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Theorem

Multiplicativity of the determinant

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Statement

Let \( K \) be a field (everything below in fact works verbatim over any commutative ring with 1), let \( n \in \mathbb{N} \), and let \( A, B \in M_n(K) \) be square matrices of the same size \( n \times n \). Then \[ \det(AB) = \det(A)\,\det(B). \] Here \( \det : M_n(K) \to K \) is the determinant, characterised as the unique function that is \( n \)-linear and alternating in the columns and satisfies \( \det(I_n) = 1 \) (equivalently, given by the Leibniz formula \( \det B = \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{j=1}^{n} B_{\sigma(j),j} \)).

Why it matters

Multiplicativity is the single most-used structural property of the determinant. It says that \( \det \) converts matrix multiplication — a complicated, noncommutative operation — into ordinary multiplication of scalars. Geometrically, over \( \mathbb{R} \), it is the statement that when you compose two linear maps, their volume-scaling factors multiply: \( |\det A| \) measures how \( A \) scales \( n \)-dimensional volume, and doing \( B \) then \( A \) scales volume by \( |\det A|\,|\det B| \).

Almost every downstream use of determinants routes through this theorem: the invertibility criterion \( \det A \neq 0 \), the similarity invariance \( \det(P^{-1}AP) = \det A \) (hence the determinant of a linear operator is well defined), the fact that \( \det : GL_n(K) \to K^\times \) is a group homomorphism with kernel \( SL_n(K) \), and the chain rule for Jacobians in the change-of-variables formula of multivariable calculus.

Hypotheses
Both matrices are square, of the same size \( n \times n \).For rectangular matrices the statement does not even parse: take \( A = (1 \;\; 0) \) (size \( 1 \times 2 \)) and \( B = A^{T} \) (size \( 2 \times 1 \)). Then \( AB = (1) \) has determinant \( 1 \) while \( BA = \left(\begin{smallmatrix}1 & 0\\ 0 & 0\end{smallmatrix}\right) \) has determinant \( 0 \), and \( \det A \), \( \det B \) are undefined. No consistent multiplicative assignment exists; the correct rectangular generalisation is the Cauchy–Binet formula. The entries lie in a commutative ring.Over the quaternions \( \mathbb{H} \) the naive Leibniz determinant \( ad - bc \) is worthless: the matrix \( M = \left(\begin{smallmatrix} i & j \\ j & i \end{smallmatrix}\right) \) has \( ad - bc = i^2 - j^2 = -1 - (-1) = 0 \), yet \( M \) is invertible (the system \( ix_1 + jx_2 = 0 \), \( jx_1 + ix_2 = 0 \) forces \( x_1 = kx_2 \) and then \( 2ix_2 = 0 \), so the kernel is trivial). Commutativity of the scalars is used both to make the Leibniz formula basis-independent and in the final step \( \det(B)\det(A) = \det(A)\det(B) \). The noncommutative repair is the Dieudonné determinant, valued in the abelianisation \( \mathbb{H}^\times / [\mathbb{H}^\times, \mathbb{H}^\times] \).
Proof

Write \( a_1, \dots, a_n \) for the columns of \( A \), \( b_1, \dots, b_n \) for the columns of \( B \), and \( e_1, \dots, e_n \) for the standard basis of \( K^n \), so that \( Ae_i = a_i \). Throughout we view \( \det \) as a function of the \( n \) columns of its argument, and we use two facts proved earlier in this unit: (L1) \( \det \) is \( n \)-linear in the columns, and (L2) \( \det \) is alternating (it vanishes whenever two columns are equal). The Leibniz formula is quoted as (L3).

