Laplace expansion

This article is about the expansion of the determinant of a square matrix as a weighted sum of determinants of sub-matrices. For the expansion of an 1/r-potential using spherical harmonical functions, see Laplace expansion (potential).

In linear algebra, the Laplace expansion, named after Pierre-Simon Laplace, also called cofactor expansion, is an expression for the determinant |B| of an n × n square matrix B that is a weighted sum of the determinants of n sub-matrices of B, each of size (n−1) × (n−1). The Laplace expansion is of theoretical interest as one of several ways to view the determinant, as well as of practical use in determinant computation.

The i, j cofactor of B is the scalar C_ij defined by

$C_{ij}\ = (-1)^{i+j} M_{ij}\,,$

where M_ij is the i, j minor matrix of B, that is, the determinant of the (n–1) × (n–1) matrix that results from deleting the i-th row and the j-th column of B.

Then the Laplace expansion is given by the following

Theorem. Suppose B = (b_ij) is an n × n matrix and fix any i, j ∈ {1, 2, ..., n}.

Then its determinant |B| is given by:

$\begin{align}|B| & {} = b_{i1} C_{i1} + b_{i2} C_{i2} + \cdots + b_{in} C_{in} \\ & {} = b_{1j} C_{1j} + b_{2j} C_{2j} + \cdots + b_{nj} C_{nj} \\ & {} = \sum_{j'=1}^{n} b_{ij'} C_{ij'} = \sum_{i'=1}^{n} b_{i'j} C_{i'j} . \end{align}$

Examples

Consider the matrix

$B = \begin{bmatrix} 1 & 2 & 3 \\ 4 & 5 & 6 \\ 7 & 8 & 9 \end{bmatrix}.$

The determinant of this matrix can be computed by using the Laplace expansion along any one of its rows or columns. For instance, an expansion along the first row yields:

$|B| = 1 \cdot \begin{vmatrix} 5 & 6 \\ 8 & 9 \end{vmatrix} - 2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 3 \cdot \begin{vmatrix} 4 & 5 \\ 7 & 8 \end{vmatrix}$

${} = 1 \cdot (-3) - 2 \cdot (-6) + 3 \cdot (-3) = 0.$

Laplace expansion along the second column yields the same result:

$|B| = -2 \cdot \begin{vmatrix} 4 & 6 \\ 7 & 9 \end{vmatrix} + 5 \cdot \begin{vmatrix} 1 & 3 \\ 7 & 9 \end{vmatrix} - 8 \cdot \begin{vmatrix} 1 & 3 \\ 4 & 6 \end{vmatrix}$

${} = -2 \cdot (-6) + 5 \cdot (-12) - 8 \cdot (-6) = 0.$

It is easy to verify that the result is correct: the matrix is singular because the sum of its first and third column is twice the second column, and hence its determinant is zero.

Proof

Suppose $B$ is an n × n matrix and $i,j\in\{1,2,\dots,n\}.$ For clarity we also label the entries of $B$ that compose its $i,j$ minor matrix $M_{ij}$ as

$(a_{st})$ for $1 \le s,t \le n-1.$

Consider the terms in the expansion of $|B|$ that have $b_{ij}$ as a factor. Each has the form

$\sgn \tau\,b_{1,\tau(1)} \cdots b_{i,j} \cdots b_{n,\tau(n)} = \sgn \tau\,b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}$

for some permutation τ ∈ S_n with $\tau(i)=j$, and a unique and evidently related permutation $\sigma\in S_{n-1}$ which selects the same minor entries as $\tau.$ Similarly each choice of $\sigma$ determines a corresponding $\tau,$ i.e. the correspondence $\sigma\leftrightarrow\tau$ is a bijection between $S_{n-1}$ and $\{\tau\in S_n\colon\tau(i)=j\}.$ The permutation $\tau$ can be derived from $\sigma$ as follows.

Define $\sigma'\in S_n$ by $\sigma'(k) = \sigma(k)$ for $1 \le k \le n-1$ and $\sigma'(n) = n$. Then $\sgn\sigma'=\sgn\sigma$ and

$\tau\,=(n,n-1,\ldots,i)\sigma'(j,j+1,\ldots,n)$

Since the two cycles can be written respectively as $n-i$ and $n-j$transpositions,

$\sgn\tau\,= (-1)^{2n-(i+j)} \sgn\sigma'\,= (-1)^{i+j} \sgn\sigma.$

And since the map $\sigma\leftrightarrow\tau$ is bijective,

$\sum_{\tau \in S_n\colon\tau(i)=j}$	$\sgn \tau\,b_{1,\tau(1)} \cdots b_{n,\tau(n)}$
	$= \sum_{\sigma \in S_{n-1}} (-1)^{i+j}\sgn\sigma\, b_{ij} a_{1,\sigma(1)} \cdots a_{n-1,\sigma(n-1)}$
	$=\ b_{ij} (-1)^{i+j} |M_{ij}|,$

from which the result follows.

