Mixing
Sum of squared inner product of vector with spokes around unit circle is constant
Let $v$ be any vector in the plane, and ${w_i}$ be $n>2$ vectors evenly spaced around the unit circle. Then it seems true that
$$sum_{i=1}^n (vcdot w_i)^2 = k |v|^2$$
where $k$ is a constant independent of $v$. My intuition is that the left-hand side, as a function of the argument $theta$ of $v$, has "too much symmetry" for its low frequency.
Is there some slick proof of this identity, using e.g. the algebraic structure of the $n$ roots of unity?
linear-algebra plane-geometry
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
add a comment |
Let $v$ be any vector in the plane, and ${w_i}$ be $n>2$ vectors evenly spaced around the unit circle. Then it seems true that
$$sum_{i=1}^n (vcdot w_i)^2 = k |v|^2$$
where $k$ is a constant independent of $v$. My intuition is that the left-hand side, as a function of the argument $theta$ of $v$, has "too much symmetry" for its low frequency.
Is there some slick proof of this identity, using e.g. the algebraic structure of the $n$ roots of unity?
linear-algebra plane-geometry
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
add a comment |
Let $v$ be any vector in the plane, and ${w_i}$ be $n>2$ vectors evenly spaced around the unit circle. Then it seems true that
$$sum_{i=1}^n (vcdot w_i)^2 = k |v|^2$$
where $k$ is a constant independent of $v$. My intuition is that the left-hand side, as a function of the argument $theta$ of $v$, has "too much symmetry" for its low frequency.
Is there some slick proof of this identity, using e.g. the algebraic structure of the $n$ roots of unity?
linear-algebra plane-geometry
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
Let $v$ be any vector in the plane, and ${w_i}$ be $n>2$ vectors evenly spaced around the unit circle. Then it seems true that
$$sum_{i=1}^n (vcdot w_i)^2 = k |v|^2$$
where $k$ is a constant independent of $v$. My intuition is that the left-hand side, as a function of the argument $theta$ of $v$, has "too much symmetry" for its low frequency.
Is there some slick proof of this identity, using e.g. the algebraic structure of the $n$ roots of unity?
linear-algebra plane-geometry
linear-algebra plane-geometry
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
asked Dec 27 '18 at 1:10
user7530user7530
34.6k759113
34.6k759113
add a comment |
add a comment |
4 Answers
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This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.
Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.
Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$
Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have
begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
add a comment |
Since the $w_i$ are evenly spaced, we can say:
$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$
Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$
Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:
$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$
Now, let's put this into a summation:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$
Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:
$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$
(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)
(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)
Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
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We can use some vector calculus here. For some constant $k in mathbb{R}$ define $f : mathbb{R}^2 to mathbb{R}$ as $$f(v) = sum_{i=1}^n (vcdot w_i)^2 - k|v|^2$$
Taking the gradient of this function yields
$$nabla f(v) = sum_{i=1}^n 2(vcdot w_i) - 2kv = 2left[sum_{i=1}^n (vcdot w_i)w_i - kvright]$$
because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.
Now we calculate the sum $sum_{i=1}^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
begin{align}
sum_{i=1}^n (vcdot w_i)w_i &= sum_{i=1}^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)begin{bmatrix}cos(phi + itheta)\ sin(phi + itheta) end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^ncos^2(phi + itheta) + v_y sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_{i=1}^nsin^2(phi + itheta)end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^nfrac{1+cos(2phi + 2itheta)}2 + v_y sum_{i=1}^nfrac12sin(2phi+2itheta)\ v_x sum_{i=1}^nfrac12sin(2phi+2itheta)+ v_y sum_{i=1}^nfrac{1-cos(2phi + 2itheta)}2end{bmatrix}\
&= frac{n}2begin{bmatrix}
v_x \ v_y
end{bmatrix}\
&= frac{n}2v
end{align}
because $sum_{i=1}^n cos(2phi + 2itheta) = sum_{i=1}^n sin(2phi + 2itheta) = 0$.
