Inner product of two vectors in complex vector space in index notation












2












$begingroup$


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



enter image description here



At which point is my argument flawed? Any help is appreciated.






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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



enter image description here



At which point is my argument flawed? Any help is appreciated.






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$












  • $begingroup$
    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














2












2








2





$begingroup$


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



enter image description here



At which point is my argument flawed? Any help is appreciated.






linear-algebra inner-product-space







share|cite|improve this question











$endgroup$




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



enter image description here



At which point is my argument flawed? Any help is appreciated.






linear-algebra inner-product-space




linear-algebra inner-product-space






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 8 '18 at 12:50









amWhy

1




1










asked Dec 8 '18 at 12:38









MathematicingMathematicing

2,44321854




2,44321854












  • $begingroup$
    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










  • $begingroup$
    @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










1 Answer
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oldest

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1












$begingroup$

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.






share|cite|improve this answer











$endgroup$













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    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      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.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        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.






        share|cite|improve this answer











        $endgroup$



        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.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 8 '18 at 14:56

























        answered Dec 8 '18 at 12:57









        Picaud VincentPicaud Vincent

        1,36439




        1,36439






























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