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Natural isomorphism to dual space of an inner product space in complex case.
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$k$ is a field and $V$ is a finite dimensional $k$ vector space that has an inner product $langle - , - rangle$.
If $k = mathbbR$, there is a natural isomorphism $phi colon V rightarrow V^*$, $v mapsto langle v, - rangle$. However, if $k = mathbbC$, $langle v, - rangle$ is not linear because $langle v, cx rangle = overlinec langle v, x rangle$. $langle -, v rangle$ is linear but in this case $V rightarrow V^*$, $v mapsto langle -, v rangle$ is not linear.
Are there no natural isomorphism between $V$ and $V^*$ when $k=mathbbC$?
linear-algebra
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
asked Jul 30 at 11:13
satoukibi
16016
add a comment |Â
up vote
3
down vote
favorite
$k$ is a field and $V$ is a finite dimensional $k$ vector space that has an inner product $langle - , - rangle$.
If $k = mathbbR$, there is a natural isomorphism $phi colon V rightarrow V^*$, $v mapsto langle v, - rangle$. However, if $k = mathbbC$, $langle v, - rangle$ is not linear because $langle v, cx rangle = overlinec langle v, x rangle$. $langle -, v rangle$ is linear but in this case $V rightarrow V^*$, $v mapsto langle -, v rangle$ is not linear.
Are there no natural isomorphism between $V$ and $V^*$ when $k=mathbbC$?
linear-algebra
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
asked Jul 30 at 11:13
satoukibi
16016
Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
1
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$k$ is a field and $V$ is a finite dimensional $k$ vector space that has an inner product $langle - , - rangle$.
If $k = mathbbR$, there is a natural isomorphism $phi colon V rightarrow V^*$, $v mapsto langle v, - rangle$. However, if $k = mathbbC$, $langle v, - rangle$ is not linear because $langle v, cx rangle = overlinec langle v, x rangle$. $langle -, v rangle$ is linear but in this case $V rightarrow V^*$, $v mapsto langle -, v rangle$ is not linear.
Are there no natural isomorphism between $V$ and $V^*$ when $k=mathbbC$?
linear-algebra
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
asked Jul 30 at 11:13
satoukibi
16016
$k$ is a field and $V$ is a finite dimensional $k$ vector space that has an inner product $langle - , - rangle$.
If $k = mathbbR$, there is a natural isomorphism $phi colon V rightarrow V^*$, $v mapsto langle v, - rangle$. However, if $k = mathbbC$, $langle v, - rangle$ is not linear because $langle v, cx rangle = overlinec langle v, x rangle$. $langle -, v rangle$ is linear but in this case $V rightarrow V^*$, $v mapsto langle -, v rangle$ is not linear.
Are there no natural isomorphism between $V$ and $V^*$ when $k=mathbbC$?
linear-algebra
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
asked Jul 30 at 11:13
satoukibi
16016
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
edited Jul 30 at 11:45
Jendrik Stelzner
7,5121937
7,5121937
asked Jul 30 at 11:13
satoukibi
16016
asked Jul 30 at 11:13
satoukibi
16016
asked Jul 30 at 11:13
satoukibi
16016
16016
Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
1
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32
add a comment |Â
Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
1
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32
Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
1
1
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32
add a comment |Â
1 Answer
1
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Take $phi: v mapsto langle -, v rangle$. As you've identified, this is not linear but it's almost linear, $phi(lambda v +u) = barlambdaphi(v) + phi(u)$.
So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $barV$, the conjugate space where scalar multiplication is given by $lambda cdot v = barlambdav$.
Note that the natural isomorphism between $V$ and its double dual still holds.
Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).
answered Jul 30 at 11:27
Daniel Mroz
851314
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Take $phi: v mapsto langle -, v rangle$. As you've identified, this is not linear but it's almost linear, $phi(lambda v +u) = barlambdaphi(v) + phi(u)$.
So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $barV$, the conjugate space where scalar multiplication is given by $lambda cdot v = barlambdav$.
Note that the natural isomorphism between $V$ and its double dual still holds.
Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).
answered Jul 30 at 11:27
Daniel Mroz
851314
add a comment |Â
up vote
0
down vote
Take $phi: v mapsto langle -, v rangle$. As you've identified, this is not linear but it's almost linear, $phi(lambda v +u) = barlambdaphi(v) + phi(u)$.
So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $barV$, the conjugate space where scalar multiplication is given by $lambda cdot v = barlambdav$.
Note that the natural isomorphism between $V$ and its double dual still holds.
Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).
answered Jul 30 at 11:27
Daniel Mroz
851314
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Take $phi: v mapsto langle -, v rangle$. As you've identified, this is not linear but it's almost linear, $phi(lambda v +u) = barlambdaphi(v) + phi(u)$.
So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $barV$, the conjugate space where scalar multiplication is given by $lambda cdot v = barlambdav$.
Note that the natural isomorphism between $V$ and its double dual still holds.
Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).
answered Jul 30 at 11:27
Daniel Mroz
851314
Take $phi: v mapsto langle -, v rangle$. As you've identified, this is not linear but it's almost linear, $phi(lambda v +u) = barlambdaphi(v) + phi(u)$.
So instead of an isomorphism between $V^*$ and $V$ we have an isomorphism between $V^*$ and $barV$, the conjugate space where scalar multiplication is given by $lambda cdot v = barlambdav$.
Note that the natural isomorphism between $V$ and its double dual still holds.
Also, be careful to remember that an isomorphism between a vector space and its dual only exists in the finite dimensional case (otherwise it's injective but not surjective), and it's not really natural as it relies on a choice of basis (contrast with the isomorphism between $V$ and $(V^*)^*$ which is independent of basis).
answered Jul 30 at 11:27
Daniel Mroz
851314
answered Jul 30 at 11:27
Daniel Mroz
851314
answered Jul 30 at 11:27
Daniel Mroz
851314
answered Jul 30 at 11:27
Daniel Mroz
851314
851314
add a comment |Â
add a comment |Â
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Generally, a space and its dual are not naturally isomorphic. A space and the dual of its dual are instead.
– xbh
Jul 30 at 11:20
Maybe $v mapsto overline langle v, cdot rangle$? 'cause $overline langle v, cx rangle = overline overline c langle v, x rangle = c overline langle v,x rangle $.
– xbh
Jul 30 at 11:25
1
What do you mean by an inner product?
– Bob
Jul 30 at 14:41
i have noticed that physics texts often have $langle x, yrangle$ conjugate-linear in $x$ and linear in $y$.
– DanielWainfleet
Jul 30 at 19:32