Assume we are working on a problem of rotating a coordinate system frame $(A)$ to another coordinate system frame $(B)$ under a rotation matrix ${}_{A}^{B}\mathcal{R}$.
As an example, consider the diagram shown below:
The rotation matrix is given as follows:
$$
{}_{A}^{B}\mathbf{\mathcal{R}}:=
\begin{bmatrix}\mathbf{i}_{A}\cdot\mathbf{i}_{B}&\mathbf{j}_{A}\cdot\mathbf{i}_{B}&\mathbf{k}_{A}\cdot\mathbf{i}_{B}\\ \mathbf{i}_{A}\cdot\mathbf{j}_{B}&\mathbf{j}_{A}\cdot\mathbf{j}_{B}&\mathbf{k}_{A}\cdot\mathbf{j}_{B}\\ \mathbf{i}_{A}\cdot\mathbf{k}_{B}&\mathbf{k}_{A}\cdot\mathbf{k}_{B}&\mathbf{k}_{A}\cdot\mathbf{k}_{B} \end{bmatrix}
$$
I know that ${}_{A}^{B}\mathbf{\mathcal{R}}$ is orthogonal, but I just noticed that:
$$
\operatorname{det}({}_{A}^{B}\mathbf{\mathcal{R}})=1
\tag1$$
I wish to know what geometric significance $(1)$ has to offer. I know that since the determinant is non-zero then this would mean that any vector in $\mathbb{R}^{3}$ of some coordinate system $(A)$ can be formed as a linear combination of the column vectors of ${}_{A}^{B}\mathbf{\mathcal{R}}$ in the coordinate system of $(B)$.
However, the fact that the determinant is exactly $1$ must have some special implication(s) which I would hope for someone to provide me some detailed explanation regarding this.
Best Answer
Assume an $n$-dimensional real vector space $V$ equiped with an inner product.
The determinant of a linear operator is a measure of the scaling effect of that operator on $n$-dimensional volumes spanned by bases. If the determinant of an operator gives the number $x$, then the volume spanned by the images of any basis relative to the volume spanned by that basis is $x$. So a determinant equal to $1$ means that the operator preseves volumes.
Of course the volume spanned by a basis is measured by choosing (on the inner product space $V$) a top-degree form $\omega$, (i.e. an alternating multilinear map that takes $n$ vectors $v_1,\dots,v_n \in V$ to a real number,) which is sometime also called a determinant, that gives on some chosen orthonormal $\mathcal B$ basis the result $1$.
Furthermore $\omega$ defines an orientation on the space $V$. So the bases on which $\omega$ gives positive values are said to have the same orientation as $\mathcal B$, while the basis one which $\omega$ gives negative values are of the opposite orientation.