Intuition for complex dot product [closed]

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I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $mathbbC^n$ are viewed as real vectors in $mathbbC^2n$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?

linear-algebra intuition

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edited Jul 16 at 4:16

Tengu

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asked Jul 16 at 2:28

J.G95

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closed as off-topic by JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos Jul 16 at 22:06

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• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos

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I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $mathbbC^n$ are viewed as real vectors in $mathbbC^2n$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?

linear-algebra intuition

share|cite|improve this question

edited Jul 16 at 4:16

Tengu

2,3391920

asked Jul 16 at 2:28

J.G95

62

closed as off-topic by JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos Jul 16 at 22:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

add a comment | 

up vote
0
down vote

favorite

up vote
0
down vote

favorite

I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $mathbbC^n$ are viewed as real vectors in $mathbbC^2n$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?

linear-algebra intuition

share|cite|improve this question

edited Jul 16 at 4:16

Tengu

2,3391920

asked Jul 16 at 2:28

J.G95

62

I am studying the complex dot product and I am trying to develop an intuition for it. The real part seems to tell me the real dot product, if the complex vectors in $mathbbC^n$ are viewed as real vectors in $mathbbC^2n$. So this would be a measure of how similar the complex vectors are. But what information do we gain from the complex part?

linear-algebra intuition

share|cite|improve this question

edited Jul 16 at 4:16

Tengu

2,3391920

asked Jul 16 at 2:28

J.G95

62

share|cite|improve this question

share|cite|improve this question

edited Jul 16 at 4:16

Tengu

2,3391920

edited Jul 16 at 4:16

Tengu

2,3391920

edited Jul 16 at 4:16

Tengu

2,3391920

2,3391920

asked Jul 16 at 2:28

J.G95

62

asked Jul 16 at 2:28

J.G95

62

asked Jul 16 at 2:28

J.G95

62

62

closed as off-topic by JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos Jul 16 at 22:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

closed as off-topic by JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos Jul 16 at 22:06

This question appears to be off-topic. The users who voted to close gave this specific reason:

• "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – JMoravitz, John Ma, Taroccoesbrocco, Chris Godsil, José Carlos Santos

If this question can be reworded to fit the rules in the help center, please edit the question.

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add a comment | 

2 Answers
2

active

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up vote
2
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A vector $v in mathbbC^n$ can be expressed as
$$
v = beginbmatrix a_1 + ib_1 \ vdots \ a_n + ib_n endbmatrix = beginbmatrix a_1 \ vdots \ a_n endbmatrix + ibeginbmatrix b_1 \ vdots \ b_n endbmatrix = Re(v)+iIm(v)
$$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
langle v,v rangle geq 0, text(equality iff $v equiv 0$)
$$
then it is sensible to define (for any $u,v in mathbbC^n$)
$$
langle u,v rangle := sum_i baru_i v_i
$$
or equivalently,
beginalign
langle u,v rangle &= big(Re(u)^top-iIm(u)^topbig)big(Re(v)+iIm(v)big)\
&= big(Re(u)^topRe(v) + Im(u)^top Im(v)big) + ibig(Re(u)^top Im(v) - Im(u)^topRe(v)big)\
&= Rebig(langle u,v ranglebig) + i Imbig(langle u,v ranglebig)
endalign

We see that,
$$
fracRebig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^topRe(v) + Im(u)^top Im(v)sqrtlangle u,u rangle langle v,v rangle = cos(theta)
$$
where $theta$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Re(v) \ Im(v) endbmatrix$.

On the other hand,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^top Im(v) - Im(u)^topRe(v)sqrtlangle u,u rangle langle v,v rangle = cos(phi)
$$
where $phi$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Im(v) \ -Re(v) endbmatrix$.

The latter real vector is the same as the one that defined $theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $mathbbC^1$, we have $phi = theta - fracpi2$ and thus,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = sin(theta)
$$

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edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

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up vote
2
down vote

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2times 2$ matrix $$beginpmatrixa&-b\b&aendpmatrix.$$

(There is another embedding into $M_2(mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=beginpmatrixv_1\v_2\ vdots\ v_nendpmatrix$$
corresponds to a $2ntimes 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $left(overlinev_1,dots,overlinev_nright).$ Then the $2times 2$ matrix, $A_v^TA_w,$ corresponds to $sum_i=1^noverlinev_iw_i$.

