The only function that behaves like a volume

In two dimensions the determinant is an area: the unit square goes to a parallelogram, and $ad - bc$ is how much room that parallelogram takes up. One dimension up the same sentence works with volume in place of area, and the formula that computes it has six terms rather than two.

At $n$ by $n$ the sum has $n!$ terms, which is $24$ at four, $120$ at five, and out of hand immediately. That growth is a hint that the formula is the wrong place to start, and the right place is to ask what properties a “volume of the image” ought to have and then to discover that they leave no choice.

Three conditions #

Write $\det(v_1, \dots, v_n)$ for a function of the $n$ columns, and ask for three things.

Multilinear. Scaling one column scales the answer by the same factor, and splitting one column into a sum splits the answer into a sum. Geometrically this is what a volume does: stretch one edge of a box by three and the box holds three times as much, whatever the other edges are doing.

Alternating. If two columns are equal the answer is zero. Geometrically: a box with two edges lying along the same direction is flat, and a flat box has no volume.

Normalised. The identity gives one. The unit cube has volume one, which is what fixes the units.

There is exactly one function with all three, in every dimension. That is the theorem this rung is about, and its proof is a computation short enough to do here.

Why the conditions leave no choice #

Write each column in the standard basis: $v_j = \sum_i a_{ij} e_i$. Multilinearity says the whole determinant expands into a sum over all ways of picking one basis vector from each column — $n^n$ terms, each of the form

$$a_{i_1 1} a_{i_2 2} \cdots a_{i_n n} \cdot \det(e_{i_1}, e_{i_2}, \dots, e_{i_n}).$$

Now the alternating condition does its work. Any term in which two of the $i$’s agree has a repeated column, so its determinant factor is zero and the term vanishes. What survives is exactly the terms where the $i$’s are all different — that is, the terms indexed by permutations, and there are $n!$ of those rather than $n^n$.

Each surviving factor $\det(e_{σ(1)}, \dots, e_{σ(n)})$ is the identity’s columns in a shuffled order. Swapping two columns of any alternating multilinear function turns its sign over — expand $\det(u + v, u + v)$, which is zero, into four terms and two of them vanish, leaving $\det(u,v) = -\det(v,u)$ — so the factor is $+1$ or $-1$ according to whether $σ$ is built from an even or an odd number of swaps. That the parity of a permutation is well defined at all is a fact worth its own essay, and it is exactly the fact the determinant needs.

Putting the pieces together gives the formula the figures compute:

$$\det A = \sum_{σ} \operatorname{sign}(σ) \, a_{σ(1) 1} a_{σ(2) 2} \cdots a_{σ(n) n}.$$

The derivation is the whole content. The formula is not a definition anybody would guess; it is what three geometric conditions force, and every one of its features — the number of terms, the signs, the products running once through each row and each column — arrived from one of the three.

Reading the formula as a rule about rows and columns #

The permutation sum has a shape worth saying in words, because it is the shape every hand computation exploits.

Each term picks exactly one entry from each row and exactly one from each column, multiplies them, and attaches a sign. So a determinant is a sum over all the ways of placing $n$ non-attacking rooks on the board, weighted by the entries they land on. Two rooks in one row would mean two factors from the same row, and the formula never does that; the ways of placing them are precisely the permutations.

Reading it that way explains several familiar facts at once. A matrix with a zero row has every term containing a factor from that row, so every term is zero. A triangular matrix has only one non-zero term — the diagonal, since any other placement must pick a zero above or below it — so its determinant is the product of the diagonal entries. And a matrix of ones and zeroes has a determinant that is a count with signs, which is where determinants that count objects come from and is the subject of a later rung.

It also makes the size of the computation vivid. Placing rooks on an eleven-by-eleven board can be done in nearly forty million ways, and evaluating the formula directly means visiting every one. Nobody does. Elimination reaches the same number with about a thousand operations, and the reason it is allowed is the row-operation rule above — which is itself a two-line consequence of the three conditions.

Signed, and what the sign is recording #

The value can be negative, and the negative is not an accident of the formula. Swapping two columns reverses the sign, and swapping two columns is what a mirror does.

So “volume” in the statement above is signed volume, and the sign is the orientation: whether the image of a right-handed frame is right-handed. This collection has an essay on that property in its own right — orientation is a sign — where the determinant is taken around a loop on a surface and the sign is the whole of what survives. Here it is enough to notice that the alternating condition is what put the sign there, and that a function of the columns which ignored their order could not have been a volume, because a reflection would then be invisible to it.

What follows for free #

Uniqueness is a lever. Any function of the columns with the three properties is the determinant, so proving something about the determinant reduces to checking three conditions on a candidate.

The product rule. Fix a matrix $B$ and consider $A \mapsto \det(AB)/\det(B)$ as a function of $A$’s columns. It is multilinear and alternating, because $A \mapsto AB$ is linear in the columns and carries repeats to repeats, and it gives one when $A$ is the identity. So it is the determinant, and $\det(AB) = \det(A)\det(B)$ falls out with no computation at all.

Transposes. $\det(A^{\mathsf T}) = \det(A)$, because the permutation sum is unchanged when every term is re-read with rows and columns exchanged — a permutation and its inverse have the same sign. So every statement about columns above is equally a statement about rows, which is why row operations may be used to compute a determinant even though the conditions were stated about columns.

