Binet-Cauchy Identity

Theorem

$\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} = \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j} + \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}$

where all of the $a, b, c, d$ are elements of a commutative ring.

Thus the identity holds for $\Z, \Q, \R, \C$.

Proof 1

Expanding the last term:

\(\ds \)

\(\)

\(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i b_j - a_j b_i} \paren {c_i d_j - c_j d_i}\)

\(\ds \)

\(=\)

\(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i c_i b_j d_j + a_j c_j b_i d_i}\)

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i d_i b_j c_j + a_j d_j b_i c_i}\)

\(\ds \)

\(=\)

\(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i c_i b_j d_j + a_j c_j b_i d_i} + \sum_{i \mathop = 1}^n a_i c_i b_i d_i\)

These new terms are the same

\(\ds \)

\(\)

\(\, \ds - \, \)

\(\ds \sum_{1 \mathop \le i \mathop < j \mathop \le n} \paren {a_i d_i b_j c_j + a_j d_j b_i c_i} - \sum_{i \mathop = 1}^n a_i d_i b_i c_i\)

\(\ds \)

\(=\)

\(\ds \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i c_i b_j d_j - \sum_{i \mathop = 1}^n \sum_{j \mathop = 1}^n a_i d_i b_j c_j\)

Completing the sums

\(\ds \)

\(=\)

\(\ds \paren {\sum_{i \mathop = 1}^n a_i c_i} \paren {\sum_{j \mathop = 1}^n b_j d_j} - \paren {\sum_{i \mathop = 1}^n a_i d_i} \paren {\sum_{j \mathop = 1}^n b_j c_j}\)

Factoring terms indexed by $i$ and $j$

Hence the result.

$\blacksquare$

Proof 2

This is a special case of the Cauchy-Binet Formula:

$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop < \cdots \mathop < j_m \le n} \map \det {\mathbf A_{j_1 j_2 \ldots j_m} } \map \det {\mathbf B_{j_1 j_2 \ldots j_m} }$

where:

$\mathbf A$ is an $m \times n$ matrix

$\mathbf B$ is an $n \times m$ matrix.

For $1 \le j_1, j_2, \ldots, j_m \le n$:

$\mathbf A_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of columns $j_1, j_2, \ldots, j_m$ of $\mathbf A$.

$\mathbf B_{j_1 j_2 \ldots j_m}$ denotes the $m \times m$ matrix consisting of rows $j_1, j_2, \ldots, j_m$ of $\mathbf B$.

In this case $m = 2$, giving:

$\ds \map \det {\mathbf A \mathbf B} = \sum_{1 \mathop \le j_1 \mathop < j_2 \mathop \le n} \map \det {\mathbf A_{j_1 j_2} } \, \map \det {\mathbf B_{j_1 j_2} }$

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Also known as

The Binet-Cauchy Identity is also known as Binet's formula.

Source of Name

This entry was named for Jacques Philippe Marie Binet and Augustin Louis Cauchy.

Historical Note

The Binet-Cauchy Identity is a special case of the Cauchy-Binet Formula, which was presented by Jacques Philippe Marie Binet and Augustin Louis Cauchy on the same day in $1812$.

Sources

• 1997: Donald E. Knuth: The Art of Computer Programming: Volume 1: Fundamental Algorithms (3rd ed.) ... (previous) ... (next): $\S 1.2.3$: Sums and Products: Exercise $30$