Cauchy–Binet formula

In linear algebra, the Cauchy–Binet formula, named after Augustin-Louis Cauchy and Jacques Philippe Marie Binet, is an identity for the determinant of the product of two rectangular matrices of transpose shapes (so that the product is well-defined and square). It generalizes the statement that the determinant of a product of square matrices is equal to the product of their determinants. The formula is valid for matrices with the entries from any commutative ring. In linear algebra, the Cauchy–Binet formula, named after Augustin-Louis Cauchy and Jacques Philippe Marie Binet, is an identity for the determinant of the product of two rectangular matrices of transpose shapes (so that the product is well-defined and square). It generalizes the statement that the determinant of a product of square matrices is equal to the product of their determinants. The formula is valid for matrices with the entries from any commutative ring. Let A be an m×n matrix and B an n×m matrix. Write for the set { 1, ..., n }, and ( [ n ] m ) {displaystyle { binom {}{m}}} for the set of m-combinations of (i.e., subsets of size m; there are ( n m ) {displaystyle { binom {n}{m}}} of them). For S ∈ ( [ n ] m ) {displaystyle Sin { binom {}{m}}} , write A,S for the m×m matrix whose columns are the columns of A at indices from S, and BS, for the m×m matrix whose rows are the rows of B at indices from S. The Cauchy–Binet formula then states Example: taking m = 2 and n = 3, and matrices A = ( 1 1 2 3 1 − 1 ) {displaystyle A={egin{pmatrix}1&1&2\3&1&-1\end{pmatrix}}} and B = ( 1 1 3 1 0 2 ) {displaystyle B={egin{pmatrix}1&1\3&1\0&2end{pmatrix}}} , the Cauchy–Binet formula gives the determinant: Indeed A B = ( 4 6 6 2 ) {displaystyle AB={egin{pmatrix}4&6\6&2end{pmatrix}}} , and its determinant is −28, which is also the value − 2 × − 2 + − 3 × 6 + − 7 × 2 {displaystyle -2 imes -2+-3 imes 6+-7 imes 2} given by the right hand side of the formula. If n < m then ( [ n ] m ) {displaystyle { binom {}{m}}} is the empty set, and the formula says that det(AB) = 0 (its right hand side is an empty sum); indeed in this case the rank of the m×m matrix AB is at most n, which implies that its determinant is zero. If n = m, the case where A and B are square matrices, ( [ n ] m ) = { [ n ] } {displaystyle { binom {}{m}}={}} (a singleton set), so the sum only involves S = , and the formula states that det(AB) = det(A)det(B). For m = 0, A and B are empty matrices (but of different shapes if n > 0), as is their product AB; the summation involves a single term S = Ø, and the formula states 1 = 1, with both sides given by the determinant of the 0×0 matrix. For m = 1, the summation ranges over the collection ( [ n ] 1 ) {displaystyle { binom {}{1}}} of the n different singletons taken from , and both sides of the formula give ∑ j = 1 n A 1 , j B j , 1 {displaystyle extstyle sum _{j=1}^{n}A_{1,j}B_{j,1}} , the dot product of the pair of vectors represented by the matrices. The smallest value of m for which the formula states a non-trivial equality is m = 2; it is discussed in the article on the Binet–Cauchy identity. Let a , b , c , d , x , y , z , w {displaystyle {oldsymbol {a}},{oldsymbol {b}},{oldsymbol {c}},{oldsymbol {d}},{oldsymbol {x}},{oldsymbol {y}},{oldsymbol {z}},{oldsymbol {w}}} be three-dimensional vectors. In the case m > 3, the right-hand side always equals 0. The following simple proof presented in relies on two facts that can be proven in several different ways: