Abstract We give a proof of the Cauchy–Binet formula for the determinant of the product of two matrices that mostly avoids explicit matrix manipulations. × × × Let k

The Cauchy–Binet formula

Andrew Putman

Abstract


We give a proof of the Cauchy–Binet formula for the determinant of the product of two matrices that mostly avoids explicit matrix manipulations.



× ×

×
Let k be a field. All matrices in this note have entries in k. Let A be an n m matrix and let B be an m n matrix. The product AB is thus an n n matrix. The Cauchy–Binet formula shows how to express the determinant of AB in terms of A and B. When n = m, it reduces to the familiar fact that det(AB) = det(A) det(B).


Stating it requires introducing some notation. Let [m] = {1, . . . , m}. For I ⊂ [m], let A_I be the n × |I| submatrix of A consisting of the rows of A lying in I. Similarly, let _IB be the |I| × n submatrix of B consisting of the columns of B lying in I.

× ×
Theorem 0.1 (Cauchy–Binet formula). Let A be an n m matrix and let B be an m n

Σ
matrix. Then

det(AB) =
I⊂[m]
|I|=n
det(A_I) det(_IB).

× →
Proof. For an r s matrix C, let φ_C : k^s k^r be the associated linear map. Thus φ_AB
equals the composition

−→ k
_kn ^φB
−→ kⁿ.

m ^φA
Letting {˙e₁, . . . , ˙e_n} be the standard basis for kⁿ, we thus have that
φ_A ◦ φ_B(˙e₁ ∧ · · · ∧ ˙e_n) = det(AB)˙e₁ ∧ · · · ∧ ˙e_n.
To express this in terms of A and B, we will have to first understand φ_B : ∧ⁿ kⁿ → ∧ⁿk^m. Let {f^˙₁, . . . , f^˙_n} be the standard basis k^m. The vector space ∧ⁿk^m thus has a basis
{f^˙_i1 ∧ · · · ∧ f^˙_in | {i₁ < · · · < i_n} ⊂ [m]}.

We claim that

Σ
φ_B(˙e₁ ∧ · · · ∧ ˙e_n) =
I={i₁<···<i_n}⊂[m]


det(_IB)f^˙_i1 ∧ · · · ∧ f^˙_in . (0.1)

1

n
To see this, fix some I = {i₁ < · · · < i_n} ⊂ [m]. Let V_I = (f^˙_i , . . . , f^˙_i ) ⊂ k^m and let

→ ∈
π_I : k^m V_I be the projection whose kernel is generated by the f^˙_j with j / I. Identifying

m ^πI
V_I with kⁿ via its natural basis, the composition

−→ k
_kn ^φB
−→ V_I

equals the linear map associated to _IB. We thus have
(π_I ◦ φ_B)_∗(˙e₁ ∧ · · · ∧ ˙e_n) = det(_IB)f^˙_i1 ∧ · · · ∧ f^˙_in .
The equation (0.1) follows.



{ · · · } ⊂
Fixing some I = i₁ < < i_n [m] again, the next step is to observe that if we again identify V_I with kⁿ via its natural basis, the composition



−→

I
V ,→ k^{m φA} kⁿ
equals the linear map associated to A_I. It follows that
φ_A(f^˙_i1 ∧ · · · ∧ f^˙_in ) = det(A_I)˙e₁ ∧ · · · ∧ ˙e_n.
Combining this with (0.1), we see that

Σ
φ_A ◦ φ_B(˙e₁ ∧ · · · ∧ ˙e_n) =

Σ
I={i₁<···_n}⊂[m]
=
I={i₁<···<i_n}⊂[m]
det(_IB)φ_A(f^˙_i1 ∧ · · · ∧ f^˙_in ) det(_IB) det(A_I)˙e₁ ∧ · · · ∧ ˙e_n.

The theorem follows.
Andrew Putman Department of Mathematics University of Notre Dame 164 Hurley Hall
Notre Dame, IN 46556
[e-mail]