T$ is injective, $dimker(T)=0$. The rank-nullity theorem says
$$
dim U=dimker(T)+dimoperatorname{range}(T)=0+n=n
$$
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
Thank you so much.
– Dima
Dec 6 '18 at 22:15
add a comment |
Your proof is basically right. You could just make it simpler and clearer.
Let ${T(u_1),T(u_2),dots,T(u_n)}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. ${u_1,u_2,dots,u_n}$ is linearly independent.
Indeed, if $alpha_1u_1+alpha_2u_2+dots+alpha_nu_n=0$, then also
$$
0=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)=
alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)
$$
forcing $alpha_1=alpha_2=dots=alpha_n=0$.
Second fact. ${u_1,u_2,dots,u_n}$ spans $V$.
Let $vin V$; then $T(v)=alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)$ for some scalars $alpha_1,alpha_2,dots,alpha_n$. This can be rewritten as
$$
T(v)=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=alpha_1u_1+alpha_2u_2+dots+alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $dimoperatorname{range}(T)=dim V=n$; since $T$ is injective, $dimker(T)=0$. The rank-nullity theorem says
$$
dim U=dimker(T)+dimoperatorname{range}(T)=0+n=n
$$
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
Your proof is basically right. You could just make it simpler and clearer.
Let ${T(u_1),T(u_2),dots,T(u_n)}$ be a basis of $V$. This is possible because $T$ is surjective.
First fact. ${u_1,u_2,dots,u_n}$ is linearly independent.
Indeed, if $alpha_1u_1+alpha_2u_2+dots+alpha_nu_n=0$, then also
$$
0=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)=
alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)
$$
forcing $alpha_1=alpha_2=dots=alpha_n=0$.
Second fact. ${u_1,u_2,dots,u_n}$ spans $V$.
Let $vin V$; then $T(v)=alpha_1T(u_1)+alpha_2T(u_2)+dots+alpha_nT(u_n)$ for some scalars $alpha_1,alpha_2,dots,alpha_n$. This can be rewritten as
$$
T(v)=T(alpha_1u_1+alpha_2u_2+dots+alpha_nu_n)
$$
and, since $T$ is injective, we obtain $v=alpha_1u_1+alpha_2u_2+dots+alpha_nu_n$.
Also the proof with the rank-nullity theorem is correct. Since $T$ is surjective, $dimoperatorname{range}(T)=dim V=n$; since $T$ is injective, $dimker(T)=0$. The rank-nullity theorem says
$$
dim U=dimker(T)+dimoperatorname{range}(T)=0+n=n
$$
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
answered Dec 6 '18 at 22:03
egregegreg
179k1485202
179k1485202
Thank you so much.
– Dima
Dec 6 '18 at 22:15
add a comment |
Thank you so much.
– Dima
Dec 6 '18 at 22:15
Thank you so much.
– Dima
Dec 6 '18 at 22:15
Thank you so much.
– Dima
Dec 6 '18 at 22:15
add a comment |
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1
rank nullity thm
– mathworker21
Nov 30 '18 at 0:02
Use
$operatorname{text}$for $operatorname{text}$, whenever "text" is, say, "span".– Shaun
Nov 30 '18 at 0:17
1
@Shaun thank you.
– Dima
Nov 30 '18 at 0:27
1
In spite of what it says in the accepted answer, your first proof is correct. That said, you should improve how you phrase it: Rather than "then there exists $T(u)$..." simply say "then $T(u)$..."
– Andrés E. Caicedo
Dec 2 '18 at 19:32
1
Well, it is, but I don't know how the rank-nullity theorem was proved in your lecture. It may have very well used a version of the result you were asked to prove, in which case your argument would have been circular. If the result was established independently then, sure, you have a proof.
– Andrés E. Caicedo
Dec 3 '18 at 4:04