Determinants opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 1 / 25
Notation The determinant of a square matrix n n A is denoted det(a) or A. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 2 / 25
Definition of Determinant The determinant of n n A = [a ij ] is defined as n n A = n! ( 1) m(p k) k=1 n i=1 a ipk (i), where p 1,..., p n! are the n! distinct bijections (one-to-one and onto functions) from {1,..., n} to {1,..., n} and m(p k ) = n 1 n i=1 i =i+1 1[p k (i) > p k (i )]. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 3 / 25
Example: n = 2 Suppose Then n! = 2, [ ] a11 a 12 A =. a 21 a 22 p 1 (1) = 1 p 1 (2) = 2 m(p 1 ) = 0 p 2 (1) = 2 p 2 (2) = 1 m(p 2 ) = 1, and A = 2 k=1 ( 1) m(p k) 2 i=1 = a 11 a 22 a 12 a 21. a ipk (i) opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 4 / 25
Example: n = 3 Find an expression for the determinant of a 11 a 12 a 13 A = a 21 a 22 a 23. a 31 a 32 a 33 opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 5 / 25
Example: n = 3 n! = 6, p 1 (1) = 1 p 1 (2) = 2 p 1 (3) = 3 m(p 1 ) = 0 p 2 (1) = 1 p 2 (2) = 3 p 2 (3) = 2 m(p 2 ) = 1 p 3 (1) = 2 p 3 (2) = 1 p 3 (3) = 3 m(p 3 ) = 1 p 4 (1) = 2 p 4 (2) = 3 p 4 (3) = 1 m(p 4 ) = 2 p 5 (1) = 3 p 5 (2) = 1 p 5 (3) = 2 m(p 5 ) = 2 p 6 (1) = 3 p 6 (2) = 2 p 6 (3) = 1 m(p 6 ) = 3, and A = a 11 a 22 a 33 a 11 a 23 a 32 a 12 a 21 a 33 +a 12 a 23 a 31 + a 13 a 21 a 32 a 13 a 22 a 31. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 6 / 25
Example: n = 1 Suppose A = [a 11 ]. Then n! = 1, p 1 (1) = 1 m(p 1 ) = 0, and A = 1 ( 1) m(p k) k=1 i=1 1 a ipk (i) = a 11. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 7 / 25
Verbal Description of Determinant omputation 1 Form all possible products of n matrix elements that contain exactly one element from each row and exactly one element from each column. 2 For each product, count the number of pairs of elements for which the lower element is to the left of the higher element. (For example, a 31 is lower than a 12 because a 31 is in the third row while a 12 is in the first row. Also, a 31 is to the left of a 12 because a 31 is in the first column while a 12 is in the second column.) If the number of such pairs is odd, multiply the product by 1. If the number of pairs is even, multiply the product by 1. 3 Add the signed products determined in steps 1 and 2. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 8 / 25
Prove that if A is an n n matrix and c R, then ca = c n A. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 9 / 25
Proof: ca = = n n! ( 1) m(p k) (ca ipk (i)) k=1 i=1 n! n ( 1) m(pk) c n a ipk (i) k=1 i=1 n! = c n = c n A. ( 1) m(p k) n k=1 i=1 a ipk (i) opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 10 / 25
Prove that if n n A is lower or upper triangular, then A = n a ii. (Note that this result implies A = n i=1 a ii for a diagonal matrix.) i=1 opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 11 / 25
Proof for A Lower Triangular: Suppose A is lower triangular. Then a ij = 0 whenever i < j. Thus, n i=1 a ip k (i) = 0 if i p k (i) > i. ecause p k is a bijection, p k (i) i i = 1,..., n p k (i) = i i = 1,..., n. Therefore, n! A = ( 1) m(p k) n n a ipk (i) = a ii. k=1 i=1 i=1 opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 12 / 25
Suppose A is an n n matrix. Prove that A = A. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 13 / 25
Proof: Let q k = p 1 k ; i.e., q k (j) = i p k (i) = j. Note that i < i, p k (i) = j > p k (i ) = j j < j, q k (j ) = i > q k (j) = i. Thus, m(p k ) = m(q k ). opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 14 / 25
Proof: It follows that A = = = n n! ( 1) m(p k) a ipk (i) k=1 i=1 n! n ( 1) m(p k) a qk (j)j k=1 j=1 n n! ( 1) m(q k) a qk (j)j k=1 j=1 = A. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 15 / 25
Find an expression for I n n m n = 0 n m m m I n n m n 0 n m m m. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 16 / 25
I n n m n 0 n m m m = opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 17 / 25
Likewise and n n n m 0 m n m m I n n n m 0 m n m m I I n n n m 0 m n m m =, =, =. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 18 / 25
More generally, it can be shown that n n m n n m 0 D m m = D and n n n m 0 m n m m D = D opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 19 / 25
Result A.18 (b): If A and are each n n matrices, then A = A. (We will not go through a proof here. Most graduate-level linear algebra textbooks contain a proof of this result.) opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 20 / 25
Use Result A.18 (b) to prove A nonsingular A 0. n n opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 21 / 25
Proof (= ): A A 1 = AA 1 = I = 1 A 0. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 22 / 25
Proof ( =): Suppose A is singular. We will show A = 0. A singular b 0 Ab = 0. Let n n be a nonsingular matrix whose first column is b. (We know such a matrix exists by Fact V5.) Then A = A = 0 because the first column of A is 0. Dividing both sides by 0 yields A = 0. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 23 / 25
Determinant of a Partitioned Matrix If A is nonsingular, A D = A D A 1. If D is nonsingular, A D = D A D 1. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 24 / 25
Proof: A D = I A D = AA 1 A D = A A 1 A D = A A D A 1 = A A D A 1 A 1 0 I = A I 0 A 1 D A 1 = A D A 1. A similar argument proves the second result. opyright c 2012 Dan Nettleton (Iowa State University) Statistics 611 25 / 25