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
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
$endgroup$
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
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
$endgroup$
$begingroup$
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
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
$endgroup$
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
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
edited Dec 8 '18 at 12:50
amWhy
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
edited Dec 8 '18 at 12:50
amWhy
1
edited Dec 8 '18 at 12:50
amWhy
1
edited Dec 8 '18 at 12:50
amWhy
1
1
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
asked Dec 8 '18 at 12:38
MathematicingMathematicing
2,44821855
2,44821855
$begingroup$
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 |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031060%2finner-product-of-two-vectors-in-complex-vector-space-in-index-notation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
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 |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031060%2finner-product-of-two-vectors-in-complex-vector-space-in-index-notation%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
$begingroup$
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