Row operations. Adding a multiple of one column to another leaves the determinant alone, since multilinearity splits it into the original plus a multiple of a determinant with a repeated column, and the second term is zero. That single line is what makes elimination a way of evaluating a determinant, and it is why the $n!$ terms of the formula never have to be visited.

A matrix is not the map #

One consequence of the uniqueness deserves separating out, because it is the reason the determinant appears in places with no matrix in sight.

A linear map between spaces has no entries until a basis is chosen, and different bases give different matrices. The determinant of those matrices is nevertheless the same number, because changing basis replaces $A$ by $P^{-1} A P$ and the product rule then gives $\det(P)^{-1} \det(A) \det(P)$, which is $\det(A)$. So the determinant belongs to the map rather than to any of its matrices — the same map in a better basis has the same determinant, however much friendlier its entries look.

That invariance is what licenses phrases like “the determinant of a rotation” or “the volume scale factor of a map”, and it is not automatic: most functions of the entries change when the basis does. The trace is the other famous survivor, and between them the two are the only ones in two dimensions — everything else invariant under change of basis is built from those, which is what the characteristic polynomial records.

The vanishing case #

A determinant of zero says the image is flat: the columns fail to span, so some direction is crushed. In two dimensions that is a parallelogram collapsed to a segment; in three, a parallelepiped collapsed to a plane or a line.

That is the same statement as “the columns are dependent”, and the same statement again as “the map cannot be undone”, and the equivalence of those three is most of a first course in the subject. What the volume reading adds is a reason rather than a proof: a map that loses a dimension must send every box to something flat, and flat things have no volume, so a single number computed from the entries can detect it.

The reading also says why the determinant is a blunt instrument. It reports one number, so it can say whether a dimension was lost and not how many; a matrix that collapses three-dimensional space to a line has the same determinant as one that collapses it to a plane. It also says nothing about which direction was lost, which is the kernel and needs the matrix rather than the number. What a map throws away is a separate measurement, and needs the rank rather than the determinant.

What the conditions look like when one is dropped #

Uniqueness invites the obvious experiment: keep two conditions and drop the third, and see what is left.

Drop normalisation and any constant multiple of the determinant survives, which is why the condition is bookkeeping rather than content — it fixes a unit and nothing else. Every theorem above is unaffected up to that constant.

Drop alternating and the field opens up enormously. The permanent — the same sum with every sign set to plus — is multilinear and normalised and not alternating, and it is a genuinely different function with genuinely different behaviour; the last rung of this ladder is about how different. So the alternating condition is where the geometry lives: it is the one that makes a repeated direction cost everything, and it is the one that produces the signs.

Drop multilinearity and there is nothing left to say, because the whole derivation was an expansion licensed by it. A function of the columns that does not respect scaling is not measuring a volume in any sense, and the $n^n$ terms never appear to be cut down to $n!$.

The experiment is worth doing because it locates the content. Two of the three conditions are almost definitions of what “measuring a box” means; the third — alternating — is the one that turns a measurement into an algebraic object with signs in it, and every consequence in the section above traces back to that one.

In as many dimensions as wanted #

Nothing above mentioned two or three except in the pictures. The conditions make sense for any $n$, the derivation is the same paragraph, and the resulting formula has $n!$ terms.

The three-dimensional case has one extra piece of luck that does not generalise, and it is worth naming so that nobody leans on it. In three dimensions the determinant of three vectors equals the scalar triple product $u \cdot (v \times w)$, and the cross product is a construction peculiar to three dimensions — it exists because the space of directions perpendicular to two given ones is one-dimensional exactly when $n$ is three. So the triple product is a third computation of the same number here and has no counterpart in dimension four, where the determinant survives intact and the cross product does not.

Higher dimensions are where the geometric reading earns its place, because there is no picture. The claim “the determinant is the volume scale factor” continues to hold in dimension eleven, where volume means the eleven-dimensional measure and no drawing exists; the three conditions still pin the function down; and the sign still records whether a frame was flipped. A reader who learned the two-by-two formula and nothing else has no way to see that. A reader who learned the conditions has the general case immediately, and gets the formula back whenever it is wanted.

What the pictures cannot show #

Every figure here is two- or three-dimensional, and the theorem is about all dimensions. Worse, the two pictures that carry the argument — the parallelepiped and the four parallelograms — are the two dimensions where the reader’s intuition about volume is already reliable, so they can illustrate the conditions and cannot establish that the conditions are the right ones for the general case.

What the figures do establish is agreement between computations. The parallelepiped’s volume is computed by the permutation sum, by the cofactor expansion and by the triple product, and the three are asserted to match; the parallelograms have their areas measured from their own drawn corners by the shoelace formula and set against what each condition predicts. That is the honest content of a picture in this subject: it checks that two routes to a number meet, and the routes it checks are the ones a reader could otherwise only take on trust.

The one thing no figure attempts is the sign in dimension four and up. Orientation there is a real property with real consequences, and it corresponds to nothing anybody can look at, which is exactly why the algebraic definition — the parity of a permutation — has to carry it.

Where the ladder goes next #

The determinant has been treated as a property of a map so far. It is also a property of the lattice of points a pair of vectors generates, and read that way it answers a different question: how much room a discrete set of points leaves. That reading turns the number into a statement about how large a region has to be before it must contain a lattice point, and the theorem there is Minkowski’s.