1
\[ (AB)e_j = A(Be_j) = Ab_j \qquad (j = 1, \dots, n), \] so the \( j \)-th column of \( AB \) is \( Ab_j \).
Definition of matrix multiplication: multiplying a matrix by \( e_j \) extracts its \( j \)-th column, and matrix multiplication is associative on the triple \( A, B, e_j \). A
2
\[ b_j = \sum_{i=1}^{n} B_{ij}\, e_i \quad\Longrightarrow\quad Ab_j = \sum_{i=1}^{n} B_{ij}\, a_i . \]
Expansion of \( b_j \) in the standard basis (its coordinates are by definition the entries \( B_{1j}, \dots, B_{nj} \)), followed by linearity of the map \( x \mapsto Ax \) and \( Ae_i = a_i \). A
3
\[ \det(AB) = \det\!\left( \sum_{i_1=1}^{n} B_{i_1,1}\, a_{i_1}, \; \dots, \; \sum_{i_n=1}^{n} B_{i_n,n}\, a_{i_n} \right) = \sum_{f} \left( \prod_{j=1}^{n} B_{f(j),j} \right) \det\!\left( a_{f(1)}, \dots, a_{f(n)} \right), \] where \( f \) ranges over all maps \( f : \{1, \dots, n\} \to \{1, \dots, n\} \).
Multilinearity (L1), applied once in each of the \( n \) column slots and expanded completely: each choice of one index \( i_j = f(j) \) per slot contributes one term, and the scalar coefficients \( B_{f(j),j} \) pull out of \( \det \). This is the key expansion of the whole proof. B
4
If \( f \) is not injective, then \( \det\!\left( a_{f(1)}, \dots, a_{f(n)} \right) = 0 \).
If \( f(p) = f(q) \) with \( p \neq q \), the matrix \( \left( a_{f(1)} \mid \cdots \mid a_{f(n)} \right) \) has two equal columns, so its determinant vanishes by the alternating property (L2). Since a map \( \{1,\dots,n\} \to \{1,\dots,n\} \) is injective iff it is bijective, only permutations \( f = \sigma \in S_n \) survive. B
5
\[ \det\!\left( a_{\sigma(1)}, \dots, a_{\sigma(n)} \right) = \operatorname{sgn}(\sigma)\, \det\!\left( a_{1}, \dots, a_{n} \right) = \operatorname{sgn}(\sigma)\, \det A \qquad (\sigma \in S_n). \]
Two ingredients, both from earlier in the unit. (i) Alternating implies antisymmetric: expanding \( 0 = \det(\dots, u+v, \dots, u+v, \dots) \) by (L1) and killing the two equal-column terms by (L2) shows that swapping two columns negates \( \det \) — note this implication holds in every characteristic, including characteristic 2 (it is the converse that fails there). (ii) Every \( \sigma \in S_n \) is a product of transpositions, and the parity of the number of transpositions is well defined by the sign homomorphism \( \operatorname{sgn} : S_n \to \{\pm 1\} \). Applying (i) once per transposition gives the factor \( \operatorname{sgn}(\sigma) \). C
6
\[ \det(AB) = \sum_{\sigma \in S_n} \left( \prod_{j=1}^{n} B_{\sigma(j),j} \right) \operatorname{sgn}(\sigma)\, \det A = \left( \sum_{\sigma \in S_n} \operatorname{sgn}(\sigma) \prod_{j=1}^{n} B_{\sigma(j),j} \right) \det A = \det(B)\, \det(A). \]
Substitute Steps 4 and 5 into the expansion of Step 3; the bracketed sum is exactly the Leibniz formula (L3) for \( \det B \). B
7
\[ \det(AB) = \det(B)\det(A) = \det(A)\det(B). \qquad \blacksquare \]
Multiplication in \( K \) is commutative — this is the one place the commutativity hypothesis is used explicitly (it is also used silently inside the Leibniz formula). A
Result
\[ \det(AB) = \det(A)\,\det(B) \qquad \text{for all } A, B \in M_n(K). \]

Reading. The determinant turns matrix multiplication into scalar multiplication. Composing two linear maps multiplies their volume-scaling factors; in particular, a product of matrices is singular exactly when at least one factor is singular.

Scope. Valid for square matrices of one common size \( n \times n \) over any field — and, with the same proof, over any commutative ring with 1 (the proof never divides). It fails to even make sense for rectangular matrices (see Cauchy–Binet) and fails over noncommutative rings such as \( \mathbb{H} \).

