Inner product of two vectors in complex vector space in index notation
$begingroup$
Suppose $vec{u}, vec{v} in left(V, mathbb{C}^{n}right)$: by construction
$$
begin{split}
vec{u} &= sum_{i=1}^{n}u_{i}vec{e}_{i}\
vec{v} &= sum_{i=1}^{n}v_{i}vec{e}_{i}
end{split}.
$$
The inner product of vectors $vec{u}$ and $vec{v}$ gives
$$
langle vec{u}, vec{v}rangle
= langle sum_{i=1}^{n}u_{i}vec{e}_{i}, sum_{i=1}^{n}v_{i}vec{e}_{i}rangle.
$$
But $langle vec{u}, vec{v}rangle = langle vec{v}, vec{u}rangle^{*}$ with $*$ being the complex conjugate, and recalling that $langle X,alpha Yrangle = alpha^{*} langle X,Yrangle$ for $alpha$ a complex scalar, this gives
$$
left(sum_{i,j=1}^{n}u_{i}^{*}v_{j}langle vec{e}_{j}, vec{e}_{i}rangle right)^{*}
=sum_{i,j=1}^{n}u_{i}v_{j}^{*}
langle vec{e}_{j}, vec{e}_{i} rangle.
$$
This is in contrast to the following
At which point is my argument flawed? Any help is appreciated.
linear-algebra inner-product-space
edited Dec 8 '18 at 12:50
amWhy
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
$endgroup$
add a comment |
$begingroup$
Suppose $vec{u}, vec{v} in left(V, mathbb{C}^{n}right)$: by construction
$$
begin{split}
vec{u} &= sum_{i=1}^{n}u_{i}vec{e}_{i}\
vec{v} &= sum_{i=1}^{n}v_{i}vec{e}_{i}
end{split}.
$$
The inner product of vectors $vec{u}$ and $vec{v}$ gives
$$
langle vec{u}, vec{v}rangle
= langle sum_{i=1}^{n}u_{i}vec{e}_{i}, sum_{i=1}^{n}v_{i}vec{e}_{i}rangle.
$$
But $langle vec{u}, vec{v}rangle = langle vec{v}, vec{u}rangle^{*}$ with $*$ being the complex conjugate, and recalling that $langle X,alpha Yrangle = alpha^{*} langle X,Yrangle$ for $alpha$ a complex scalar, this gives
$$
left(sum_{i,j=1}^{n}u_{i}^{*}v_{j}langle vec{e}_{j}, vec{e}_{i}rangle right)^{*}
=sum_{i,j=1}^{n}u_{i}v_{j}^{*}
langle vec{e}_{j}, vec{e}_{i} rangle.
$$
This is in contrast to the following
At which point is my argument flawed? Any help is appreciated.
linear-algebra inner-product-space
edited Dec 8 '18 at 12:50
amWhy
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
$endgroup$
$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48
add a comment |
$begingroup$
Suppose $vec{u}, vec{v} in left(V, mathbb{C}^{n}right)$: by construction
$$
begin{split}
vec{u} &= sum_{i=1}^{n}u_{i}vec{e}_{i}\
vec{v} &= sum_{i=1}^{n}v_{i}vec{e}_{i}
end{split}.
$$
The inner product of vectors $vec{u}$ and $vec{v}$ gives
$$
langle vec{u}, vec{v}rangle
= langle sum_{i=1}^{n}u_{i}vec{e}_{i}, sum_{i=1}^{n}v_{i}vec{e}_{i}rangle.
$$
But $langle vec{u}, vec{v}rangle = langle vec{v}, vec{u}rangle^{*}$ with $*$ being the complex conjugate, and recalling that $langle X,alpha Yrangle = alpha^{*} langle X,Yrangle$ for $alpha$ a complex scalar, this gives
$$
left(sum_{i,j=1}^{n}u_{i}^{*}v_{j}langle vec{e}_{j}, vec{e}_{i}rangle right)^{*}
=sum_{i,j=1}^{n}u_{i}v_{j}^{*}
langle vec{e}_{j}, vec{e}_{i} rangle.
$$
This is in contrast to the following
At which point is my argument flawed? Any help is appreciated.
