What is the difference between the isomorphisms between $V$, $V^*$, and $V$, $V^**$?
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Recently, I am studying dual space with the book Linear Algebra by Stephen, Arnold, and Lawrence. In the book, it says that $V$ and $V^*$ are isomorphic. Also, it says that $V$ and $V^**$ are isomorphic in a "natural" way, which means that there is an isomorphism between $V$ and $V^**$ which is independent of the choice of bases. So, my question is
If I denote the relation between $V$ and $V^**$ by $V=V^**$, can I say $V=V^*$?
linear-algebra vector-spaces dual-spaces
edited Sep 5 at 8:49
Christoph
11k1240
asked Sep 5 at 8:40
Ivy
327312
add a comment |Â
up vote
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Recently, I am studying dual space with the book Linear Algebra by Stephen, Arnold, and Lawrence. In the book, it says that $V$ and $V^*$ are isomorphic. Also, it says that $V$ and $V^**$ are isomorphic in a "natural" way, which means that there is an isomorphism between $V$ and $V^**$ which is independent of the choice of bases. So, my question is
If I denote the relation between $V$ and $V^**$ by $V=V^**$, can I say $V=V^*$?
linear-algebra vector-spaces dual-spaces
edited Sep 5 at 8:49
Christoph
11k1240
asked Sep 5 at 8:40
Ivy
327312
1
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
1
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Recently, I am studying dual space with the book Linear Algebra by Stephen, Arnold, and Lawrence. In the book, it says that $V$ and $V^*$ are isomorphic. Also, it says that $V$ and $V^**$ are isomorphic in a "natural" way, which means that there is an isomorphism between $V$ and $V^**$ which is independent of the choice of bases. So, my question is
If I denote the relation between $V$ and $V^**$ by $V=V^**$, can I say $V=V^*$?
linear-algebra vector-spaces dual-spaces
edited Sep 5 at 8:49
Christoph
11k1240
asked Sep 5 at 8:40
Ivy
327312
Recently, I am studying dual space with the book Linear Algebra by Stephen, Arnold, and Lawrence. In the book, it says that $V$ and $V^*$ are isomorphic. Also, it says that $V$ and $V^**$ are isomorphic in a "natural" way, which means that there is an isomorphism between $V$ and $V^**$ which is independent of the choice of bases. So, my question is
If I denote the relation between $V$ and $V^**$ by $V=V^**$, can I say $V=V^*$?
linear-algebra vector-spaces dual-spaces
linear-algebra vector-spaces dual-spaces
edited Sep 5 at 8:49
Christoph
11k1240
asked Sep 5 at 8:40
Ivy
327312
edited Sep 5 at 8:49
Christoph
11k1240
asked Sep 5 at 8:40
Ivy
327312
edited Sep 5 at 8:49
Christoph
11k1240
edited Sep 5 at 8:49
Christoph
11k1240
edited Sep 5 at 8:49
Christoph
11k1240
11k1240
asked Sep 5 at 8:40
Ivy
327312
asked Sep 5 at 8:40
Ivy
327312
asked Sep 5 at 8:40
Ivy
327312
327312
1
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
1
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52
add a comment |Â
1
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
1
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52
1
1
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
1
1
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52
add a comment |Â
1 Answer
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First of all, the isomorphisms $Vcong V^*$ and $Vcong V^**$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.
For an isomorphism $Vto V^*$ you have to specify how a vector $vin V$ should act as a linear form $Vto K$. It turns out that this amounts to specifying a non-degenerate bilinear form $langle-,-ranglecolon Vtimes Vto K$ and send $vin V$ to $langle v,-ranglein V^*$, i.e., the linear form sending $w$ to $langle v,wrangle$. Another way to achieve the same thing is by picking a basis $e_1,dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $langle e_i, e_jrangle = delta_ij$. In this case you get a dual basis $e_i^* = langle e_i, -rangle$.
To summarize, in order to construct an isomorphism $Vto V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.
