Mathematics_1_oneside.pdf

# Product the determinant of the product of two

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Product. The determinant of the product of two matrices equals the Theorem 10.9 product of their determinants, i.e., det( A · B ) = det( A ) · det( B ) . P ROOF . Let A and B be two n × n matrices. If A does not have full rank, then rank( A ) < n and Lemma 10.3 implies det( A ) = 0 and thus det( A ) · det( B ) = 0. On the other hand by Theorem 6.23 rank( AB ) rank( A ) < n and hence det( AB ) = 0. If A has full rank, then the columns of A form a basis of R n and we find for the columns of AB , [ AB ] j = n i = 1 b i j a i . Consequently, Lemma 10.5 and Theorem 10.6 immediately imply det( AB ) = det( a 1 ,..., a n ) X σ S n sgn( σ ) n Y i = 1 b σ ( i ) , i = det( A ) · det( B ) as claimed. Singular matrix. Let A be an n × n matrix. Then the following are Theorem 10.10 equivalent: (1) det( A ) = 0. (2) The columns of A are linearly dependent. (3) A does not have full rank. (4) A is singular. P ROOF . The equivalence of (2), (3) and (4) has already been shown in Section 6.3 . Implication (2) (1) is stated in Lemma 10.3 . For implica- tion (1) (4) see Problem 10.14 . This finishes the proof. An n × n matrix A is invertible if and only if det( A ) 6= 0. Corollary 10.11 We can use the determinant to estimate the rank of a matrix.

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10.4 E VALUATION OF THE D ETERMINANT 78 Rank of a matrix. The rank of an m × n matrix A is r if and only if there Theorem 10.12 is an r × r subdeterminant fl fl fl fl fl fl fl a i 1 j 1 ... a i 1 j r . . . . . . . . . a i r j 1 ... a i r j r fl fl fl fl fl fl fl 6= 0 but all ( r + 1) × ( r + 1) subdeterminants vanish. P ROOF . By Gauß elimination we can find an invertible r × r submatrix but not an invertible ( r + 1) × ( r + 1) submatrix. Inverse matrix. The determinant of the inverse of a regular matrix is Theorem 10.13 the reciprocal value of the determinant of the matrix, i.e., det( A - 1 ) = 1 det( A ) . P ROOF . See Problem 10.15 . Finally we return to the volume of a parallelepiped which we used as motivation for the definition of the determinant. Since we have no formal definition of the volume yet, we state the last theorem without proof. Volume. Let a 1 ,..., a n R n . Then the volume of the n -dimensional Theorem 10.14 parallelepiped created by these vectors is given by the absolute value of the determinant, Vol( a 1 ,..., a n ) = fl fl det( a 1 ,..., a n ) fl fl . 10.4 Evaluation of the Determinant Leibniz formula ( 10.1 ) provides an explicit expression for evaluating the determinant of a matrix. For small matrices one may expand sum and products and finds an easy to use scheme, known as Sarrus’ rule (see Problems 10.17 and 10.18 ): fl fl fl fl a 11 a 12 a 21 a 22 fl fl fl fl = a 11 a 22 - a 21 a 12 . fl fl fl fl fl fl a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33 fl fl fl fl fl fl = a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 - a 31 a 22 a 13 - a 32 a 23 a 11 - a 33 a 21 a 12 . (10.2) For larger matrices Leibniz formula ( 10.1 ) expands to much longer expressions. For an n × n matrix we find a sum of n ! products of n factors. However, for triangular matrices this formula reduces to the product of the diagonal entries, see Problem 10.19 .
10.4 E VALUATION OF THE D ETERMINANT 79 Triangular matrix.

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