linear-algebra inner-product-space
edited Dec 8 '18 at 12:50
amWhy
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
$endgroup$
Suppose $vec{u}, vec{v} in left(V, mathbb{C}^{n}right)$: by construction
$$
begin{split}
vec{u} &= sum_{i=1}^{n}u_{i}vec{e}_{i}\
vec{v} &= sum_{i=1}^{n}v_{i}vec{e}_{i}
end{split}.
$$
The inner product of vectors $vec{u}$ and $vec{v}$ gives
$$
langle vec{u}, vec{v}rangle
= langle sum_{i=1}^{n}u_{i}vec{e}_{i}, sum_{i=1}^{n}v_{i}vec{e}_{i}rangle.
$$
But $langle vec{u}, vec{v}rangle = langle vec{v}, vec{u}rangle^{*}$ with $*$ being the complex conjugate, and recalling that $langle X,alpha Yrangle = alpha^{*} langle X,Yrangle$ for $alpha$ a complex scalar, this gives
$$
left(sum_{i,j=1}^{n}u_{i}^{*}v_{j}langle vec{e}_{j}, vec{e}_{i}rangle right)^{*}
=sum_{i,j=1}^{n}u_{i}v_{j}^{*}
langle vec{e}_{j}, vec{e}_{i} rangle.
$$
This is in contrast to the following
At which point is my argument flawed? Any help is appreciated.
linear-algebra inner-product-space
linear-algebra inner-product-space
edited Dec 8 '18 at 12:50
amWhy
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
edited Dec 8 '18 at 12:50
amWhy
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
edited Dec 8 '18 at 12:50
amWhy
192k28225439
edited Dec 8 '18 at 12:50
amWhy
192k28225439
edited Dec 8 '18 at 12:50
amWhy
192k28225439
192k28225439
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821854
2,44821854
$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48
add a comment |
$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48
$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
You must take care that there are two different conventions:
In some cases the hermitian inner product is defined as $langle u,v rangle=u^tbar{v}$,
in other cases it is defined as $langle u,v rangle=bar{u}^t v=u^*v$ (mostly in quantum mechanics, Dirac's bra-ket notations)
Further details Alternative definitions, notations and remarks
It is the origin of the confusion, in your first statement you use the first convention, in the last one, the second one.
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
$endgroup$
add a comment |
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You must take care that there are two different conventions:
In some cases the hermitian inner product is defined as $langle u,v rangle=u^tbar{v}$,
in other cases it is defined as $langle u,v rangle=bar{u}^t v=u^*v$ (mostly in quantum mechanics, Dirac's bra-ket notations)
Further details Alternative definitions, notations and remarks
It is the origin of the confusion, in your first statement you use the first convention, in the last one, the second one.
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
$endgroup$
add a comment |
$begingroup$
You must take care that there are two different conventions:
In some cases the hermitian inner product is defined as $langle u,v rangle=u^tbar{v}$,
in other cases it is defined as $langle u,v rangle=bar{u}^t v=u^*v$ (mostly in quantum mechanics, Dirac's bra-ket notations)
Further details Alternative definitions, notations and remarks
It is the origin of the confusion, in your first statement you use the first convention, in the last one, the second one.
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
$endgroup$
add a comment |
$begingroup$
You must take care that there are two different conventions:
In some cases the hermitian inner product is defined as $langle u,v rangle=u^tbar{v}$,
in other cases it is defined as $langle u,v rangle=bar{u}^t v=u^*v$ (mostly in quantum mechanics, Dirac's bra-ket notations)
Further details Alternative definitions, notations and remarks
It is the origin of the confusion, in your first statement you use the first convention, in the last one, the second one.
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
$endgroup$
You must take care that there are two different conventions:
In some cases the hermitian inner product is defined as $langle u,v rangle=u^tbar{v}$,
in other cases it is defined as $langle u,v rangle=bar{u}^t v=u^*v$ (mostly in quantum mechanics, Dirac's bra-ket notations)
Further details Alternative definitions, notations and remarks
It is the origin of the confusion, in your first statement you use the first convention, in the last one, the second one.
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
edited Dec 8 '18 at 14:56
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
answered Dec 8 '18 at 12:57
Picaud VincentPicaud Vincent
1,49439
1,49439
add a comment |
add a comment |
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$begingroup$
What have you done in the last expression?
$endgroup$
– Aniruddha Deshmukh
Dec 8 '18 at 12:47
$begingroup$
@AniruddhaDeshmukh I have merely applied the skew symmetry property.
$endgroup$
– Mathematicing
Dec 8 '18 at 12:48