Mixing
How to prove that two subspaces are equal if and only if their annihilators are equal
$begingroup$
By definition, $W_1=W_2 rightarrow W_1^0=W_2^0$, $W_1, W_2 subseteq V$ are subspaces of $V$. But how does one show the reverse implication? $V$ is not assumed to be finite dimensional.
The annihilator is defined as $S^0= {fin V^*| f(s)=0 forall sin S}$, where $Ssubseteq V, S$ is a subset of vector space V.
linear-algebra vector-spaces
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
$endgroup$
add a comment |
$begingroup$
By definition, $W_1=W_2 rightarrow W_1^0=W_2^0$, $W_1, W_2 subseteq V$ are subspaces of $V$. But how does one show the reverse implication? $V$ is not assumed to be finite dimensional.
The annihilator is defined as $S^0= {fin V^*| f(s)=0 forall sin S}$, where $Ssubseteq V, S$ is a subset of vector space V.
linear-algebra vector-spaces
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
$endgroup$
$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33
add a comment |
$begingroup$
By definition, $W_1=W_2 rightarrow W_1^0=W_2^0$, $W_1, W_2 subseteq V$ are subspaces of $V$. But how does one show the reverse implication? $V$ is not assumed to be finite dimensional.
The annihilator is defined as $S^0= {fin V^*| f(s)=0 forall sin S}$, where $Ssubseteq V, S$ is a subset of vector space V.
linear-algebra vector-spaces
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
$endgroup$
By definition, $W_1=W_2 rightarrow W_1^0=W_2^0$, $W_1, W_2 subseteq V$ are subspaces of $V$. But how does one show the reverse implication? $V$ is not assumed to be finite dimensional.
The annihilator is defined as $S^0= {fin V^*| f(s)=0 forall sin S}$, where $Ssubseteq V, S$ is a subset of vector space V.
linear-algebra vector-spaces
linear-algebra vector-spaces
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
edited Nov 16 '15 at 20:30
grayQuant
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
asked Nov 16 '15 at 20:22
grayQuantgrayQuant
1,11621749
1,11621749
$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33
add a comment |
$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33
$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
To prove $ W_1^0=W_2^0 Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 neq W_2$ then there is some vector $alpha in W_1 setminus W_2$. Now suppose $B_{W_2} = {w_1,...,w_k}$ be a basis of $W_2$. Then $ B_{W_2}cup { alpha }$ is linearly independent since $alphanotin W_2$. Extend $ B_{W_2}cup { alpha }$ to a basis of $V$ by $B_V={w_1,...,w_k,alpha,w_{k+2},...,w_n}.$
Now define a (unique) linear functional $f:Vrightarrow Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k quad; f(alpha)=1_{Bbb F} ; $and $f(w_j)=1_{Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 ; forall win W_2 $ (since $forall w in W_2 ;, w=a_1w_1+...+a_kw_k ; $for some $a_1,...,a_k in Bbb F$) i.e $f in W_2^0$ but $fnotin W_1^0$ since $alpha in W_1 $ and $f(alpha) neq 0$. Hence $ W_1^0 neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim;V=n$)
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
$endgroup$
add a comment |
$begingroup$
I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=phi_1,...,phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ phi_i(v_j)=delta_{ij}.$$
This means that $phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{ell})=W_2$$
Then, by definition
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0 implies W_1=span(v_k,...,v_{ell})=W_2.$$
This completes the proof.
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
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2 Answers
2
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2 Answers
2
active
oldest
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$begingroup$
To prove $ W_1^0=W_2^0 Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 neq W_2$ then there is some vector $alpha in W_1 setminus W_2$. Now suppose $B_{W_2} = {w_1,...,w_k}$ be a basis of $W_2$. Then $ B_{W_2}cup { alpha }$ is linearly independent since $alphanotin W_2$. Extend $ B_{W_2}cup { alpha }$ to a basis of $V$ by $B_V={w_1,...,w_k,alpha,w_{k+2},...,w_n}.$
Now define a (unique) linear functional $f:Vrightarrow Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k quad; f(alpha)=1_{Bbb F} ; $and $f(w_j)=1_{Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 ; forall win W_2 $ (since $forall w in W_2 ;, w=a_1w_1+...+a_kw_k ; $for some $a_1,...,a_k in Bbb F$) i.e $f in W_2^0$ but $fnotin W_1^0$ since $alpha in W_1 $ and $f(alpha) neq 0$. Hence $ W_1^0 neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim;V=n$)
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
$endgroup$
add a comment |
$begingroup$
To prove $ W_1^0=W_2^0 Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 neq W_2$ then there is some vector $alpha in W_1 setminus W_2$. Now suppose $B_{W_2} = {w_1,...,w_k}$ be a basis of $W_2$. Then $ B_{W_2}cup { alpha }$ is linearly independent since $alphanotin W_2$. Extend $ B_{W_2}cup { alpha }$ to a basis of $V$ by $B_V={w_1,...,w_k,alpha,w_{k+2},...,w_n}.$
