linear algebra
You can calculate the volume of a parallelepiped by $|(A \times B) \cdot C|$, where $A$, $B$ and $C$ are vectors. I wonder does the order matter? If it does how, is it determined. I know I can just put it in a matrix and calculate the determinant but I would like to know how it is in this case.
Thanks!
Maybe a picture helps (from Wikipedia):
If $a=(1,2,x)$, $b=(-2,1,0)$, $c=(1,1,3)$, then this is your parallelpiped.
You can convince yourself, that its volume is given by the determinant:
$$\left|\det\left(\begin{matrix}1 & -2 & 1\\2 & 1 & 1\\x & 0 & 3\end{matrix}\right)\right|$$
We should really start with the description for the volume of a parallelotope.
Definition: Let $\Pi$ be an $n$-dimensional parallelotope defined by edge vectors $\mathcal{B}=\left\{ \mathbf{v}_{1},\ \cdots,\ \mathbf{v}_{n}\right\}$. We define the volume of the parallelotope recursively as follows:
For a 1-dimensional parallelotope, we have $\mathrm{vol}(\Pi)$.
For an $n$-dimensional parallelotope with $n>1$, we define the base of the parllelotope to be the volume of the $(n-1)$-dimensional parallelotope defined by edges $\mathcal{B}\backslash\left\{\mathbf{v}_{n}\right\}$, and we define the height to be the length of the component of $\mathbf{v}_{n}$ orthogonal to every vector in the base. Then we define $$\mathrm{vol}\left(\Pi\right)=\mathrm{base}\times\mathrm{height}.$$
You should convince yourself that this is the just a simple definition for the intuitive notion of the volume for a paralleotope. Our main theorem relating the volume to the determinant follows.
Theorem: Let $\Pi$ be an $m$-dimensional parallelotope defined by edge vectors $\mathcal{B}=\left\{ \mathbf{v}_{1},\ \dots,\ \mathbf{v}_{m}\right\}$, where $\mathbf{v}_{i}\in\mathbb{R}^{n}$ for $n\ge m$. That is, we are looking at an $m$-dimensional parallelotope embedded inside $n$-dimensional space. Suppose that $A$ is the $m\times n$ matrix with row vectors $\mathcal{B}$, given by $$A=\begin{pmatrix}\mathbf{v}_{1}^{\mathrm{T}}\\ \vdots\\ \mathbf{v}_{m}^{\mathrm{T}}\end{pmatrix}$$ Then the $m$-dimensional volume of the parallelotope is given by $$\left[\mathrm{vol}\left(\Pi\right)\right]^{2}=\det\left(AA^{\mathrm{T}}\right).$$
Proof: Note that $AA^{\mathrm{T}}$ is an $m\times m$ square matrix. Suppose that $m=1$. Then $$\det AA^{\mathrm{T}}=\det\left(\mathbf{v}_{1}^{\mathrm{T}}\mathbf{v}_{1}\right)=\mathbf{v}_{1}\cdot\mathbf{v}_{1}=\|\mathbf{v}_{1}\|^{2}=\left[\mathrm{vol}_{1}(\mathbf{v}_{1})\right]^{2},$$ so the proposition holds for $m=1$. Now we induct on $m$. Suppose that the proposition holds for $m\ge 1$ and consider $m+1$. Letting $A_{m}$ denote the matrix containing the rows $\mathbf{v}_{1}$ to $\mathbf{v}_{m}$, we can write $A=A_{m+1}$ as
$$A=\begin{pmatrix}A_{m}\\ \mathbf{v}_{m+1}^{\mathrm{T}} \end{pmatrix}$$
We may decompose $\mathbf{v}_{m+1}$ orthogonally as $$\mathbf{v}_{m+1}=\mathbf{v}_{\perp}+\mathbf{v}_{\parallel},$$ where $\mathbf{v}_{\perp}$ lies in the orthogonal complement of the base (this is our height), i.e. $\mathbf{v}_{\perp}\cdot\mathbf{v}_{i}=0$ for $1\le i\le m$, and where $\mathbf{v}_{\parallel}\in\mathrm{span}\left\{ \mathbf{v}_{1},\ \cdots,\ \mathbf{v}_{m}\right\}$. Suppose that $$\mathbf{v}_{\parallel}=c_{1}\mathbf{v}_{1}+\cdots+c_{m}\mathbf{v}_{m}.$$ We apply a sequence of elementary row operations to $A$, adding a multiple $-c_{i}$ of row $i$ to row $m+1$ for each $1\le i\le m$. Write the resulting matrix as $B$, and we have $$B=\begin{pmatrix}A_{m}\\ \mathbf{v}_{\perp}^{\mathrm{T}} \end{pmatrix}=E_{m}\cdots E_{1}A,$$
