Determinants Beifang Chen 1 Motivation Determinant is a function that each square real matrix A is assigned a real number, denoted det A, satisfying certain properties If A is a 3 3 matrix, writing A [u, v, w], we require the absolute value of the determinant det A to be the volume of the parallelepiped spanned by the vectors u, v, w Since volume is linear in each side of the parallelepiped, it is required that the determinant function det A satisfies the following properties: (a) Linearity in Each Column: det[u 1 + u 2, v, w] det[u 1, v, w] + det[u 2, v, w], det[cu, v, w] c det[u, v, w]; det[u, v 1 + v 2, w] det[u, v 1, w] + det[u, v 2, w], det[u, cv, w] c det[u, v, w]; det[u, v, w 1 + w 2 ] det[u, v, w 1 ] + det[u, v, w 2 ], det[u, v, cw] c det[u, v, w] (b) Vanishing Property: If two columns are the same, then the determinant is zero, ie, det[u, u, w] det[u, v, u] det[u, v, v] 0 (Actually, applying (a), it is easy to see that (b) is equivalent to requiring that det[u, v, w] 0 whenever u, v, w are linearly dependent) (c) Regularity: Determinant of identity matrix is 1 (The volume of unit cube is 1) u u v v+w O v 2v 3v O w Figure 1: Linearity of area in each column Definition 11 For an n n matrix A [a ij ], the determinant of A is a real number det A, satisfying the properties: 1 Linearity: det[v 1,, au i + bv i,, v n ] a det[v 1,, u i,, v n ] + b det[v 1,, v i,, v n ]; 1
2 Vanishing Property: or equivalently, det[v 1,, v i,, v i,, v n ] 0; 3 Regularity: det I n 1 det[v 1,, v i,, v j,, v n ] det[v 1,, v j,, v i,, v n ] 2 Determinant of 1 1 and 2 2 matrices For 1 1 matrices [a 11 ] we must have det[a 11 ] a 11 det[1] a 11 1 a 11 The determinant for a 2 2 matrix A [a ij ] must be defined as a 11 a 12 a 21 a 22 a 11a 22 a 12 a 21 In fact, write A [a 1, a 2 ] Then Thus a 1 a 11 e 1 + a 21 e 2, a 2 a 12 e 1 + a 22 e 2 det A det[a 11 e 1 + a 21 e 2, a 2 ] a 11 det[e 1, a 2 ] + a 21 det[e 2, a 2 ] a 11 a 12 det[e 1, e 1 ] + a 11 a 22 det[e 1, e 2 ] + a 21 a 12 det[e 2, e 1 ] + a 21 b 22 det[e 2, e 2 ] Since det[e 1, e 2 ] 1, det[e 1, e 1 ] det[e 2, e 2 ] 0, and det[e 2, e 1 ] 1, we then have 3 Determinant of 3 3 matrices det A a 11 a 22 a 12 a 21 The determinant for 3 3 matrices must be defined as det A a 11 a 12 a 13 a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 11 a 23 a 32 a 12 a 21 a 33 a 31 a 22 a 31 4 Properties of determinant Determinant can be equivalently defined by the same properties on rows instead of columns following, we state such properties in detail 1 Addition: a i1 + b i1 a in + b in + b i1 b in In the 2