Now define a (unique) linear functional $f:Vrightarrow Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k quad; f(alpha)=1_{Bbb F} ; $and $f(w_j)=1_{Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 ; forall win W_2 $ (since $forall w in W_2 ;, w=a_1w_1+...+a_kw_k ; $for some $a_1,...,a_k in Bbb F$) i.e $f in W_2^0$ but $fnotin W_1^0$ since $alpha in W_1 $ and $f(alpha) neq 0$. Hence $ W_1^0 neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim;V=n$)
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
$endgroup$
add a comment |
$begingroup$
To prove $ W_1^0=W_2^0 Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 neq W_2$ then there is some vector $alpha in W_1 setminus W_2$. Now suppose $B_{W_2} = {w_1,...,w_k}$ be a basis of $W_2$. Then $ B_{W_2}cup { alpha }$ is linearly independent since $alphanotin W_2$. Extend $ B_{W_2}cup { alpha }$ to a basis of $V$ by $B_V={w_1,...,w_k,alpha,w_{k+2},...,w_n}.$
Now define a (unique) linear functional $f:Vrightarrow Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k quad; f(alpha)=1_{Bbb F} ; $and $f(w_j)=1_{Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 ; forall win W_2 $ (since $forall w in W_2 ;, w=a_1w_1+...+a_kw_k ; $for some $a_1,...,a_k in Bbb F$) i.e $f in W_2^0$ but $fnotin W_1^0$ since $alpha in W_1 $ and $f(alpha) neq 0$. Hence $ W_1^0 neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim;V=n$)
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
$endgroup$
To prove $ W_1^0=W_2^0 Rightarrow W_1=W_2 $ we proceed by contraposition; suppose $ W_1 neq W_2$ then there is some vector $alpha in W_1 setminus W_2$. Now suppose $B_{W_2} = {w_1,...,w_k}$ be a basis of $W_2$. Then $ B_{W_2}cup { alpha }$ is linearly independent since $alphanotin W_2$. Extend $ B_{W_2}cup { alpha }$ to a basis of $V$ by $B_V={w_1,...,w_k,alpha,w_{k+2},...,w_n}.$
Now define a (unique) linear functional $f:Vrightarrow Bbb F $ such that $f(w_i)=0$ for $ i=1,...,k quad; f(alpha)=1_{Bbb F} ; $and $f(w_j)=1_{Bbb F}$ for $j=k+2,...,n$. Then clearly $f(w)=0 ; forall win W_2 $ (since $forall w in W_2 ;, w=a_1w_1+...+a_kw_k ; $for some $a_1,...,a_k in Bbb F$) i.e $f in W_2^0$ but $fnotin W_1^0$ since $alpha in W_1 $ and $f(alpha) neq 0$. Hence $ W_1^0 neq W_2^0$.
(We assume that $V$ is finite dimensional with $dim;V=n$)
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
edited Oct 16 '17 at 12:40
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
answered Oct 16 '17 at 10:45
Arpan DasArpan Das
1078
1078
add a comment |
add a comment |
$begingroup$
I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=phi_1,...,phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ phi_i(v_j)=delta_{ij}.$$
This means that $phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{ell})=W_2$$
Then, by definition
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0 implies W_1=span(v_k,...,v_{ell})=W_2.$$
This completes the proof.
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=phi_1,...,phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ phi_i(v_j)=delta_{ij}.$$
This means that $phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{ell})=W_2$$
Then, by definition
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0 implies W_1=span(v_k,...,v_{ell})=W_2.$$
This completes the proof.
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
add a comment |
$begingroup$
I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=phi_1,...,phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ phi_i(v_j)=delta_{ij}.$$
This means that $phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{ell})=W_2$$
Then, by definition
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0 implies W_1=span(v_k,...,v_{ell})=W_2.$$
This completes the proof.
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
$endgroup$
I assume that here we are dealing with a finite dimensional vector space $V$. In this case, there exists some basis $B=v_1,...,v_n$ for $V$. We define the dual basis to be $B'=phi_1,...,phi_n$, where the bases satisfy the Kroenecker Delta condition. That is
$$ phi_i(v_j)=delta_{ij}.$$
This means that $phi_i(v_j)=1$ if and only if $i=j$. So consider then that we can express $W_1,W_2$ in terms of the basis vectors for $V$. So let
$$ W_1=span(v_k,...,v_{ell})=W_2$$
Then, by definition
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0.$$
So, if we are given that $W_1^0=W_2^0$, we know that they are spanned by the same elements of the dual basis. And so:
$$ W_1^0=span(phi_1,...phi_{k-1},phi_{ell+1},...,phi_n)=W_2^0 implies W_1=span(v_k,...,v_{ell})=W_2.$$
This completes the proof.
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
answered Nov 16 '15 at 20:29
Antonios-Alexandros RobotisAntonios-Alexandros Robotis
10.6k41741
10.6k41741
add a comment |
add a comment |
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$begingroup$
What is your definition of the annihilator?
$endgroup$
– Bernard
Nov 16 '15 at 20:27
$begingroup$
@Bernard, added the definition.
$endgroup$
– grayQuant
Nov 16 '15 at 20:32
$begingroup$
That's what I thought, but I wondered if you were hiding an underlying structure.
$endgroup$
– Bernard
Nov 16 '15 at 20:33