Corollaries & converses
  • \( \det(A_1 A_2 \cdots A_k) = \det(A_1)\det(A_2)\cdots\det(A_k) \) and \( \det(A^k) = (\det A)^k \) for all \( k \in \mathbb{N} \), by induction.
  • If \( A \) is invertible then \( \det(A)\det(A^{-1}) = \det(I_n) = 1 \), so \( \det A \in K^\times \) and \( \det(A^{-1}) = (\det A)^{-1} \). In particular an invertible matrix never has determinant \( 0 \).
  • Similarity invariance: \( \det(P^{-1}AP) = \det(P)^{-1}\det(A)\det(P) = \det A \). Hence \( \det \) of a linear operator on a finite-dimensional space is well defined, independent of basis.
  • \( \det : GL_n(K) \to K^\times \) is a group homomorphism; its kernel \( SL_n(K) = \{ A : \det A = 1 \} \) is a normal subgroup, and \( GL_n(K)/SL_n(K) \cong K^\times \).
  • \( \det(AB) = \det(BA) \) always — even when \( AB \neq BA \).
  • Over a field, \( \det(AB) = 0 \) iff \( \det A = 0 \) or \( \det B = 0 \) (a field has no zero divisors).
  • Converse fails: multiplicativity does not characterise \( \det \). The maps \( A \mapsto \det(A)^2 \), \( A \mapsto 1 \), and (over \( \mathbb{R} \)) \( A \mapsto |\det A| \) are all multiplicative on \( M_n \) or \( GL_n \) but are not the determinant. (A theorem of Frobenius-type does say every polynomial multiplicative map \( M_n(\mathbb{C}) \to \mathbb{C} \) is a power of \( \det \).)
Fails without
  • Squareness dropped. \( A = (1 \;\; 0) \in M_{1\times 2}(\mathbb{R}) \), \( B = A^{T} \in M_{2\times 1}(\mathbb{R}) \): then \( \det(AB) = \det(1) = 1 \) but \( \det(BA) = \det\!\left(\begin{smallmatrix}1 & 0\\ 0 & 0\end{smallmatrix}\right) = 0 \), and neither \( \det A \) nor \( \det B \) is defined. Any attempt to define determinants of rectangular matrices so that multiplicativity survives is doomed by this pair; the honest substitute is Cauchy–Binet: for \( A \in M_{m\times n} \), \( B \in M_{n\times m} \) with \( m \le n \), \( \det(AB) = \sum_{S} \det(A_{\cdot,S})\det(B_{S,\cdot}) \), summing over the \( m \)-element subsets \( S \) of columns/rows. When \( m \gt n \) the sum is empty and \( \det(AB) = 0 \).
  • Commutativity dropped. Over \( \mathbb{H} \), the matrix \( M = \left(\begin{smallmatrix} i & j \\ j & i \end{smallmatrix}\right) \) is invertible but its naive Leibniz determinant is \( i^2 - j^2 = 0 \); had a multiplicative scalar determinant existed and extended the diagonal case, \( M \cdot M^{-1} = I \) would force \( 0 \cdot \det(M^{-1}) = 1 \), a contradiction. No scalar-valued multiplicative determinant on \( M_2(\mathbb{H}) \) restricting to the usual one exists.
  • The formula does not transfer to sums. Multiplicativity tempts the false analogue \( \det(A+B) = \det A + \det B \): with \( A = B = I_2 \), \( \det(A+B) = \det(2I_2) = 4 \neq 2 = \det A + \det B \).
Common errors
  • The additivity trap. Writing \( \det(A+B) = \det A + \det B \). The determinant is multiplicative, not additive; it is linear in each column separately, never in the whole matrix.
  • Scalar pull-out. Writing \( \det(cA) = c\det A \). Correct is \( \det(cA) = c^{n}\det A \): the scalar multiplies each of the \( n \) columns.
  • Rectangular abuse. Applying the theorem to \( \det(AB) \) where \( A \) is \( m \times n \), \( B \) is \( n \times m \), \( m \neq n \). The factors have no determinant; use Cauchy–Binet.
  • Zero-product confusion. From \( \det(AB) = 0 \) concluding \( A = 0 \) or \( B = 0 \). The correct conclusion (over a field) is \( \det A = 0 \) or \( \det B = 0 \); nonzero singular matrices abound.
  • Circularity in "elementary matrix" proofs. Proving multiplicativity by writing \( A \) as a product of elementary matrices silently assumes \( A \) invertible and uses \( \det(EX) = \det(E)\det(X) \) for each elementary \( E \) — legitimate, but only if the singular case and the elementary-matrix identities are proved separately; students often omit both.
  • Transposes for free. Using \( \det(A^{T}B) = \det(A)\det(B) \) while believing it follows from this theorem alone; it also needs the separate theorem \( \det(A^{T}) = \det(A) \).
Discussion

