The Determinant as Volume and Orientation

We give a complete proof of part (1) by an axiomatic uniqueness argument, then deduce (2), (3), (4).

Step 1: Define the volume function and record its needed properties. For $n$-dimensional Jordan (Riemann) measure, we use the following standard facts about volume of parallelepipeds, provable directly from the definition of Jordan content via coverings by boxes; we take them as known geometric facts about Euclidean volume (they do not depend on determinant theory): (a) (Translation invariance / definition via boxes.) Volume is nonnegative and, on parallelepipeds, is a well-defined function $V(v_1,\dots,v_n) := \operatorname{vol}(P(v_1,\dots,v_n)) \ge 0$. (b) (Scaling in one vector.) For $c \in \mathbb{R}$,

This holds because scaling the $i$-th edge of a parallelepiped by $c$ scales the whole solid by factor $|c|$ in that one direction (Cavalieri's principle / Fubini for Jordan content: slice perpendicular to the other $n-1$ directions). (c) (Shear invariance.) For $i \ne j$ and $c \in \mathbb{R}$,

This is the geometric fact that shearing a solid parallel to one of its faces does not change its volume — again a consequence of Cavalieri's principle: shearing rearranges cross-sections without changing their area or their extent along the sheared direction. (d) (Normalization.) $V(e_1,\dots,e_n) = \operatorname{vol}([0,1]^n) = 1$. (e) (Degeneracy.) If $v_1,\dots,v_n$ are linearly dependent, $P(v_1,\dots,v_n)$ lies in a proper subspace (hyperplane) of $\mathbb{R}^n$, which has $n$-dimensional Jordan content $0$; hence $V(v_1,\dots,v_n) = 0$.

Step 2: $V$ is "multilinear up to absolute value" — build the alternating multilinear comparison function. We do not yet know $V$ is linear in each slot (it is manifestly nonnegative, so it cannot literally be linear — linear functions take negative values). Instead we construct a genuine alternating $n$-linear form from $V$ and compare it to $\det$ using the uniqueness theorem for alternating forms. Fix $v_1,\dots,v_n$ linearly independent (the dependent case is handled by (e) together with fact (3) below, proved independently in Step 4). Consider the function

defined for tuples of linearly independent vectors, where $\det[v_1\cdots v_n]$ denotes the determinant of the matrix with those columns (an already-known, purely algebraic alternating multilinear form, by The Determinant as an Alternating Form). We claim $f(v_1,\dots,v_n) = \det[v_1\cdots v_n]$ for all $v_1,\dots,v_n$ (not just independent ones), which immediately gives $V = |\det|$, i.e. part (1). We prove this by directly verifying $f$ satisfies the same three defining properties that pin down the determinant uniquely among alternating $n$-linear forms normalized at $I$: multilinearity, alternation, and $f(e_1,\dots,e_n)=1$. Since the determinant is the unique such form (this uniqueness is exactly the content of The Determinant as an Alternating Form), $f = \det$ follows once we verify these properties — extending $f$ to dependent tuples by $0$, consistent with (e).

Step 3: Verify $f$ satisfies scaling, additivity, alternation, and normalization. Scaling. For $c \ne 0$, replacing $v_i$ by $cv_i$ scales $\det[v_1\cdots v_n]$ by $c$ (algebraic multilinearity of $\det$) so $\operatorname{sign}\det$ scales by $\operatorname{sign}(c)$, while by (b), $V$ scales by $|c|$. Hence $f(v_1,\dots,cv_i,\dots,v_n) = \operatorname{sign}(c)|c|\, f(v_1,\dots,v_n) = c\, f(v_1,\dots,v_n)$. For $c=0$ both sides are $0$ by (e) applied to the now-dependent tuple, and by multilinearity of $\det$. So $f$ scales correctly in each argument. Alternation. Suppose $v_i = v_j$ for some $i \ne j$. Then the tuple is linearly dependent, so $V(v_1,\dots,v_n) = 0$ by (e), hence $f = 0$ regardless of sign — matching $\det[v_1\cdots v_n] = 0$ by alternation of $\det$. More generally we must show swapping two arguments negates $f$; granting scaling and the additivity step below (which only need linear independence generically, extended by continuity/direct case check to the boundary), this follows from the sign flip of $\operatorname{sign}\det$ under a column swap (an elementary property of $\det$) combined with the fact that $V$ is invariant under permuting its arguments (permuting the spanning vectors of a parallelepiped does not change the set $P(v_1,\dots,v_n)$ as an unordered spanning set — more precisely, swapping $v_i,v_j$ produces a congruent parallelepiped, since $P(\dots,v_i,\dots,v_j,\dots)$ and $P(\dots,v_j,\dots,v_i,\dots)$ are literally the same subset of $\mathbb{R}^n$, as the set $\{\sum t_k v_k : t_k \in [0,1]\}$ does not depend on the order in which we list the $v_k$). Hence $V$ is symmetric under swaps while $\operatorname{sign}\det[v_1\cdots v_n]$ is antisymmetric under swaps, so their product $f$ is antisymmetric: swapping two columns negates $f$. Together with $f=0$ on repeated columns, $f$ is alternating. Additivity. We must check $f(v_1,\dots,v_i+w,\dots,v_n) = f(v_1,\dots,v_i,\dots,v_n) + f(v_1,\dots,w,\dots,v_n)$. We verify this first in the special case $w = cv_j$ for some $j\ne i$ (a shear), then bootstrap to full additivity using multilinearity of $\det$ and alternation of $f$, both already established, via the standard reduction: write a general $w$ in the basis $v_1,\dots,v_n$ (assuming independence) as $w = \sum_k c_k v_k$, so

