How to prove invariance of dot-product to rotation of coordinate system

Question

Using the definition of a dot-product as the sum of the products of the various components, how do you prove that the dot product will remain the same when the coordinate system rotates?

Preferably an intuitive proof please, explainable to a high-school student. Thanks in advance.

Answer 1

First you should show that for any two vectors $v$ and $w$ in $\mathbb{R}^n$ (taking $n=3$ if necessary) $v\cdot w = |v||w|\operatorname{cos}\theta $, where $\theta$ is the (smaller) angle between both vectors.

This is a very geometric fact and you can probably prove it to them if they know the cosines law. First observe that:

$$||v-w||^2 = ||v||^2+||w||^2-2||v||w||\operatorname{cos}\theta.$$

This comes from the the fact that the vectors $v$, $w$ and $(v-w)$ form a triangle (draw it). On the other hand,

$$||v-w||^2 = (v-w)\cdot(v-w)=v\cdot v -2v\cdot w + w\cdot w .$$

The result follows immediately. After this you only need to observe that rotations don't affect lenghts or angles, then by the formula above the dot product is invariant under rotations.

Answer 2

If $v$ and $u$ are vectors then the dot product is $vu^t$ or equivalently $v^tu$, $uv^t$ or $u^tv$ where $x^t$ is the transpose of $x$, i.e. switch the rows and columns, so that

$$(x_1, x_2, x_3, ...)^t = \begin{pmatrix} x_1\\ x_2\\ x_3\\ .\\ .\\ \end{pmatrix}$$ and the dot product of $u$ and $v$ is then seen to be

$$(u_1, u_2, u_3, ...)\begin{pmatrix} v_1\\ v_2\\ v_3\\ .\\ .\\ \end{pmatrix} = u_1v_1 + u_2v_2+ u_3v_3 + \ldots$$

If you now rotate the coordinate system with a rotation matrix $R$ then the components of a vector $u$ become $uR$, and he components of the transposed vector $v^t$ become $R^tv^t$ (where $R^t$ is the transpose of $R$ given by $R^t_{i, j} = R_{j, i}$), so that the dot product in the new coordinates is $uRR^tv^t$.

A property or rotations is that their matrices are orthogonal and their transpose is equal to their inverse so that $R^t = R^{-1}$, so the scalar product is = $uRR^{-1}v^t$ and $RR^{-1} = I$ (the identity matrix), so that $uRR^tv^t = uRR^{-1}v^t = uIv^t = uv^t$, i.e. the dot product is invariant under rotation.

That may be a bit much at high school level, so perhaps a demonstration in 2-D would help.

A rotation matrix has the form

$$\begin{pmatrix} \cos\theta& -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix}$$ where $\theta$ is the angle of rotation.

$u$ rotated is $$(u_x, u_y)\begin{pmatrix} \cos\theta& -\sin\theta\\ \sin\theta & \cos\theta \end{pmatrix} = ([u_x]\cos\theta + u_y \sin\theta], [-u_x\sin\theta + u_y\cos\theta])$$ and similarly for $v$, so that $$uv^t = ([u_x\cos\theta +u_y\sin\theta]\cdot[v_x \cos\theta + v_y \sin\theta] + [-u_x\sin\theta + u_y\cos\theta]\cdot[-v_x\sin\theta + v_y\cos\theta])$$

and when you multiply it out and gather the terms you end up with $(u_xv_x + u_yv_y)(\cos^2\theta + \sin^2\theta)$, i.e. the same $(u_xv_x + u_yv_y)$ as you have without the rotation.