The geometric content is best seen over \( \mathbb{R} \): a linear map \( T : \mathbb{R}^n \to \mathbb{R}^n \) scales the volume of every measurable set by the constant factor \( |\det T| \), and preserves or reverses orientation according to the sign of \( \det T \). Composition of maps must therefore compose scaling factors multiplicatively — the theorem is the exact algebraic expression of this fact, and it is why the Jacobian determinant obeys a chain rule, \( \det D(g \circ f)(x) = \det Dg(f(x)) \cdot \det Df(x) \), which in turn drives the change-of-variables formula for multiple integrals.

Historically, the product theorem was presented to the Institut de France in 1812 independently by Jacques Binet and Augustin-Louis Cauchy, on the same day; Cauchy's memoir also contained what is now called the Cauchy–Binet formula, the correct generalisation to products of rectangular matrices. Special cases had appeared earlier in the work of Lagrange and in Gauss's Disquisitiones Arithmeticae (1801) in the language of quadratic forms and their composition — Gauss's observation that the "determinant" of a composed form multiplies is essentially the \( n = 2 \) case.

Structurally, the theorem says \( \det \) is a monoid homomorphism \( (M_n(K), \cdot) \to (K, \cdot) \), restricting to a surjective group homomorphism \( \det : GL_n(K) \to K^\times \) with kernel \( SL_n(K) \). This one fact organises a large amount of linear algebra and group theory: it makes the determinant a similarity invariant (so eigenvalue products, characteristic polynomials, and determinants of abstract operators are well defined), it exhibits \( SL_n \) as a normal subgroup with \( GL_n/SL_n \cong K^\times \), and in Lie theory it globalises the trace identity \( \det(\exp X) = \exp(\operatorname{tr} X) \), pairing the multiplicative theorem with the additivity of trace.

The proof given above, read structurally, is the uniqueness theorem for alternating forms in disguise: for fixed \( A \), the map \( X \mapsto \det(AX) \) is \( n \)-linear and alternating in the columns of \( X \), hence equals \( c \cdot \det X \) with \( c = \det(A I_n) = \det A \). The coordinate-free formulation lives in the exterior algebra: \( \Lambda^n(K^n) \) is a free \( K \)-module of rank 1, any endomorphism \( f \) of \( K^n \) induces \( \Lambda^n f \), which is multiplication by the scalar \( \det f \), and functoriality \( \Lambda^n(f \circ g) = \Lambda^n f \circ \Lambda^n g \) is the theorem, with no computation at all. This argument works over an arbitrary commutative ring, which is why the theorem holds there. Over noncommutative division rings, \( \Lambda^n \) breaks down along with the Leibniz formula, and the surviving invariant is Dieudonné's determinant \( GL_n(D) \to D^\times/[D^\times, D^\times] \), which is multiplicative by construction but only after abelianising the target.

Common misconceptions. Multiplicativity does not imply any formula for \( \det(A+B) \) — no useful general one exists. Also, \( \det(AB) = \det(BA) \) does not mean \( AB \) and \( BA \) are similar (they always share a characteristic polynomial when \( A, B \) are square, and are similar when a factor is invertible, but e.g. \( A = \left(\begin{smallmatrix}0 & 1\\ 0 & 0\end{smallmatrix}\right), B = \left(\begin{smallmatrix}0 & 0\\ 0 & 1\end{smallmatrix}\right) \) give \( AB \neq 0 \) nilpotent and \( BA = 0 \), which are not similar).

Worked examples

Example 1 (powers without powering). Let \( A = \begin{pmatrix} 1 & 2 \\ 3 & 4 \end{pmatrix} \in M_2(\mathbb{R}) \). Compute \( \det(A^{3}) \) without computing \( A^{3} \).