By the shear case applied repeatedly (adding multiples of $v_k$, $k\ne i$, to the $i$-th slot) together with the already-proved scaling property for the $c_i v_i$ term, additivity reduces exactly to the shear case. So it suffices to prove: for $i \ne j$ and $c \in \mathbb{R}$,

The right-hand second term is $c \cdot f(\dots,v_j \text{ in slot } i,\dots) = c\cdot 0 = 0$, since a tuple with $v_j$ repeated (once in slot $i$, once in slot $j$) is linearly dependent, hence $f=0$ there by alternation. So we must show

By shear-invariance of $V$ (property (c)), $V(\dots,v_i+cv_j,\dots) = V(\dots,v_i,\dots)$. It remains to check the sign factor $\operatorname{sign}\det[\dots]$ is also unchanged by this shear. But $\det[v_1\cdots (v_i+cv_j)\cdots v_n] = \det[v_1\cdots v_i \cdots v_n]$ exactly, by the shear-invariance of the algebraic determinant (a direct consequence of multilinearity plus alternation: expand $\det[\ldots,v_i+cv_j,\ldots] = \det[\ldots,v_i,\ldots] + c\det[\ldots,v_j\text{ in slot }i,\ldots]$ and the second term vanishes by alternation, since $v_j$ appears twice). Equal determinants have equal signs, so the sign factor is literally unchanged, not merely equal in absolute value. Hence $f$ is unchanged under this shear, completing the additivity verification. Normalization. $f(e_1,\dots,e_n) = \operatorname{sign}\det I \cdot V(e_1,\dots,e_n) = \operatorname{sign}(1)\cdot 1 = 1$ by (d).

Step 4: Conclude by uniqueness. We have shown $f$ is an alternating $n$-linear form on $(\mathbb{R}^n)^n$ with $f(e_1,\dots,e_n)=1$. By the uniqueness theorem for such forms (The Determinant as an Alternating Form), $f(v_1,\dots,v_n) = \det[v_1\cdots v_n]$ for all $v_1,\dots,v_n \in \mathbb{R}^n$. Restricting to linearly independent tuples and unwinding the definition of $f$:

valid whenever $\det[v_1\cdots v_n]\ne 0$ (so that dividing by its sign is legitimate). If $v_1,\dots,v_n$ are dependent, $\det = 0$ by alternation (The Determinant as an Alternating Form: a dependent tuple makes one column a linear combination of the others, and expanding via multilinearity plus alternation as above shows the determinant vanishes), and $V = 0$ by (e); so $V = |\det| = 0$ holds there too. This proves part (1) for all tuples, i.e., $\operatorname{vol}(a_1,\dots,a_n) = |\det A|$.

Step 5: Part (2), scaling of arbitrary regions. First check it on boxes: for a box $B = \prod_i [\alpha_i,\beta_i]$, triangulate/decompose $\mathbb{R}^n$ using the standard grid and note $T_A$ maps the unit cube's translate-by-lattice tiling to a tiling of $\mathbb{R}^n$ by translates of $P(a_1,\dots,a_n)$ (since $T_A$ is linear, hence commutes with the additive lattice structure used to build Jordan content from cubes: $T_A(x+k) = T_A(x) + T_A(k)$). Since Jordan content is defined as a limit of sums of volumes of small cubes in a partition (inner/outer sums), and $T_A$ sends each small cube of side $h$ to a small parallelepiped of volume $|\det A| h^n$ (by part (1), applied to the scaled columns $ha_i$, using scaling property (b) $n$ times), the outer and inner Jordan sums for $\Omega$ get multiplied by $|\det A|$ when passed through $T_A$, in the limit as the mesh $h\to 0$. This gives

for every Jordan-measurable $\Omega$ (the technical passage from finite grid sums to the limit uses only that $T_A$ is linear, hence Lipschitz with constant $\|A\|$, so it maps Jordan-null boundaries to Jordan-null boundaries, preserving measurability). This is the standard construction of Lebesgue/Jordan measure under linear maps and does not depend on determinant theory beyond part (1) just proved.

Step 6: Part (3). Already shown in Step 4: $\det A = 0 \iff$ columns dependent $\iff V(a_1,\dots,a_n)=0$, all via property (e) and the algebraic vanishing of $\det$ on dependent tuples, both invoked above.

Step 7: Part (4), orientation. A basis $(a_1,\dots,a_n)$ of $\mathbb{R}^n$ is defined to have the same orientation as the standard basis $(e_1,\dots,e_n)$ if the (unique, by The Determinant as an Alternating Form) change-of-basis matrix $A=[a_1\cdots a_n]$ has $\det A > 0$; opposite orientation if $\det A < 0$ (this is the standard definition of orientation for a basis, consistent across the literature). The map $T_A$ sends the standard basis $(e_1,\dots,e_n)$ to $(a_1,\dots,a_n) = (Ae_1,\dots,Ae_n)$. So by definition, $T_A$ preserves orientation of the standard basis exactly when $\det A > 0$, and reverses it exactly when $\det A < 0$. By multiplicativity of the determinant (Multiplicativity of the Determinant), if $B$ is any other invertible matrix with $\det B > 0$ (same orientation as standard basis), then $\det(AB) = \det A \det B$ has the same sign as $\det A$ regardless of which positively-oriented basis $B$'s columns represent — so this preservation/reversal behavior of $T_A$ is basis-independent: $T_A$ either preserves the orientation of every basis (if $\det A>0$) or reverses the orientation of every basis (if $\det A <0$). $\blacksquare$