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The Jordan content of the parallelepiped spanned by the columns of a square real matrix is the absolute value of its determinant

Statement

Let $n\ge1$ . For column vectors $v_1,\ldots,v_n\in\mathbb R^n$ , let $A=[v_1\ \cdots\ v_n]$ and $P(v_1,\ldots,v_n):=\left\{\sum_{j=1}^n t_jv_j:0\le t_j\le1\right\}.$ Then $P(v_1,\ldots,v_n)$ is Jordan measurable and $\operatorname{cont}(P(v_1,\ldots,v_n))=|\det A|.$ This includes the singular case, when the content is zero.

Facts & Assumptions

Given: The spanning vectors, their column matrix $A$ , and the unit cube $Q=[0,1]^n$ .

[L1]

A linear map scales the content of every bounded Jordan set by the absolute determinant of its matrix (A linear endomorphism of $\mathbb R^n$ sends bounded Jordan sets to bounded Jordan sets and scales their content by the absolute determinant).

[L2]

The unit cube is a rectangle of volume $1$ (Axis-parallel rectangles in $\mathbb{R}^m$ and their volume).

Proof

technique · direct

1.1

The matrix map $T(x)=Ax$ sends $Q$ exactly onto $P(v_1,\ldots,v_n)$ .

L1given

2.1

Apply [L1] to $Q$ and use [L2] to obtain the stated content formula: $\operatorname{cont}(P)=|\det A|\operatorname{cont}(Q)=|\det A|.$ If the columns are dependent, [L1] simultaneously supplies Jordan measurability and the zero-content conclusion.

L1L2step 1.1∎

Depends on

Used by

The content of a concrete three-dimensional parallelepiped computed from its spanning matrix Example

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 121 results over 19 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

A. Leibman, Multidimensional Real Analysis, Lemma 5.5.4 (standard reference, not scraped)