1
\[ \det A = 1 \cdot 4 - 2 \cdot 3 = -2. \]
The \( 2 \times 2 \) determinant formula (Leibniz for \( n = 2 \)). A
2
\[ \det(A^{3}) = \det(A \cdot A \cdot A) = \det(A)\det(A)\det(A) = (\det A)^{3}. \]
Multiplicativity, applied twice (once to \( A \cdot A^2 \), once to \( A \cdot A \)). A
3
\[ \det(A^{3}) = (-2)^{3} = -8. \]
Arithmetic in \( \mathbb{R} \). A
\[ \det(A^{3}) = -8. \]

Reading. One \( 2 \times 2 \) determinant and one cube replace nine dot products and a \( 3 \)-fold matrix product; for \( \det(A^{100}) \) the saving is decisive: \( \det(A^{100}) = 2^{100} \).

Example 2 (similarity invariance, from scratch). Let \( P = \begin{pmatrix} 1 & 2 \\ 1 & 3 \end{pmatrix} \), \( A = \begin{pmatrix} 4 & 1 \\ 2 & 1 \end{pmatrix} \), and \( M = P A P^{-1} \). Compute \( \det M \) without ever computing \( P^{-1} \) or \( M \).

1
\[ \det P = 1 \cdot 3 - 2 \cdot 1 = 1, \qquad \det A = 4 \cdot 1 - 1 \cdot 2 = 2. \]
The \( 2 \times 2 \) determinant formula. Since \( \det P = 1 \neq 0 \), \( P \) is invertible, so \( M \) is defined. A
2
\[ P P^{-1} = I_2 \;\Longrightarrow\; \det(P)\det(P^{-1}) = \det(I_2) = 1 \;\Longrightarrow\; \det(P^{-1}) = \frac{1}{\det P} = 1. \]
The theorem applied to the product \( P P^{-1} \), plus \( \det(I_2) = 1 \) (normalisation axiom). This is the standard derivation of \( \det(P^{-1}) = (\det P)^{-1} \). B
3
\[ \det M = \det\!\left( P \cdot (A P^{-1}) \right) = \det(P)\,\det(A)\,\det(P^{-1}) = 1 \cdot 2 \cdot 1 = 2. \]
Multiplicativity applied twice (matrix multiplication is associative, so the bracketing is harmless), then Steps 1–2. A
\[ \det\!\left( P A P^{-1} \right) = \det A = 2. \]

Reading. Conjugation never changes the determinant — the computation above is the whole proof that determinants are similarity invariants, and it required no entry of \( M \).

