# What is the 3-volume of the 3-parallelepiped defined by $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}$?

We have $\left\{\vec{v_1},\vec{v_2},\vec{v_3}\right\}=\left\{\begin{bmatrix}1\\0\\0\\0\end{bmatrix},\begin{bmatrix}1\\1\\1\\1\end{bmatrix},\begin{bmatrix}1\\2\\3\\4\end{bmatrix}\right\}$

QR-factorization gives $\text{det }(A^TA)=\text{det }(R^TR)=\text{det }(R^2)$ such that $\sqrt{\text{det }(A^TA)}$ yields the volume of our desired parallelepiped. Letting $A=\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}$,

$\sqrt{\text{det }(A^TA)}=\sqrt{\begin{pmatrix}1&0&0&0\\1&1&1&1\\1&2&3&4\end{pmatrix}\begin{pmatrix}1&1&1\\0&1&2\\0&1&3\\0&1&4\end{pmatrix}}=\sqrt{\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}}$

Expanding down the first column of $\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}$ using Laplace yields $\text{det }\begin{pmatrix}1&1&1\\1&4&10\\1&10&30\end{pmatrix}=6$, such that $\sqrt{\text{det }(A^TA)}=\sqrt{6}$, which is the volume of the desired parallelepiped.

Is there a quicker and more energy-efficient algorithm to solve this? I know that for $3\times3$ matrices we can use $\text{det }A=\vec{u}\cdot(\vec{v}\times\vec{w})$, given $\left\{\vec{u},\vec{v},\vec{w}\right\}$ span the columns of $A$. Is there an analog here?

• is not the volume zero?
– abel
Apr 7 '15 at 22:47
• The 4-volume, yes. But they are linearly independent, therefore they span a 3-volume. Apr 7 '15 at 22:50


Naturally, there's "good fortune" in this example: After "column reduction", the first column was a standard basis vector orthogonal to the remaining columns.

Alternatively, there's a "cross product" for $(n - 1)$ vectors in $\Reals^{n}$ whose definition generalizes the usual cross product in $\Reals^{3}$: Assemble your vectors into an $(n - 1) \times n$ matrix, prepend the standard basis vectors into the first row (obtaining a square matrix) and take the determinant. The magnitude of the resulting vector is the volume of the parallelipiped of the original set of vectors. (See below.) Here you'd have $$\left\lvert\begin{array}{@{}cccc@{}} \Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\ 1 & 0 & 0 & 0 \\ 1 & 1 & 1 & 1 \\ 1 & 2 & 3 & 4 \\ \end{array}\right\rvert = \left\lvert\begin{array}{@{}cccc@{}} \Basis_{1} & \Basis_{2} & \Basis_{3} & \Basis_{4} \\ 1 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 \\ 0 & 0 & 1 & 2 \\ \end{array}\right\rvert = \Basis_{2} - 2\Basis_{3} + \Basis_{4},\qquad \|\Basis_{2} - 2\Basis_{3} + \Basis_{4}\| = \sqrt{6}.$$ (I leave to you the assessment of whether this is more energy-efficient for your purposes, though again I think there's a favorable comparison in this example.)

This "generalized cross product" has the advertised property for reasons analogous to the situation with the ordinary cross product:

• If $(v_{j})_{j=1}^{n-1}$ and $w$ are vectors in $\Reals^{n}$ and $v_{1} \times \dots \times v_{n-1}$ denotes the "cross product" described above, then $$\left\lvert\begin{array}{@{}c@{}} w \\ v_{1} \\ \vdots \\ v_{n-1} \\ \end{array}\right\rvert = (v_{1} \times \dots \times v_{n-1}) \cdot w. \tag{2}$$

• By (2), the cross product $v_{1} \times \dots \times v_{n-1}$ is orthogonal to each $v_{j}$, and therefore orthogonal to the parallelipiped spanned by the $(v_{j})$.

• The preceding item and the fact that the determinant in (2) is the $n$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ and $w$ implies that the $(n - 1)$-dimensional volume of the parallelipiped spanned by the $(v_{j})$ is the magnitude of their cross product.

• @BenjaminLoya: Yes. expanding about the second row. :) (As you probably noticed, there's an overall sign error in the vector, though the magnitude is correct.) Apr 8 '15 at 10:37