where each $E_{i}$ is an elementary matrix adding a multiple of one row to another. Notice that the above operation corresponds to shearing the parallelotope so that the last edge is perpendicular to the base. We see that these operations do not change the determinant as $$\det\left(BB^{\mathrm{T}}\right)=\det\left(E_{m}\cdots E_{1}\left(AA^{\mathrm{T}}\right)E_{1}^{\mathrm{T}}\cdots E_{m}^{\mathrm{T}}\right)=\det\left(AA^{\mathrm{T}}\right).$$
Through block-multiplication, we get $BB^{\mathrm{T}}$ as $$BB^{\mathrm{T}}=\begin{pmatrix}A_{m}\\ \mathbf{v}_{\perp}^{\mathrm{T}} \end{pmatrix}\begin{pmatrix}A_{m}^{\mathrm{T}} & \mathbf{v}_{\perp}\end{pmatrix}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & A_{m}\mathbf{v}_{\perp}\\ \mathbf{v}_{\perp}^{\mathrm{T}}A_{m}^{\mathrm{T}} & \mathbf{v}_{\perp}^{\mathrm{T}}\mathbf{v}_{\perp} \end{pmatrix}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & A_{m}\mathbf{v}_{\perp}\\ \left(A_{m}\mathbf{v}_{\perp}\right)^{\mathrm{T}} & \|\mathbf{v}_{\perp}\|^{2} \end{pmatrix}.$$
Now notice that $$A_{m}\mathbf{v}_{\perp}=\begin{pmatrix}\mathbf{v}_{1}^{\mathrm{T}}\\ \vdots\\ \mathbf{v}_{m}^{\mathrm{T}} \end{pmatrix}\mathbf{v}_{\perp}=\begin{pmatrix}\mathbf{v}_{1}\cdot\mathbf{v}_{\perp}\\ \vdots\\ \mathbf{v}_{m}\cdot\mathbf{v}_{\perp} \end{pmatrix}=\mathbf{0}.$$ Therefore we have $$BB^{\mathrm{T}}=\begin{pmatrix}A_{m}A_{m}^{\mathrm{T}} & \mathbf{0}\\ \mathbf{0}^{\mathrm{T}} & \|\mathbf{v}_{\perp}\|^{2} \end{pmatrix}$$
Taking the determinant, we therefore have $$\det\left(BB^{\mathrm{T}}\right)=\|\mathbf{v}_{\perp}\|^{2}\det\left(A_{m}A_{m}^{\mathrm{T}}\right).$$
By definition, $\|\mathbf{v}_{\perp}\|$ is the height of the parallelotope, and by the induction hypothesis, $\det\left(A_{m}A_{m}^{\mathrm{T}}\right)$ is the square of the base. Therefore the result follows. $\square$
As a corollary, we can easily verify that the volume is independent of which $n-1$ vectors you choose to define your base, as it should be.
Corollary: The volume of a parallelotope is invariant of which base and height you choose.
Proof: Choosing a different base and height corresponds to permuting the rows of the matrix $A$. This does not change the magnitude of the resulting determinant. $\square$
Finally, note that the singular values are the absolute values of $A$ are precisely the square roots of the eigenvalues of $AA^\mathrm{T}$. Since the product of the eigenvalues is the determinant, it follows that the product of the singular values of $A$ gives the square root of $\det(AA^\mathrm{T})$, which is precisely the volume.
Best Answer
If you know that the scalar triple product is equal to the determinant of a matrix whose rows are the components of the vectors, and if you recall the effects of operations on the rows of a matrix, then you can show that swapping any two of the vectors $A,B,C$ in the scalar triple product $(A \times B) \cdot C$ will swap the corresponding rows of the matrix and therefore will flip the sign of the determinant but will not change the magnitude of the determinant. Hence the interchange of any two vectors (which could be $B$ and $C$ or could be $A$ and $C$, not just $A$ and $B$) will likewise flip the sign of the scalar triple product but will not change its magnitude.
Any reordering of the three vectors $A$, $B$, and $C$ can be accomplished by either one or two interchanges of two vectors. For example, to get from $(A,B,C)$ to $(B,C,A)$, swap the first two vectors, then the last two. Hence of the six possible ways to order the three vectors $A$, $B$, and $C$, three orderings will give you positive scalar triple products and three will give you negative scalar triple products, but all scalar triple products will have the same magnitude.