Now for an isomorphism $Vto V^**$, you have to specify how a vector $vin V$ acts as a linear form $V^*to K$. So given a linear form $omegain V^*$, how does $vin V$ act on $omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $omega$ to the vector $v$. Hence, the isomorphism $Vto V^**$ sends $vin V$ to the linear form $V^*to K$ with $omegamapsto omega(v)$.
We see that we did not need to make any choices in writing down this isomorphism $Vto V^**$.
To answer your question, if you use "$V=V^**$" to denote the identification of $V$ and $V^**$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $Vto V^*$ requires you to make some (arbitrary) choices.
answered Sep 5 at 9:06
Christoph
11k1240
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
First of all, the isomorphisms $Vcong V^*$ and $Vcong V^**$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.
For an isomorphism $Vto V^*$ you have to specify how a vector $vin V$ should act as a linear form $Vto K$. It turns out that this amounts to specifying a non-degenerate bilinear form $langle-,-ranglecolon Vtimes Vto K$ and send $vin V$ to $langle v,-ranglein V^*$, i.e., the linear form sending $w$ to $langle v,wrangle$. Another way to achieve the same thing is by picking a basis $e_1,dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $langle e_i, e_jrangle = delta_ij$. In this case you get a dual basis $e_i^* = langle e_i, -rangle$.
To summarize, in order to construct an isomorphism $Vto V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.
Now for an isomorphism $Vto V^**$, you have to specify how a vector $vin V$ acts as a linear form $V^*to K$. So given a linear form $omegain V^*$, how does $vin V$ act on $omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $omega$ to the vector $v$. Hence, the isomorphism $Vto V^**$ sends $vin V$ to the linear form $V^*to K$ with $omegamapsto omega(v)$.
We see that we did not need to make any choices in writing down this isomorphism $Vto V^**$.
To answer your question, if you use "$V=V^**$" to denote the identification of $V$ and $V^**$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $Vto V^*$ requires you to make some (arbitrary) choices.
answered Sep 5 at 9:06
Christoph
11k1240
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
add a comment |Â
up vote
2
down vote
First of all, the isomorphisms $Vcong V^*$ and $Vcong V^**$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.
For an isomorphism $Vto V^*$ you have to specify how a vector $vin V$ should act as a linear form $Vto K$. It turns out that this amounts to specifying a non-degenerate bilinear form $langle-,-ranglecolon Vtimes Vto K$ and send $vin V$ to $langle v,-ranglein V^*$, i.e., the linear form sending $w$ to $langle v,wrangle$. Another way to achieve the same thing is by picking a basis $e_1,dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $langle e_i, e_jrangle = delta_ij$. In this case you get a dual basis $e_i^* = langle e_i, -rangle$.
To summarize, in order to construct an isomorphism $Vto V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.
Now for an isomorphism $Vto V^**$, you have to specify how a vector $vin V$ acts as a linear form $V^*to K$. So given a linear form $omegain V^*$, how does $vin V$ act on $omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $omega$ to the vector $v$. Hence, the isomorphism $Vto V^**$ sends $vin V$ to the linear form $V^*to K$ with $omegamapsto omega(v)$.
We see that we did not need to make any choices in writing down this isomorphism $Vto V^**$.
To answer your question, if you use "$V=V^**$" to denote the identification of $V$ and $V^**$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $Vto V^*$ requires you to make some (arbitrary) choices.
answered Sep 5 at 9:06
Christoph
11k1240
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
add a comment |Â
up vote
2
down vote
up vote
2
down vote
First of all, the isomorphisms $Vcong V^*$ and $Vcong V^**$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.
For an isomorphism $Vto V^*$ you have to specify how a vector $vin V$ should act as a linear form $Vto K$. It turns out that this amounts to specifying a non-degenerate bilinear form $langle-,-ranglecolon Vtimes Vto K$ and send $vin V$ to $langle v,-ranglein V^*$, i.e., the linear form sending $w$ to $langle v,wrangle$. Another way to achieve the same thing is by picking a basis $e_1,dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $langle e_i, e_jrangle = delta_ij$. In this case you get a dual basis $e_i^* = langle e_i, -rangle$.