Problems
  1. Let \( A, B \in M_4(\mathbb{R}) \) with \( \det A = 3 \) and \( \det B = -2 \). Compute \( \det(AB) \), \( \det(BA) \), \( \det(A^{3}B^{2}) \), and \( \det(2A) \).
    Solution By the theorem, \( \det(AB) = 3 \cdot (-2) = -6 \) and \( \det(BA) = (-2) \cdot 3 = -6 \) (equal even if \( AB \neq BA \)). Iterating, \( \det(A^{3}B^{2}) = (\det A)^{3}(\det B)^{2} = 27 \cdot 4 = 108 \). Finally \( \det(2A) = 2^{4} \det A = 16 \cdot 3 = 48 \): this last one uses multilinearity in each of the \( 4 \) columns, not the product theorem — writing \( \det(2A) = \det(2I_4 \cdot A) = \det(2I_4)\det(A) = 2^4 \cdot 3 \) routes it through the theorem correctly.
  2. Prove that similar matrices have equal determinants: if \( B = P^{-1} A P \) with \( P \in GL_n(K) \), then \( \det B = \det A \). Deduce that all matrices representing a fixed linear operator \( T : V \to V \) (with \( \dim V = n \)) in different bases share one determinant.
    Solution First, \( \det(P^{-1}) = (\det P)^{-1} \): apply the theorem to \( P^{-1} P = I_n \) to get \( \det(P^{-1})\det(P) = 1 \). Then \[ \det B = \det(P^{-1} A P) = \det(P^{-1})\,\det(A)\,\det(P) = (\det P)^{-1} (\det P)\, \det A = \det A, \] using the theorem twice and commutativity of \( K \) to cancel. If \( T \) is a linear operator and \( A, B \) are its matrices in two bases, the change-of-basis theorem gives \( B = P^{-1} A P \) for the transition matrix \( P \), so \( \det B = \det A \); hence \( \det T := \det A \) is well defined.
  3. Let \( A \in M_n(\mathbb{R}) \) satisfy \( A^{k} = I_n \) for some \( k \geq 1 \). Show that \( \det A = \pm 1 \), and that \( \det A = 1 \) whenever \( k \) is odd. What can you say over \( \mathbb{C} \)?
    Solution By induction using the theorem, \( \det(A^{k}) = (\det A)^{k} \). Since \( A^{k} = I_n \), \( (\det A)^{k} = \det(I_n) = 1 \). Set \( d = \det A \in \mathbb{R} \). Then \( |d|^{k} = 1 \), and since \( t \mapsto t^{k} \) is strictly increasing on \( t \gt 0 \) with value 1 only at \( t = 1 \), we get \( |d| = 1 \), i.e. \( d = \pm 1 \). If \( k \) is odd, \( (-1)^{k} = -1 \neq 1 \) rules out \( d = -1 \), so \( d = 1 \). Over \( \mathbb{C} \), the same computation shows \( \det A \) is a \( k \)-th root of unity: \( \det A \in \{ e^{2\pi i m/k} : 0 \le m \lt k \} \), and every such root occurs (take \( A = \operatorname{diag}(e^{2\pi i m/k}, 1, \dots, 1) \)).
  4. Let \( A, B \in M_n(K) \), \( K \) a field, with \( AB = I_n \). Prove that \( BA = I_n \) (a one-sided inverse of a square matrix is two-sided). You may use the adjugate identity \( A \operatorname{adj}(A) = \det(A) I_n \) from this unit.
    Solution Applying the theorem to \( AB = I_n \): \( \det(A)\det(B) = \det(I_n) = 1 \), so \( \det A \neq 0 \). By the adjugate identity, \( A \cdot \left( \det(A)^{-1} \operatorname{adj}(A) \right) = I_n \) and likewise \( \left( \det(A)^{-1} \operatorname{adj}(A) \right) A = I_n \), so \( A \) is invertible with a genuine two-sided inverse \( A^{-1} \). Now multiply \( AB = I_n \) on the left by \( A^{-1} \): \( B = A^{-1} \). Therefore \( BA = A^{-1} A = I_n \). (Note how the determinant converts a one-sided algebraic condition into the scalar statement \( \det A \neq 0 \), which is side-blind.)
  5. (Structure.) Show that \( \det : GL_n(K) \to K^\times \) is a surjective group homomorphism, that \( SL_n(K) = \ker(\det) \) is a normal subgroup of \( GL_n(K) \), and that \( GL_n(K)/SL_n(K) \cong K^\times \).
    Solution Homomorphism: for \( A, B \in GL_n(K) \), \( AB \) is invertible with inverse \( B^{-1}A^{-1} \), and \( \det(AB) = \det(A)\det(B) \) by the theorem; moreover \( \det A \in K^\times \) since \( \det(A)\det(A^{-1}) = 1 \), so \( \det \) really lands in \( K^\times \). Surjectivity: for \( \lambda \in K^\times \), the matrix \( D_\lambda = \operatorname{diag}(\lambda, 1, \dots, 1) \) is invertible (inverse \( D_{\lambda^{-1}} \)) and \( \det D_\lambda = \lambda \) (expand along the first column, or use that the determinant of a diagonal matrix is the product of its diagonal entries). Kernel: \( \ker(\det) = \{ A \in GL_n(K) : \det A = 1 \} = SL_n(K) \) by definition, and the kernel of any group homomorphism is normal: for \( A \in SL_n(K) \), \( P \in GL_n(K) \), \( \det(P A P^{-1}) = \det(P)\det(A)\det(P)^{-1} = \det A = 1 \), so \( P A P^{-1} \in SL_n(K) \). Quotient: the first isomorphism theorem applied to the surjective homomorphism \( \det \) gives \( GL_n(K)/SL_n(K) = GL_n(K)/\ker(\det) \cong \operatorname{im}(\det) = K^\times \).