The Laplace expansion is a formula for calculating determinants that is calculationally more efficient than the Leibniz formula but less efficient than the decomposition into triangular matrices. See if you like the proof below or the one that heavily relies on permutations.
In fact, we have already dealt with the Laplace expansion when we said that is a linear function of column
Our task is to analyze the coefficients They will be identified as determinants of certain submatrices of up to the sign.
For simplicity, let us start with
Definition. For any denote the matrix resulting from after deleting row and column is of size Its determinant is called -minor.
Step 1. We want to show that By the cross-out rule depends only on the elements of When studying we can assume without loss of generality that
Then (2) becomes This allows us to assert that satisfies Axioms 1-3:
1) is a homogeneous function of any row of because has this property,
2) Adding one of the rows of to another row does not change because stays the same, and
3) If then and
Since, again, is a function of elements of only, it follows that
Step 2. To analyze as above, we can assume that
(2) becomes Here where is the second unit column-vector and are columns of Let be the result of permuting the first and second rows of This permutation changes the sign of and does not change (the matrix for is the same as for ). Hence,
For is the same as for in Step 1, so (3) and (4) imply
Step We can assume that
(2) becomes where and is the -th unit column-vector. As in Step 2, we can reduce this case to Step 1. Permute rows and then permute rows and and so on. In total we need permutations leading to changes in sign. Instead of (4) we get
where is the result of permutations. The element in the upper left corner of is 1 and therefore is the same as for in Step 1. The conclusion is that
To consider (1), we reduce the case of the -th column to the case of the first column. For this, we permute columns and then columns and and so on. In total we need permutations. Denoting the new matrix we have To we can apply what we know about (2). This will lead to multiplying (5) by The result is
Thus we have derived the Laplace expansion:
Theorem. For one has an expansion by column
The meaning of this expansion is that the calculation of the determinant of is reduced to the calculation of the determinants of matrices of lower dimension. Instead of expansions by column, one can use expansions by row, whichever is convenient. The expression is called a cofactor of the element
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