Therefore, for $k = frac{n}2$ we have $nabla f equiv 0$ so $f = text{const}$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_{i=1}^n (vcdot w_i)^2 = frac{n}2|v|^2$$
holds.
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
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Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.
Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binom{cos(ka)}{cos(ka)} text{
associated with} w_k:=e^{i k a}in mathbb{C} text{where} a:=2 pi/n. tag{1}$$
We have the following identities :
$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace{1+1+cdots +1}_{n text{times}}=n. tag{2}$$
and, besides, by squaring relationship $sum_{k=1}^n w_k equiv 0$ :
$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag{3}$$
From (2) and (3), we can conclude that :
$$begin{cases}
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
end{cases}.tag{4}$$
Now, we are able to establish the following identity :
$$sum_{k=1}^n (Vcdot w_k)^2 = frac{n}{2}|V|^2 tag{5}$$
by converting its LHS into the classical shape of a quadratic form :
$$sum_{k=1}^n (V^TW_k)(W_k^TV)=V^T(sum_{k=1}^n W_kW_k^T)V = V^TMV$$
(here $W_k$ and $V$ are considered as column vectors)
where the associated matrix is :
$$M=begin{pmatrix}sum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2end{pmatrix}=begin{pmatrix}n/2&0\0&n/2end{pmatrix}$$
(by using relationships (4)).
We can therefore conclude that the LHS of (5) is indeed equal to its RHS.
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
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4 Answers
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4 Answers
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This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.
Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.
Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$
Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have
begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
add a comment |
This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.
Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.
Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$
Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have
begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
add a comment |
This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.
Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.
Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$
Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have
begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
This is similar to Noble's answer but I wanted to highlight the geometric aspect of the problem a little more. First we note that
$$v cdot w_i = || v || cos(theta) $$
where $theta$ is the angle between $v$ and $w_i$. Since $||v||^2$ is on both sides of the equation, we can assume $||v|| = 1$.
Now consider a unit vector $v$ and a regular polygon with $n$ sides centered at the origin. Connect the vertices of the polygon to the center. $v$ lies between two of the radii. Now going counter clockwise, $v$ makes the following angles with the $w_i:$
$$x, x + frac{2 pi}n, x + 2 cdot frac{2 pi}n, cdots, x + (n-1) cdot frac{2 pi}n $$
for some $x$ as shown in the image at the bottom of the post.
Therefore, the sum that we want is
$$ sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2.$$
Using Euler's formula $cos(t) = frac{e^{it} + e^{-it}}2$, we have
begin{align}
sum_{k=0}^{n-1} cos left( x + k , frac{2 pi}n right)^2 &= frac{1}4 sum_{k=0}^{n-1} left(e^{2ileft(x + k, frac{2pi}n right)} + e^{-2ileft(x + k, frac{2pi}n right)} + 2right) \
&= frac{n}2 + (e^{2ix} + e^{-2ix}) sum_{k=0}^{n-1} omega^k
end{align}
where $omega$ is a non trivial $n$th root of unity. Then we can easily calculate that
$$ sum_{k=0}^{n-1} omega^k = frac{omega^n-1}{omega - 1} = 0$$
which proves the claim for $k = frac{n}2.$
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
answered Dec 27 '18 at 2:07
Sandeep SilwalSandeep Silwal
5,84311236
5,84311236
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
add a comment |
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
This is indeed a simple solution. I have also attempted to give a geometrical flavor in my (more recent) solution, but directed by the quest for a quadratic form.
