Laplace expansion

Summary

In linear algebra, the Laplace expansion, named after Pierre-Simon Laplace, also called cofactor expansion, is an expression of the determinant of an n × n-matrix B as a weighted sum of minors, which are the determinants of some (n − 1) × (n − 1)-submatrices of B. Specifically, for every i, the Laplace expansion along the ith row is the equality where is the entry of the ith row and jth column of B, and is the determinant of the submatrix obtained by removing the ith row and the jth column of B. Similarly, the Laplace expansion along the jth column is the equality (Each identity implies the other, since the determinants of a matrix and its transpose are the same.)

The coefficient of in the above sum is called the cofactor of in B.

The Laplace expansion is often useful in proofs, as in, for example, allowing recursion on the size of matrices. It is also of didactic interest for its simplicity and as one of several ways to view and compute the determinant. For large matrices, it quickly becomes inefficient to compute when compared to Gaussian elimination.

Examples

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Consider the matrix

 

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:

 

Laplace expansion along the second column yields the same result:

 

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

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Suppose   is an n × n matrix and   For clarity we also label the entries of   that compose its   minor matrix   as

  for  

Consider the terms in the expansion of   that have   as a factor. Each has the form

 

for some permutation τSn with  , and a unique and evidently related permutation   which selects the same minor entries as τ. Similarly each choice of σ determines a corresponding τ i.e. the correspondence   is a bijection between   and   Using Cauchy's two-line notation, the explicit relation between   and   can be written as

 

where   is a temporary shorthand notation for a cycle  . This operation decrements all indices larger than j so that every index fits in the set {1,2,...,n-1}

The permutation τ can be derived from σ as follows. Define   by   for   and  . Then   is expressed as

 

Now, the operation which apply   first and then apply   is (Notice applying A before B is equivalent to applying inverse of A to the upper row of B in two-line notation)

 

where   is temporary shorthand notation for  .

the operation which applies   first and then applies   is

 

above two are equal thus,

 
 

where   is the inverse of   which is  .

Thus

 

Since the two cycles can be written respectively as   and   transpositions,

 

And since the map   is bijective,

 

from which the result follows. Similarly, the result holds if the index of the outer summation was replaced with  .[1]

Laplace expansion of a determinant by complementary minors

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Laplace's cofactor expansion can be generalised as follows.

Example

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Consider the matrix

 

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 {1, 2, 3, 4}, namely let   be the aforementioned set.

By defining the complementary cofactors to be

 
 

and the sign of their permutation to be

 

The determinant of A can be written out as

 

where   is the complementary set to  .

In our explicit example this gives us

 

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.

General statement

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Let   be an n × n matrix and   the set of k-element subsets of {1, 2, ... , n},   an element in it. Then the determinant of   can be expanded along the k rows identified by   as follows:

 

where   is the sign of the permutation determined by   and  , equal to  ,   the square minor of   obtained by deleting from   rows and columns with indices in   and   respectively, and   (called the complement of  ) defined to be   ,   and   being the complement of   and   respectively.

This coincides with the theorem above when  . The same thing holds for any fixed k columns.

Computational expense

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The Laplace expansion is computationally inefficient for high-dimension matrices, with a time complexity in big O notation of O(n!). Alternatively, using a decomposition into triangular matrices as in the LU decomposition can yield determinants with a time complexity of O(n3).[2] The following Python code implements the Laplace expansion:

def determinant(M):
    # Base case of recursive function: 1x1 matrix
    if len(M) == 1: 
        return M[0][0]

    total = 0
    for column, element in enumerate(M[0]):
        # Exclude first row and current column.
        K = [x[:column] + x[column + 1 :] for x in M[1:]]
        s = 1 if column % 2 == 0 else -1 
        total += s * element * determinant(K)
    return total

See also

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References

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  1. ^ Walter, Dan; Tytun, Alex (1949). "Elementary problem 834". American Mathematical Monthly. 56 (6). American Mathematical Society: 409. doi:10.2307/2306289. JSTOR 2306289.
  2. ^ Stoer Bulirsch: Introduction to Numerical Mathematics
  • David Poole: Linear Algebra. A Modern Introduction. Cengage Learning 2005, ISBN 0-534-99845-3, pp. 265–267 (restricted online copy, p. 265, at Google Books)
  • Harvey E. Rose: Linear Algebra. A Pure Mathematical Approach. Springer 2002, ISBN 3-7643-6905-1, pp. 57–60 (restricted online copy, p. 57, at Google Books)