Construct the determinant from permutations

Try

Take the matrix with rows (1,2,0), (2,5,1), (0,1,3). Choose one entry in each column, using each row exactly once. Choosing all diagonal entries gives product 1·5·3=15. Choosing rows 1,3,2 for columns 1,2,3 gives product 1·1·1=1. These selections will enter with different signs.

Why impose both restrictions? If a selected row could repeat, we would be multiplying data from only part of the coordinate system. A permutation specifies a complete assignment: σ(j) is the row selected in column j. A 3×3 matrix has six such assignments. Draw one line through each selected entry in your own matrix and check that no two selections share a row or column.

Predict

An inversion in a permutation list is a pair of positions i before j whose values are in decreasing order. The list 231 has two inversions, namely 2 before 1 and 3 before 1. Give it sign +1. The list 321 has three inversions and sign −1. In general the sign is (−1) raised to the inversion count.

Predict whether swapping two equal columns should change a quantity intended to measure independent directions. They already describe a collapsed configuration, so the desired value is zero. We will prove this vanishing directly. In characteristic two, where −1=1, the weaker equation “swapping negates the value” cannot force zero by itself. That distinction matters even though ordinary real-number examples conceal it.

Experiment

Use the explorer to inspect the six selections for the fixed matrix. Before changing a row, compute the predicted total from the table below. Then compare a row swap, adding a multiple of another row, scaling a row, and duplicating a row. The displayed examples test the definition; general row-operation laws will be proved separately rather than inferred from five successful experiments.

Where do the six signed terms come from?

Pick exactly one entry from every row and every column. The column sequence records the chosen column in rows 1, 2 and 3. Its inversion count determines the sign.

Matrix changeOriginal matrixSwap rows 1 and 2Row 2 minus twice row 1Double row 1Replace row 3 with row 1 Inspect column sequence1, 2, 31, 3, 22, 1, 32, 3, 13, 1, 23, 2, 1 Reset

No row operation: compare all six signed terms. Predict the total before comparing the rows below.

Matrix entries; ● marks the three selected entries

Row	Col 1	Col 2	Col 3
1	1 ●	2	0
2	2	5 ●	1
3	0	1	3 ●

Selected entries: 1 × 5 × 3 = 15. There are 0 inversions, so the sign is +1 and the signed term is 15.

An inversion is a pair of row positions i<j whose selected column numbers satisfy σ(i)>σ(j). For example, (2,3,1) has two inversions, (2,1) and (3,1), so its sign is positive.

Every allowed column sequence, including zero products

Columns	Inversions	Sign	Entry product	Signed term
1, 2, 3 ●	0	+1	1 × 5 × 3 = 15	15
1, 3, 2	1	−1	1 × 1 × 1 = 1	-1
2, 1, 3	1	−1	2 × 2 × 3 = 12	-12
2, 3, 1	2	+1	2 × 1 × 0 = 0	0
3, 1, 2	2	+1	0 × 2 × 1 = 0	0
3, 2, 1	3	−1	0 × 5 × 0 = 0	0

Determinant = (15) + (-1) + (-12) + (0) + (0) + (0) = 2.

Compare the original and row-swapped totals. Then inspect the repeated-row case and pair terms by swapping their selected entries in rows 1 and 3. Equal products have opposite signs. A finite table verifies this example; the lesson proves the general rule. This bounded integer explorer uses exact arithmetic and temporary controls; it does not record completion.

Convention bridge. The explorer lists the selected columns in row order, whereas our written table lists the selected rows in column order. A selection represented by σ in our convention is represented by σ⁻¹ in the explorer. The selected entries are identical. The sign-composition proof below gives sgn(σ⁻¹)=sgn(σ), so both sums agree. For the symmetric starting matrix some displayed products coincide, but on a changed matrix do not identify the two permutation lists without taking the inverse.

Selected rows by column	Inversions	Selected product	Signed contribution
123	0	1·5·3=15	15
132	1	1·1·1=1	−1
213	1	2·2·3=12	−12
231	2	2·1·0=0	0
312	2	0·2·1=0	0
321	3	0·5·0=0	0

The total is 15−1−12=2. The cancellation is part of the calculation: multiplying the diagonal alone would give the wrong answer. For an n×n matrix A over a field K, define

D(A)=Σ over permutations σ of sgn(σ) times ∏ from j=1 to n of A at row σ(j), column j.

Here the signs ±1 are interpreted as field elements. For n=0 there is one permutation of the empty index set and its product is 1, so the determinant of the empty matrix is 1. This convention will make later empty-minor and product formulas consistent.

Understand

First justify the signs independently of determinants. Work with the formal polynomial V(x₁,…,xₙ)=∏ over i before j of (xⱼ−xᵢ), with integer coefficients. A formal polynomial is an algebraic expression with indeterminates; we are not evaluating all its variables at some coincident numerical values. On replacing its variables by xσ(1),…,xσ(n), each unordered pair still occurs once. Exactly the inversion pairs reverse the order of a difference. Reordering the finitely many factors therefore gives

V(xσ(1),…,xσ(n))=sgn(σ)V(x₁,…,xₙ).