share|cite|improve this answer

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

A vector $v in mathbbC^n$ can be expressed as
$$
v = beginbmatrix a_1 + ib_1 \ vdots \ a_n + ib_n endbmatrix = beginbmatrix a_1 \ vdots \ a_n endbmatrix + ibeginbmatrix b_1 \ vdots \ b_n endbmatrix = Re(v)+iIm(v)
$$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
langle v,v rangle geq 0, text(equality iff $v equiv 0$)
$$
then it is sensible to define (for any $u,v in mathbbC^n$)
$$
langle u,v rangle := sum_i baru_i v_i
$$
or equivalently,
beginalign
langle u,v rangle &= big(Re(u)^top-iIm(u)^topbig)big(Re(v)+iIm(v)big)\
&= big(Re(u)^topRe(v) + Im(u)^top Im(v)big) + ibig(Re(u)^top Im(v) - Im(u)^topRe(v)big)\
&= Rebig(langle u,v ranglebig) + i Imbig(langle u,v ranglebig)
endalign

We see that,
$$
fracRebig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^topRe(v) + Im(u)^top Im(v)sqrtlangle u,u rangle langle v,v rangle = cos(theta)
$$
where $theta$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Re(v) \ Im(v) endbmatrix$.

On the other hand,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^top Im(v) - Im(u)^topRe(v)sqrtlangle u,u rangle langle v,v rangle = cos(phi)
$$
where $phi$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Im(v) \ -Re(v) endbmatrix$.

The latter real vector is the same as the one that defined $theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $mathbbC^1$, we have $phi = theta - fracpi2$ and thus,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = sin(theta)
$$

share|cite|improve this answer

edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

add a comment | 

up vote
2
down vote

A vector $v in mathbbC^n$ can be expressed as
$$
v = beginbmatrix a_1 + ib_1 \ vdots \ a_n + ib_n endbmatrix = beginbmatrix a_1 \ vdots \ a_n endbmatrix + ibeginbmatrix b_1 \ vdots \ b_n endbmatrix = Re(v)+iIm(v)
$$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
langle v,v rangle geq 0, text(equality iff $v equiv 0$)
$$
then it is sensible to define (for any $u,v in mathbbC^n$)
$$
langle u,v rangle := sum_i baru_i v_i
$$
or equivalently,
beginalign
langle u,v rangle &= big(Re(u)^top-iIm(u)^topbig)big(Re(v)+iIm(v)big)\
&= big(Re(u)^topRe(v) + Im(u)^top Im(v)big) + ibig(Re(u)^top Im(v) - Im(u)^topRe(v)big)\
&= Rebig(langle u,v ranglebig) + i Imbig(langle u,v ranglebig)
endalign

We see that,
$$
fracRebig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^topRe(v) + Im(u)^top Im(v)sqrtlangle u,u rangle langle v,v rangle = cos(theta)
$$
where $theta$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Re(v) \ Im(v) endbmatrix$.

On the other hand,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^top Im(v) - Im(u)^topRe(v)sqrtlangle u,u rangle langle v,v rangle = cos(phi)
$$
where $phi$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Im(v) \ -Re(v) endbmatrix$.

The latter real vector is the same as the one that defined $theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $mathbbC^1$, we have $phi = theta - fracpi2$ and thus,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = sin(theta)
$$

share|cite|improve this answer

edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

add a comment | 

up vote
2
down vote

up vote
2
down vote

A vector $v in mathbbC^n$ can be expressed as
$$
v = beginbmatrix a_1 + ib_1 \ vdots \ a_n + ib_n endbmatrix = beginbmatrix a_1 \ vdots \ a_n endbmatrix + ibeginbmatrix b_1 \ vdots \ b_n endbmatrix = Re(v)+iIm(v)
$$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
langle v,v rangle geq 0, text(equality iff $v equiv 0$)
$$
then it is sensible to define (for any $u,v in mathbbC^n$)
$$
langle u,v rangle := sum_i baru_i v_i
$$
or equivalently,
beginalign
langle u,v rangle &= big(Re(u)^top-iIm(u)^topbig)big(Re(v)+iIm(v)big)\
&= big(Re(u)^topRe(v) + Im(u)^top Im(v)big) + ibig(Re(u)^top Im(v) - Im(u)^topRe(v)big)\
&= Rebig(langle u,v ranglebig) + i Imbig(langle u,v ranglebig)
endalign

We see that,
$$
fracRebig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^topRe(v) + Im(u)^top Im(v)sqrtlangle u,u rangle langle v,v rangle = cos(theta)
$$
where $theta$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Re(v) \ Im(v) endbmatrix$.

On the other hand,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^top Im(v) - Im(u)^topRe(v)sqrtlangle u,u rangle langle v,v rangle = cos(phi)
$$
where $phi$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Im(v) \ -Re(v) endbmatrix$.