To summarize, in order to construct an isomorphism $Vto V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.
Now for an isomorphism $Vto V^**$, you have to specify how a vector $vin V$ acts as a linear form $V^*to K$. So given a linear form $omegain V^*$, how does $vin V$ act on $omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $omega$ to the vector $v$. Hence, the isomorphism $Vto V^**$ sends $vin V$ to the linear form $V^*to K$ with $omegamapsto omega(v)$.
We see that we did not need to make any choices in writing down this isomorphism $Vto V^**$.
To answer your question, if you use "$V=V^**$" to denote the identification of $V$ and $V^**$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $Vto V^*$ requires you to make some (arbitrary) choices.
answered Sep 5 at 9:06
Christoph
11k1240
First of all, the isomorphisms $Vcong V^*$ and $Vcong V^**$ only hold (in general) for finite dimensional $K$-vector spaces. If $V$ does not admit a finite basis, the isomorphisms may fail.
For an isomorphism $Vto V^*$ you have to specify how a vector $vin V$ should act as a linear form $Vto K$. It turns out that this amounts to specifying a non-degenerate bilinear form $langle-,-ranglecolon Vtimes Vto K$ and send $vin V$ to $langle v,-ranglein V^*$, i.e., the linear form sending $w$ to $langle v,wrangle$. Another way to achieve the same thing is by picking a basis $e_1,dots,e_n$ and declaring it to be orthonormal, hence defining a bilinear form with $langle e_i, e_jrangle = delta_ij$. In this case you get a dual basis $e_i^* = langle e_i, -rangle$.
To summarize, in order to construct an isomorphism $Vto V^*$, you have to either choose a basis of $V$ or a non-degenerate bilinear form.
Now for an isomorphism $Vto V^**$, you have to specify how a vector $vin V$ acts as a linear form $V^*to K$. So given a linear form $omegain V^*$, how does $vin V$ act on $omega$ to produce an element of $K$? In this situation it is just natural to just apply the linear form $omega$ to the vector $v$. Hence, the isomorphism $Vto V^**$ sends $vin V$ to the linear form $V^*to K$ with $omegamapsto omega(v)$.
We see that we did not need to make any choices in writing down this isomorphism $Vto V^**$.
To answer your question, if you use "$V=V^**$" to denote the identification of $V$ and $V^**$ under the above natural isomorphism, you can not write "$V=V^*$" in the same spirit, since any isomorphism $Vto V^*$ requires you to make some (arbitrary) choices.
answered Sep 5 at 9:06
Christoph
11k1240
answered Sep 5 at 9:06
Christoph
11k1240
answered Sep 5 at 9:06
Christoph
11k1240
answered Sep 5 at 9:06
Christoph
11k1240
11k1240
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
add a comment |Â
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
So, there are different kinds of isomorphism. Is that right?
– Ivy
Sep 5 at 10:28
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
I'm not sure what that means. In the end an isomorphisms is just a linear bijection in this case. The difference are the ingredients needed to define them.
– Christoph
Sep 5 at 10:36
1
1
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
The difference is that every isomorphism between $V$ and $V^*$ involves an arbitrary choice of basis or of blinear form - none is more "natural" than the rest. However, the isomorphism between $V$ and $V^**$ that maps $v in V$ to $v'' in V^**$ where $v''(omega) = omega(v)$ does not involve any arbitrary choices. There are, of course other "unnatural" isomorphisms between $V$ and $V^**$ - for example, we could define $v''$ by $v''(omega) = 2omega(v)$, but the factor of 2 is then an arbitrary choice.
– gandalf61
Sep 5 at 11:27
add a comment |Â
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1
Small MathJax tip : you can write an entire equation between two $ symbols, no need to separate every symbol ;) And by the way, if you want to write something in italic, just put it between two asterisks *.
– Arnaud D.
Sep 5 at 8:48
1
Note that we have quite a few questions on the subject already. You might be interested in these two for example : math.stackexchange.com/questions/579739/…, math.stackexchange.com/questions/1900179/…
– Arnaud D.
Sep 5 at 8:52