Apply two substitutions successively. If the second list substitution uses τ after σ, the selected variable at position j is xσ(τ(j)). The scalar multiplier is sgn(τ)sgn(σ), while direct substitution gives sgn(σ∘τ). The polynomial V is nonzero over the integers, so the two integer multipliers must agree. Thus permutation signs multiply under composition. This argument does not use determinants or assume that the target field has characteristic different from two.

Since σ∘σ⁻¹ is the identity, the composition law gives sgn(σ⁻¹)=sgn(σ). Reindexing the permutation sum by τ=σ⁻¹ converts ∏ over j of A at row σ(j), column j into ∏ over i of A at row i, column τ(i). This proves the row-selection and column-selection formulas agree. Applying the same reindexing to the transposed matrix, whose row i, column j entry is A at row j, column i, also proves D(Aᵀ)=D(A). This is transpose invariance derived from the formula, not an extra assumption.

An adjacent swap reverses parity: swapping neighboring entries changes their mutual inversion and leaves the total contribution involving every other entry unchanged. Any transposition of positions r before s is a composition of 2(s−r)−1 adjacent swaps, an odd number; hence its sign is −1. Sorting a list by swapping adjacent inverted pairs reduces its inversion count by one each time and eventually reaches the increasing list. These facts connect the inversion definition with changes of basis-list order.

Now prove column multilinearity. Every product in D uses exactly one entry from a chosen column. Replace that column by αu+βv. Distributing its one selected entry splits the product into α times its u version and β times its v version. Summing proves linearity in that column, with every other column held fixed. The same argument applies at each position.

Next prove alternation, meaning D is zero whenever two columns are equal. Suppose columns r and s agree. Pair each permutation σ with σ∘(r s), which exchanges their selected rows. The paired permutations are different and the pairing is its own inverse. The selected products agree, but the signs are opposite. Each pair therefore contributes zero. This is cancellation of pairs in the field, not division by 2. In characteristic two the signs become equal, yet a term plus itself is zero, so the proof still works.

Normalization is immediate: for the identity matrix I only the identity permutation selects entirely nonzero entries. Its contribution is 1. Every other permutation selects a zero somewhere, giving D(I)=1.

These three properties uniquely determine D. To prove that, let F be any alternating multilinear function of n columns with F(e₁,…,eₙ)=1. We first derive its swap rule: put u+v into two positions. Alternation makes the result zero, and multilinearity expands it into four terms. The terms containing u twice or v twice vanish. The other two sum to zero, so exchanging u and v negates F. This proof remains valid in characteristic two.

Expand each column as a linear combination of the standard basis. Multilinearity expands F into a finite sum indexed by all lists of selected basis indices. A list with repeated indices gives equal basis columns and hence value zero. The remaining lists are exactly the permutations. By the adjacent-swap rule and inversion sorting, the value on eσ(1),…,eσ(n) is sgn(σ) times F(e₁,…,eₙ), hence sgn(σ). The expansion is precisely the permutation formula. This establishes uniqueness on every input, rather than just agreement on a few matrices. Without imposing normalization, exactly the same expansion gives F(A)=F(e₁,…,eₙ)D(A). This includes the case F(e₁,…,eₙ)=0 and requires no division by that value. For n=0, there is only the empty column list and normalization fixes its value as 1; no column conditions remain.

The dependence test now follows. A zero column gives zero by linearity. If a list of columns is dependent, a nonzero relation coefficient can be inverted in the field to express its column through the others. Expand in that position: each resulting term repeats one of the other columns, so every term vanishes. We have proved dependence implies zero determinant. The converse is a separate theorem and must not be inferred merely by reversing this implication.

Apply

Build the six-term table for the changed matrix with rows (2,0,1), (1,3,0), (0,2,1). Only selections 123 and 231 contribute: their products are 6 and 2, both with positive sign. Thus D=8 over the rationals. Reduce the matrix over F₂ and F₃ and check that the values become 0 and 2 respectively. The formula respects reduction because it uses only finite sums, products and integer signs.

To test the characteristic-two warning, take H(u,v)=u₁v₁ on F₂². It is bilinear and H(v,u)=−H(u,v), since multiplication commutes and minus equals plus. But H(e₁,e₁)=1, so it is not alternating. Adding H to the two-dimensional determinant preserves normalization because H(e₁,e₂)=0, and preserves the swap equation, but destroys equal-column vanishing. This changed case shows exactly why the uniqueness theorem assumes alternation, not only antisymmetry.

Continue

Record the selected-row convention, the integer sign argument and the field in every calculation. A six-term experiment provides a check, while the multilinear expansion and paired cancellation justify arbitrary dimensions. The number of terms grows as n!, so the definition is a proof tool rather than a promise of an efficient large-matrix algorithm.

Return to permutations and their composition if the index manipulations are unfamiliar. The following determinant arguments can now derive row operations, cofactors and multiplicativity from a proved characterization instead of using those identities to justify the characterization itself.