The latter real vector is the same as the one that defined $theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $mathbbC^1$, we have $phi = theta - fracpi2$ and thus,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = sin(theta)
$$

share|cite|improve this answer

edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

A vector $v in mathbbC^n$ can be expressed as
$$
v = beginbmatrix a_1 + ib_1 \ vdots \ a_n + ib_n endbmatrix = beginbmatrix a_1 \ vdots \ a_n endbmatrix + ibeginbmatrix b_1 \ vdots \ b_n endbmatrix = Re(v)+iIm(v)
$$

As explained on Wikipedia, if we want to endow this vector space with an inner-product that induces a positive-definite norm,
$$
langle v,v rangle geq 0, text(equality iff $v equiv 0$)
$$
then it is sensible to define (for any $u,v in mathbbC^n$)
$$
langle u,v rangle := sum_i baru_i v_i
$$
or equivalently,
beginalign
langle u,v rangle &= big(Re(u)^top-iIm(u)^topbig)big(Re(v)+iIm(v)big)\
&= big(Re(u)^topRe(v) + Im(u)^top Im(v)big) + ibig(Re(u)^top Im(v) - Im(u)^topRe(v)big)\
&= Rebig(langle u,v ranglebig) + i Imbig(langle u,v ranglebig)
endalign

We see that,
$$
fracRebig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^topRe(v) + Im(u)^top Im(v)sqrtlangle u,u rangle langle v,v rangle = cos(theta)
$$
where $theta$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Re(v) \ Im(v) endbmatrix$.

On the other hand,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = fracRe(u)^top Im(v) - Im(u)^topRe(v)sqrtlangle u,u rangle langle v,v rangle = cos(phi)
$$
where $phi$ is, real-geometrically, the angle between $beginbmatrix Re(u) \ Im(u) endbmatrix$ and $beginbmatrix Im(v) \ -Re(v) endbmatrix$.

The latter real vector is the same as the one that defined $theta$, but rotated $90$ degrees about the axis normal to the "complex plane" subspace. In the simple case of $mathbbC^1$, we have $phi = theta - fracpi2$ and thus,
$$
fracImbig(langle u,v ranglebig)sqrtlangle u,u rangle langle v,v rangle = sin(theta)
$$

share|cite|improve this answer

edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 6:49

edited Jul 16 at 6:49

edited Jul 16 at 6:49

answered Jul 16 at 6:38

jnez71

2,176519

answered Jul 16 at 6:38

jnez71

2,176519

answered Jul 16 at 6:38

jnez71

2,176519

2,176519

add a comment | 

add a comment | 

up vote
2
down vote

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2times 2$ matrix $$beginpmatrixa&-b\b&aendpmatrix.$$

(There is another embedding into $M_2(mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=beginpmatrixv_1\v_2\ vdots\ v_nendpmatrix$$
corresponds to a $2ntimes 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $left(overlinev_1,dots,overlinev_nright).$ Then the $2times 2$ matrix, $A_v^TA_w,$ corresponds to $sum_i=1^noverlinev_iw_i$.

share|cite|improve this answer

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

add a comment | 

up vote
2
down vote

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2times 2$ matrix $$beginpmatrixa&-b\b&aendpmatrix.$$

(There is another embedding into $M_2(mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=beginpmatrixv_1\v_2\ vdots\ v_nendpmatrix$$
corresponds to a $2ntimes 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $left(overlinev_1,dots,overlinev_nright).$ Then the $2times 2$ matrix, $A_v^TA_w,$ corresponds to $sum_i=1^noverlinev_iw_i$.

share|cite|improve this answer

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

add a comment | 

up vote
2
down vote

up vote
2
down vote

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2times 2$ matrix $$beginpmatrixa&-b\b&aendpmatrix.$$

(There is another embedding into $M_2(mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=beginpmatrixv_1\v_2\ vdots\ v_nendpmatrix$$
corresponds to a $2ntimes 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $left(overlinev_1,dots,overlinev_nright).$ Then the $2times 2$ matrix, $A_v^TA_w,$ corresponds to $sum_i=1^noverlinev_iw_i$.

share|cite|improve this answer

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

The standard embedding of the complex numbers as matrices sends $a+bi$ to the $2times 2$ matrix $$beginpmatrixa&-b\b&aendpmatrix.$$

(There is another embedding into $M_2(mathbb R)$ which is just the conjugate.)

Then a complex vector: $$v=beginpmatrixv_1\v_2\ vdots\ v_nendpmatrix$$
corresponds to a $2ntimes 2$ real matrix, $A_v$. The interesting fact is that $A_v^T$ corresponds to the horizontal vector $left(overlinev_1,dots,overlinev_nright).$ Then the $2times 2$ matrix, $A_v^TA_w,$ corresponds to $sum_i=1^noverlinev_iw_i$.

share|cite|improve this answer

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

share|cite|improve this answer

share|cite|improve this answer

edited Jul 16 at 13:08

edited Jul 16 at 13:08

edited Jul 16 at 13:08

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

answered Jul 16 at 2:59

Thomas Andrews

128k10144285

128k10144285

add a comment | 

add a comment |