Sum of squared inner product of vector with spokes around unit circle is constant

Question

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 $$\sum_{i=1}^n (v\cdot w_i)^2 = k \|v\|^2$$ 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?

Answer 1

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^{2i\left(x + k\, \frac{2\pi}n \right)} + e^{-2i\left(x + k\, \frac{2\pi}n \right)} + 2\right) \\ &= \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 $k = \frac{n}2.$

Answer 2

Since the $w_i$ are evenly spaced, we can say:

$$w_i=\cos(\phi+i\theta)\hat i+\sin(\phi+i\theta)\hat j$$

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, $n\theta=2\pi$

Now, $v\cdot w_i=v_x\cos(\phi+i\theta)+v_y\sin(\phi+i\theta)$, so:

$$(v\cdot w_i)^2=v_x^2\cos^2(\phi+i\theta)+v_y^2\sin^2(\phi+i\theta)+2v_xv_y\cos(\phi+i\theta)\sin(\phi+i\theta) \\ =v_x^2\left(\frac{\cos(2\phi+2i\theta)+1}{2}\right)+v_y^2\left(\frac{1-\cos(2\phi+2i\theta)}{2}\right)+v_xv_y\sin(2\phi+2i\theta) \\ =\frac{v_x^2+v_y^2}{2}+\cos(2\phi+2i\theta)\left(\frac{v_x^2-v_y^2}{2}\right)+v_xv_y\sin(2\phi+2i\theta)$$

Now, let's put this into a summation:

$$\sum_{i=1}^n (v\cdot w_i)^2=\frac{n}{2}(v_x^2+v_y^2)+\left(\frac{v_x^2-v_y^2}{2}\right)\left[\sum_{i=1}^n\cos(2\phi+2i\theta)\right]+v_xv_y\left[\sum_{i=1}^n\sin(2\phi+2i\theta)\right]$$

Now, I will use $j=\sqrt{-1}$. The first summation and second summation are related: They are the real and imaginary part, respectively, of $\sum_{i=1}^n e^{2j\phi+2ij\theta}=e^{2j\phi}\sum_{i=1}^n e^{2ij\theta}$. If we let $\omega=e^{2j\theta}$, the sum becomes $e^{2j\phi}\sum_{i=1}^n \omega^i$, which is a geometric series:

$$\sum_{i=1}^n \omega^i=\omega\frac{\omega^n-1}{\omega-1}=\omega\frac{e^{2jn\theta}-1}{\omega-1}=\omega\frac{e^{4\pi j}-1}{\omega-1}=\omega\frac{1-1}{\omega-1}=0$$

(Note that I used $n\theta=2\pi$ and $e^{4\pi j}=\cos(4\pi)+j\sin(4\pi)=1$ in the above derivation.)

(Also, notice that this proof fails when $\omega=e^{2j\theta}=1$ because of a division-by-zero error from above. More specifically, this occurs when $2\theta=\frac{4\pi}{n}$ is a multiple of $2\pi$, 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:

$$\sum_{i=1}^n (v\cdot w_i)^2=\frac{n}{2}(v_x^2+v_y^2)+\left(\frac{v_x^2-v_y^2}{2}\right)[0]+v_xv_y[0]\rightarrow \sum_{i=1}^n (v\cdot w_i)^2=\frac{n}{2}\|v\|^2$$

Answer 3

We can use some vector calculus here. For some constant $k \in \mathbb{R}$ define $f : \mathbb{R}^2 \to \mathbb{R}$ as $$f(v) = \sum_{i=1}^n (v\cdot w_i)^2 - k\|v\|^2$$

Taking the gradient of this function yields $$\nabla f(v) = \sum_{i=1}^n 2(v\cdot w_i) - 2kv = 2\left[\sum_{i=1}^n (v\cdot w_i)w_i - kv\right]$$

because $\nabla (v \cdot w_i) = w_i$ and $\nabla \|v\|^2 = 2v$.

Now we calculate the sum $\sum_{i=1}^n (v\cdot w_i)w_i$. Using the notation from the answer by @Noble Mushtak, we get \begin{align} \sum_{i=1}^n (v\cdot w_i)w_i &= \sum_{i=1}^n \big(v_x\cos(\phi+i\theta)+v_y\sin(\phi+i\theta)\big)\begin{bmatrix}\cos(\phi + i\theta)\\ \sin(\phi + i\theta) \end{bmatrix}\\ &= \begin{bmatrix}v_x \sum_{i=1}^n\cos^2(\phi + i\theta) + v_y \sum_{i=1}^n\sin(\phi+i\theta)\cos(\phi + i\theta)\\ v_x \sum_{i=1}^n\sin(\phi+i\theta)\cos(\phi + i\theta)+ v_y \sum_{i=1}^n\sin^2(\phi + i\theta)\end{bmatrix}\\ &= \begin{bmatrix}v_x \sum_{i=1}^n\frac{1+\cos(2\phi + 2i\theta)}2 + v_y \sum_{i=1}^n\frac12\sin(2\phi+2i\theta)\\ v_x \sum_{i=1}^n\frac12\sin(2\phi+2i\theta)+ v_y \sum_{i=1}^n\frac{1-\cos(2\phi + 2i\theta)}2\end{bmatrix}\\ &= \frac{n}2\begin{bmatrix} v_x \\ v_y \end{bmatrix}\\ &= \frac{n}2v \end{align} because $\sum_{i=1}^n \cos(2\phi + 2i\theta) = \sum_{i=1}^n \sin(2\phi + 2i\theta) = 0$.

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 $$\sum_{i=1}^n (v\cdot w_i)^2 = \frac{n}2\|v\|^2$$ holds.

Answer 4

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 :

$$\sum \|W_k\|^2 = \sum\cos(ka)^2+\sum\sin(ka)^2=\underbrace{1+1+\cdots +1}_{n \ \text{times}}=n. \tag{2}$$

and, besides, by squaring relationship $\sum_{k=1}^n w_k \equiv 0$ :

$$\sum\cos(ka)^2-\sum\sin(ka)^2+2i \sum \sin(ka)\cos(ka) = 0. \tag{3}$$

From (2) and (3), we can conclude that :

$$\begin{cases} \sum_k\cos(ka)^2=\sum_k\sin(ka)^2=n/2\\ \sum_k \sin(ka)\cos(ka) = 0 \end{cases}.\tag{4}$$

Now, we are able to establish the following identity :

$$\sum_{k=1}^n (V\cdot w_k)^2 = \frac{n}{2}\|V\|^2 \tag{5}$$

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 :

$$M=\begin{pmatrix}\sum_k\cos(ka)^2&\sum_k \sin(ka)\cos(ka)\\ \sum_k \sin(ka)\cos(ka)&\sum_k\sin(ka)^2\end{pmatrix}=\begin{pmatrix}n/2&0\\0&n/2\end{pmatrix}$$

(by using relationships (4)).

We can therefore conclude that the LHS of (5) is indeed equal to its RHS.