– Jean Marie
Dec 28 '18 at 11:28
add a comment |
Since the $w_i$ are evenly spaced, we can say:
$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$
Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$
Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:
$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$
Now, let's put this into a summation:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$
Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:
$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$
(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)
(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)
Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
add a comment |
Since the $w_i$ are evenly spaced, we can say:
$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$
Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$
Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:
$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$
Now, let's put this into a summation:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$
Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:
$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$
(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)
(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)
Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
add a comment |
Since the $w_i$ are evenly spaced, we can say:
$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$
Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$
Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:
$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$
Now, let's put this into a summation:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$
Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:
$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$
(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)
(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)
Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
Since the $w_i$ are evenly spaced, we can say:
$$w_i=cos(phi+itheta)hat i+sin(phi+itheta)hat j$$
Also, since they are evenly spaced around the full circle, the distances between each $w_i$, which is $theta$ radians, times the number of vectors, which is $n$, must be a full circle: That is, $ntheta=2pi$
Now, $vcdot w_i=v_xcos(phi+itheta)+v_ysin(phi+itheta)$, so:
$$(vcdot w_i)^2=v_x^2cos^2(phi+itheta)+v_y^2sin^2(phi+itheta)+2v_xv_ycos(phi+itheta)sin(phi+itheta) \ =v_x^2left(frac{cos(2phi+2itheta)+1}{2}right)+v_y^2left(frac{1-cos(2phi+2itheta)}{2}right)+v_xv_ysin(2phi+2itheta) \ =frac{v_x^2+v_y^2}{2}+cos(2phi+2itheta)left(frac{v_x^2-v_y^2}{2}right)+v_xv_ysin(2phi+2itheta)$$
Now, let's put this into a summation:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)left[sum_{i=1}^ncos(2phi+2itheta)right]+v_xv_yleft[sum_{i=1}^nsin(2phi+2itheta)right]$$
Now, I will use $j=sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $sum_{i=1}^n e^{2jphi+2ijtheta}=e^{2jphi}sum_{i=1}^n e^{2ijtheta}$. If we let $omega=e^{2jtheta}$, the sum becomes $e^{2jphi}sum_{i=1}^n omega^i$, which is a geometric series:
$$sum_{i=1}^n omega^i=omegafrac{omega^n-1}{omega-1}=omegafrac{e^{2jntheta}-1}{omega-1}=omegafrac{e^{4pi j}-1}{omega-1}=omegafrac{1-1}{omega-1}=0$$
(Note that I used $ntheta=2pi$ and $e^{4pi j}=cos(4pi)+jsin(4pi)=1$ in the above derivation.)
(Also, notice that this proof fails when $omega=e^{2jtheta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2theta=frac{4pi}{n}$ is a multiple of $2pi$, which is the case when there are $n=1$ or $n=2$ points.)
Thus, this summation is $0$, so both its real and imaginary parts are 0. This gives us:
$$sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}(v_x^2+v_y^2)+left(frac{v_x^2-v_y^2}{2}right)[0]+v_xv_y[0]rightarrow sum_{i=1}^n (vcdot w_i)^2=frac{n}{2}|v|^2$$
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
edited Dec 27 '18 at 1:57
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
answered Dec 27 '18 at 1:37
Noble MushtakNoble Mushtak
15.2k1735
15.2k1735
add a comment |
add a comment |
We can use some vector calculus here. For some constant $k in mathbb{R}$ define $f : mathbb{R}^2 to mathbb{R}$ as $$f(v) = sum_{i=1}^n (vcdot w_i)^2 - k|v|^2$$
Taking the gradient of this function yields
$$nabla f(v) = sum_{i=1}^n 2(vcdot w_i) - 2kv = 2left[sum_{i=1}^n (vcdot w_i)w_i - kvright]$$
because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.
Now we calculate the sum $sum_{i=1}^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
begin{align}
sum_{i=1}^n (vcdot w_i)w_i &= sum_{i=1}^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)begin{bmatrix}cos(phi + itheta)\ sin(phi + itheta) end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^ncos^2(phi + itheta) + v_y sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_{i=1}^nsin^2(phi + itheta)end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^nfrac{1+cos(2phi + 2itheta)}2 + v_y sum_{i=1}^nfrac12sin(2phi+2itheta)\ v_x sum_{i=1}^nfrac12sin(2phi+2itheta)+ v_y sum_{i=1}^nfrac{1-cos(2phi + 2itheta)}2end{bmatrix}\
&= frac{n}2begin{bmatrix}
v_x \ v_y
end{bmatrix}\
&= frac{n}2v
end{align}
because $sum_{i=1}^n cos(2phi + 2itheta) = sum_{i=1}^n sin(2phi + 2itheta) = 0$.
