Rotation matrix in R^3

Question

Rotation matrices in $R^3$ are given by $$S = \begin{pmatrix} \hat e'_1.\hat e_1 & \hat e'_1.\hat e_2 & \hat e'_1.\hat e_3 \\ \hat e'_2.\hat e_1 & \hat e'_2.\hat e_2 & \hat e'_2.\hat e_3 \\ \hat e'_3.\hat e_1 & \hat e'_3.\hat e_2 & \hat e'_3.\hat e_3 \end{pmatrix} $$ where {$\hat e_1, \hat e_2, \hat e_3$} are orthogonal unit vectors in original space and {$\hat e'_1, \hat e'_2 ,\hat e'_3$} are orthogonal unit vectors in rotated space.

I quote the following reasoning from a text book that couldn't understand.

It is useful to make one observation about the elements of $S$, namely $S_{\mu\nu} = \hat e'_\mu.\hat e_\nu$. This dot product is the projection of $\hat e'_\mu$ onto the $\hat e_\nu$ direction, and is therefore the change in $x_\nu$ that is produced by a unit change in $x'_\mu$.Since the relation between the coordinates is linear, we can identify $\hat e'_\mu.\hat e_\nu$ as $\frac {\partial x_\nu}{\partial x'_\mu}$, so our transformation matrix S can be written in the alternate form

$$S = \begin{pmatrix} \frac {\partial x_1}{\partial x'_1} & \frac {\partial x_2}{\partial x'_1} & \frac {\partial x_3}{\partial x'_1} \\ \frac {\partial x_1}{\partial x'_2} & \frac {\partial x_1}{\partial x'_2} & \frac {\partial x_1}{\partial x'_2} \\ \frac {\partial x_1}{\partial x'_3} & \frac {\partial x_1}{\partial x'_3} & \frac {\partial x_1}{\partial x'_3} \end{pmatrix} $$

I cannot absorb that a dot product is the same as a partial derivative. Any help will be appreciated.

Answer 1

For any $m \times n$ matrix $A$, the coordinate representation of the linear transformation $$T_A:x \mapsto A x$$ is $$T(x)_i = (Ax)_i = \sum_{k = 1}^n A_{ik} x_k, \qquad 1 \leq i \leq m.$$ In other words, the $i$th entry of the transformed vector $Ax$ is just the usual dot product of the $i$th row of $A$ (regarded as a vector) with $x$.

On the other hand, differentiating the above coordinate formula gives that the partial derivative of $T_i$ is $$\frac{\partial T(x)_i}{\partial x_j} = \frac{\partial}{\partial x_j}\left(\sum_{k = 1}^n A_{ik} x_k\right) = A_{ij}.$$

Answer 2

For $R^2$ it is easy to visualize. You can extend it to $R^3$. The thing to note is that the projection of one unit vector on another unit vector is the cosine of the angle between them.

$e^{'}_v.{e_{\mu}} = |e^{'}_v|.|e_{\mu}|\cos{\theta} = \cos{\theta}$ , where $\theta$ is the angle between unit vectors $e^{'}_v$ and $e_{\mu}$.

Now, refer to the image attached.

The partial derivatives come into play when we do something similar in $R^3$.