Laplace expansion of a determinant by complementary minors

Laplaces cofactor expansion can be generalised as follows.

Example

Consider the matrix

$A = \begin{bmatrix} 1 & 2 & 3 & 4 \\ 5 & 6 & 7 & 8 \\ 9 & 10 & 11 & 12 \\ 13 & 14 & 15 & 16 \end{bmatrix}.$

The determinant of this matrix can be computed by using the Laplace's cofactor expansion along the first two rows as follows. Firstly note that there are 6 sets of two distinct numbers in $\left\{1,2,3,4\right\}$, namely let $S=\left\{\{1,2\},\{1,3\},\{1,4\},\{2,3\},\{2,4\},\{3,4\}\right\}$ be the aformentioned set.

By defining the complementary cofactors to be

$b_{\{j,k\}}=\begin{vmatrix} a_{1j} & a_{1k} \\ a_{2j} & a_{2k} \end{vmatrix}$,

$c_{\{j,k\}}=\begin{vmatrix} a_{3j} & a_{3k} \\ a_{4j} & a_{4k} \end{vmatrix}$,

and the sign of their permutation to be

$\varepsilon^{\{i,j\},\{p,q\}}=\mbox{sgn}\begin{bmatrix} 1 & 2 & 3 & 4 \\ i & j & p & q \end{bmatrix}$.

The determinant of A can be written out as

$|A| = \sum_{H \in S} \varepsilon^{H,H^\prime}b_{H}c_{H^\prime},$

where $H^{\prime}$ is the complemenatary set to $H$.

In our explicit example this gives us

$|A| = b_{\{1,2\}}c_{\{3,4\}} -b_{\{1,3\}}c_{\{2,4\}} +b_{\{1,4\}}c_{\{2,3\}} +b_{\{2,3\}}c_{\{1,4\}} -b_{\{2,4\}}c_{\{1,3\}} +b_{\{3,4\}}c_{\{1,2\}}$

${} = \begin{vmatrix} 1 & 2 \\ 5 & 6 \end{vmatrix} \cdot \begin{vmatrix} 11 & 12 \\ 15 & 16 \end{vmatrix} - \begin{vmatrix} 1 & 3 \\ 5 & 7 \end{vmatrix} \cdot \begin{vmatrix} 10 & 12 \\ 14 & 16 \end{vmatrix} + \begin{vmatrix} 1 & 4 \\ 5 & 8 \end{vmatrix} \cdot \begin{vmatrix} 10 & 11 \\ 14 & 15 \end{vmatrix} + \begin{vmatrix} 2 & 3 \\ 6 & 7 \end{vmatrix} \cdot \begin{vmatrix} 9 & 12 \\ 13 & 16 \end{vmatrix} - \begin{vmatrix} 2 & 4 \\ 6 & 8 \end{vmatrix} \cdot \begin{vmatrix} 9 & 11 \\ 13 & 15 \end{vmatrix} + \begin{vmatrix} 3 & 4 \\ 7 & 8 \end{vmatrix} \cdot \begin{vmatrix} 9 & 10 \\ 13 & 14 \end{vmatrix}$

${} = -4 \cdot (-4) -(-8) \cdot (-8) +(-12) \cdot (-4) +(-4) \cdot (-12) -(-8) \cdot (-8) +(-4) \cdot (-4)$

${} = 16 - 64 + 48 + 48 - 64 + 16 = 0.$

As above, It is easy to verify that the result is correct: the matrix is singular because the sum of its first and third column is twice the second column, and hence its determinant is zero.

Computational expense

The Laplace expansion is computationally inefficient for high dimension because for NxN matrices, the computational effort scales with N!. Therefore, the Laplace expansion is not suitable for large N. Using a decomposition into triangular matrices as in the LU decomposition, one can determine determinants with effort N³/3.

References

• David Poole: Linear Algebra. A Modern Introduction. Cengage Learning 2005, ISBN 0-534-99845-3, p. 265-267 (restricted online copy at Google Books)
• Harvey E. Rose: Linear Algebra. A Pure Mathematical Approach. Springer 2002, ISBN 3-7643-6905-1, p. 57-60 (restricted online copy at Google Books)

See also

• Leibniz formula for determinants

External links

• Laplace expansion at PlanetMath