Therefore, for $k = frac{n}2$ we have $nabla f equiv 0$ so $f = text{const}$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_{i=1}^n (vcdot w_i)^2 = frac{n}2|v|^2$$
holds.
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
add a comment |
We can use some vector calculus here. For some constant $k in mathbb{R}$ define $f : mathbb{R}^2 to mathbb{R}$ as $$f(v) = sum_{i=1}^n (vcdot w_i)^2 - k|v|^2$$
Taking the gradient of this function yields
$$nabla f(v) = sum_{i=1}^n 2(vcdot w_i) - 2kv = 2left[sum_{i=1}^n (vcdot w_i)w_i - kvright]$$
because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.
Now we calculate the sum $sum_{i=1}^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
begin{align}
sum_{i=1}^n (vcdot w_i)w_i &= sum_{i=1}^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)begin{bmatrix}cos(phi + itheta)\ sin(phi + itheta) end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^ncos^2(phi + itheta) + v_y sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_{i=1}^nsin^2(phi + itheta)end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^nfrac{1+cos(2phi + 2itheta)}2 + v_y sum_{i=1}^nfrac12sin(2phi+2itheta)\ v_x sum_{i=1}^nfrac12sin(2phi+2itheta)+ v_y sum_{i=1}^nfrac{1-cos(2phi + 2itheta)}2end{bmatrix}\
&= frac{n}2begin{bmatrix}
v_x \ v_y
end{bmatrix}\
&= frac{n}2v
end{align}
because $sum_{i=1}^n cos(2phi + 2itheta) = sum_{i=1}^n sin(2phi + 2itheta) = 0$.
Therefore, for $k = frac{n}2$ we have $nabla f equiv 0$ so $f = text{const}$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_{i=1}^n (vcdot w_i)^2 = frac{n}2|v|^2$$
holds.
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
add a comment |
We can use some vector calculus here. For some constant $k in mathbb{R}$ define $f : mathbb{R}^2 to mathbb{R}$ as $$f(v) = sum_{i=1}^n (vcdot w_i)^2 - k|v|^2$$
Taking the gradient of this function yields
$$nabla f(v) = sum_{i=1}^n 2(vcdot w_i) - 2kv = 2left[sum_{i=1}^n (vcdot w_i)w_i - kvright]$$
because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.
Now we calculate the sum $sum_{i=1}^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
begin{align}
sum_{i=1}^n (vcdot w_i)w_i &= sum_{i=1}^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)begin{bmatrix}cos(phi + itheta)\ sin(phi + itheta) end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^ncos^2(phi + itheta) + v_y sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_{i=1}^nsin^2(phi + itheta)end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^nfrac{1+cos(2phi + 2itheta)}2 + v_y sum_{i=1}^nfrac12sin(2phi+2itheta)\ v_x sum_{i=1}^nfrac12sin(2phi+2itheta)+ v_y sum_{i=1}^nfrac{1-cos(2phi + 2itheta)}2end{bmatrix}\
&= frac{n}2begin{bmatrix}
v_x \ v_y
end{bmatrix}\
&= frac{n}2v
end{align}
because $sum_{i=1}^n cos(2phi + 2itheta) = sum_{i=1}^n sin(2phi + 2itheta) = 0$.
Therefore, for $k = frac{n}2$ we have $nabla f equiv 0$ so $f = text{const}$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_{i=1}^n (vcdot w_i)^2 = frac{n}2|v|^2$$
holds.
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
We can use some vector calculus here. For some constant $k in mathbb{R}$ define $f : mathbb{R}^2 to mathbb{R}$ as $$f(v) = sum_{i=1}^n (vcdot w_i)^2 - k|v|^2$$
Taking the gradient of this function yields
$$nabla f(v) = sum_{i=1}^n 2(vcdot w_i) - 2kv = 2left[sum_{i=1}^n (vcdot w_i)w_i - kvright]$$
because $nabla (v cdot w_i) = w_i$ and $nabla |v|^2 = 2v$.
Now we calculate the sum $sum_{i=1}^n (vcdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get
begin{align}
sum_{i=1}^n (vcdot w_i)w_i &= sum_{i=1}^n big(v_xcos(phi+itheta)+v_ysin(phi+itheta)big)begin{bmatrix}cos(phi + itheta)\ sin(phi + itheta) end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^ncos^2(phi + itheta) + v_y sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)\ v_x sum_{i=1}^nsin(phi+itheta)cos(phi + itheta)+ v_y sum_{i=1}^nsin^2(phi + itheta)end{bmatrix}\
&= begin{bmatrix}v_x sum_{i=1}^nfrac{1+cos(2phi + 2itheta)}2 + v_y sum_{i=1}^nfrac12sin(2phi+2itheta)\ v_x sum_{i=1}^nfrac12sin(2phi+2itheta)+ v_y sum_{i=1}^nfrac{1-cos(2phi + 2itheta)}2end{bmatrix}\
&= frac{n}2begin{bmatrix}
v_x \ v_y
end{bmatrix}\
&= frac{n}2v
end{align}
because $sum_{i=1}^n cos(2phi + 2itheta) = sum_{i=1}^n sin(2phi + 2itheta) = 0$.
Therefore, for $k = frac{n}2$ we have $nabla f equiv 0$ so $f = text{const}$. Plugging in $v = 0$ yields $f equiv 0$ so your formula
$$sum_{i=1}^n (vcdot w_i)^2 = frac{n}2|v|^2$$
holds.
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
answered Dec 27 '18 at 11:39
mechanodroidmechanodroid
27k62446
27k62446
add a comment |
add a comment |
Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.
Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binom{cos(ka)}{cos(ka)} text{
associated with} w_k:=e^{i k a}in mathbb{C} text{where} a:=2 pi/n. tag{1}$$
We have the following identities :
$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace{1+1+cdots +1}_{n text{times}}=n. tag{2}$$
and, besides, by squaring relationship $sum_{k=1}^n w_k equiv 0$ :
$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag{3}$$
From (2) and (3), we can conclude that :
$$begin{cases}
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
end{cases}.tag{4}$$
Now, we are able to establish the following identity :
$$sum_{k=1}^n (Vcdot w_k)^2 = frac{n}{2}|V|^2 tag{5}$$
by converting its LHS into the classical shape of a quadratic form :
$$sum_{k=1}^n (V^TW_k)(W_k^TV)=V^T(sum_{k=1}^n W_kW_k^T)V = V^TMV$$
(here $W_k$ and $V$ are considered as column vectors)
where the associated matrix is :
$$M=begin{pmatrix}sum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2end{pmatrix}=begin{pmatrix}n/2&0\0&n/2end{pmatrix}$$
(by using relationships (4)).
We can therefore conclude that the LHS of (5) is indeed equal to its RHS.
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
add a comment |
Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.
Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binom{cos(ka)}{cos(ka)} text{
associated with} w_k:=e^{i k a}in mathbb{C} text{where} a:=2 pi/n. tag{1}$$
We have the following identities :
$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace{1+1+cdots +1}_{n text{times}}=n. tag{2}$$
and, besides, by squaring relationship $sum_{k=1}^n w_k equiv 0$ :
$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag{3}$$
From (2) and (3), we can conclude that :
$$begin{cases}
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
end{cases}.tag{4}$$
Now, we are able to establish the following identity :
$$sum_{k=1}^n (Vcdot w_k)^2 = frac{n}{2}|V|^2 tag{5}$$
by converting its LHS into the classical shape of a quadratic form :
$$sum_{k=1}^n (V^TW_k)(W_k^TV)=V^T(sum_{k=1}^n W_kW_k^T)V = V^TMV$$
(here $W_k$ and $V$ are considered as column vectors)
where the associated matrix is :
$$M=begin{pmatrix}sum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2end{pmatrix}=begin{pmatrix}n/2&0\0&n/2end{pmatrix}$$
(by using relationships (4)).
We can therefore conclude that the LHS of (5) is indeed equal to its RHS.
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
add a comment |
Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.
Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binom{cos(ka)}{cos(ka)} text{
associated with} w_k:=e^{i k a}in mathbb{C} text{where} a:=2 pi/n. tag{1}$$
We have the following identities :
$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace{1+1+cdots +1}_{n text{times}}=n. tag{2}$$
and, besides, by squaring relationship $sum_{k=1}^n w_k equiv 0$ :
$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag{3}$$
From (2) and (3), we can conclude that :
$$begin{cases}
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
end{cases}.tag{4}$$
Now, we are able to establish the following identity :
$$sum_{k=1}^n (Vcdot w_k)^2 = frac{n}{2}|V|^2 tag{5}$$
by converting its LHS into the classical shape of a quadratic form :
$$sum_{k=1}^n (V^TW_k)(W_k^TV)=V^T(sum_{k=1}^n W_kW_k^T)V = V^TMV$$
(here $W_k$ and $V$ are considered as column vectors)
where the associated matrix is :
$$M=begin{pmatrix}sum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2end{pmatrix}=begin{pmatrix}n/2&0\0&n/2end{pmatrix}$$
(by using relationships (4)).
We can therefore conclude that the LHS of (5) is indeed equal to its RHS.
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
Here is a solution that is based on expressing the LHS of the given expression under the classical "shape" of a quadratic form by looking for its associated matrix.
Let us begin by notations. Taking one of the "spikes" (vectors) as directing the $x$-axis, we can assume that :
$$W_k:=binom{cos(ka)}{cos(ka)} text{
associated with} w_k:=e^{i k a}in mathbb{C} text{where} a:=2 pi/n. tag{1}$$
We have the following identities :
$$sum |W_k|^2 = sumcos(ka)^2+sumsin(ka)^2=underbrace{1+1+cdots +1}_{n text{times}}=n. tag{2}$$
and, besides, by squaring relationship $sum_{k=1}^n w_k equiv 0$ :
$$sumcos(ka)^2-sumsin(ka)^2+2i sum sin(ka)cos(ka) = 0. tag{3}$$
From (2) and (3), we can conclude that :
$$begin{cases}
sum_kcos(ka)^2=sum_ksin(ka)^2=n/2\
sum_k sin(ka)cos(ka) = 0
end{cases}.tag{4}$$
Now, we are able to establish the following identity :
$$sum_{k=1}^n (Vcdot w_k)^2 = frac{n}{2}|V|^2 tag{5}$$
by converting its LHS into the classical shape of a quadratic form :
$$sum_{k=1}^n (V^TW_k)(W_k^TV)=V^T(sum_{k=1}^n W_kW_k^T)V = V^TMV$$
(here $W_k$ and $V$ are considered as column vectors)
where the associated matrix is :
$$M=begin{pmatrix}sum_kcos(ka)^2&sum_k sin(ka)cos(ka)\
sum_k sin(ka)cos(ka)&sum_ksin(ka)^2end{pmatrix}=begin{pmatrix}n/2&0\0&n/2end{pmatrix}$$
(by using relationships (4)).
We can therefore conclude that the LHS of (5) is indeed equal to its RHS.
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
edited Dec 31 '18 at 19:28
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
answered Dec 27 '18 at 17:49
Jean MarieJean Marie
28.8k41949
28.8k41949
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
add a comment |
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
1
1
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
Interesting solution. I was thinking along the same likes actually. Starting from $v^TMv$ you can note that $M$ is a covariance matrix and a covariance matrix's eigenvectors lie in the direction that the data 'spans' which in this case are roots of unity so these eigenvectors are also equally spaced. Thus $M$ is rotationally invariant so it must be a multiple of the identity. (Not really sure about my last claim but it seems plausible...)
– Sandeep Silwal
Dec 28 '18 at 15:06
add a comment |
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