2 Scalar multiplication: ca i1 ca in c 3 If two rows are the same, then determinant is 0, ie, 0 4 Switching two rows changes sign of determinant, ie, a j1 a jn a j1 a jn 5 Adding a multiple of one row to another row does not change determinant, ie, a j1 + ca i1 a jn + ca in a j1 a jn Property 4 and Property 5 follow from Properties 1-3 Proposition 41 For triangular matrix, a 11 a 12 a 13 a 1n 0 a 22 a 23 a 2n 0 0 a 33 a 3n 0 0 0 a nn a 11 a 22 a nn Proof If all a 11, a 22,, a nn are nonzero, then the matrix A can be reduced to a diagonal matrix whose diagonal entries are a 11, a 22,, a nn Thus det A det ( Diag[a 11, a 22,, a nn ] ) a 11 a 22 a nn 3
If some of a 11, a 22,, a nn are zero, we need to show det A 0 Let i be the smallest index such that a ii 0 If i 1, then the first column of A is zero Hence det A 0 So we may assume i 2 Since a 11, a 22,, a i 1,i 1 are nonzero, applying column operations the determinant of A equals the following: a 11 0 0 0 a 1n 0 a 22 0 0 a 2n 0 0 a i 1,i 1 0 a i 1,n 0 0 0 0 a in 0 0 0 0 a nn 0 Example 41 2 3 4 5 4 3 R 1 R 3 5 4 3 2 3 4 0 0 14 0 1 2 R 5 5R 1 R 2 2R 1 R 2 R 3 0 6 2 0 1 2 0 1 2 0 0 14 R 2 6R 3 14 Example 42 The Vandermonde determinant is a 1 a 2 a 3 R 2 a 1 R 1 a 2 1 a 2 2 a 2 R 3 a 1 R 2 0 a 2 a 1 a 3 a 1 3 0 a 2 2 a 1a 2 a 2 3 a 1a 3 0 a 2 a 1 a 3 a 1 R 3 a 2 R 2 0 a 2 (a 2 a 1 ) a 3 (a 3 a 1 ) 0 a 2 a 1 a 3 a 1 0 0 a 3 (a 3 a 1 ) a 2 (a 3 a 1 ) 0 a 2 a 1 a 3 a 1 0 0 (a 3 a 1 )(a 3 a 2 ) 5 Laplace expansion formula For a 3 3 matrix A [a ij ], by the addition property, a 11 a 12 a 13 a 11 0 0 (a 2 a 1 )(a 3 a 1 )(a 3 a 2 ) + 0 a 12 0 + 0 0 a 13 4
Note that a 11 0 0 1 0 0 a 11 a 1 0 0 11 0 a 22 a 23 0 a 32 a 33 1 0 0 a 11 a 22 0 1 a 23 0 0 a 33 + a 1 0 0 11a 32 0 0 a 23 0 1 a 33 1 0 0 a 11 a 22 0 1 0 0 0 a 33 a 1 0 0 11a 32 0 1 a 33 0 0 a 23 a a 11 a 22 a 33 a 11 a 23 a 32 a 22 a 23 11 a 32 a 33 ; 0 a 12 0 a 12 0 0 a 22 a 21 a 23 a 32 a 31 a 33 a 12 a 21 a 23 a 31 a 33 ; Thus a 11 a 12 a 13 0 0 a 13 a 11 a 22 a 23 a 32 a 33 a 13 0 0 a 23 a 21 a 22 a 33 a 31 a 32 a 13 a 21 a 22 a 31 a 32 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 Theorem 51 (Laplace Expansion) Let A [a ij ] be an n n matrix Then for any fixed i, det A n ( 1) i+j a ij det A ij, (51) j1 where A ij is the (n 1) (n 1) submatrix obtained from A by deleting the ith row and the jth column The identity (51) is known as the Laplace expansion along the ith row The determinant det A ij is called the minor of the entry a ij We define the cofactor of a ij as C ij ( 1) i+j det A ij The classical adjoint of A is the matrix C 11 C 21 C n1 C 12 C 22 C n2 [ ] T adj A : C ij C 1n C 2n C nn Example 51 5 4 2 1 2 3 1 2 5 7 3 9 1 2 1 4 1 2 0 5 2 3 1 2 1 2 0 3 3 1 0 2 7 0 9 5 0 1 3 1 2 1 2 5 1 2 3 3 1 2 7 9 5 1 7 + 45 38 5
Theorem 52 For any n n matrix A [a ij ], In other words, If det A 0, then A is invertible and a i1 C j1 + a i2 C j2 + + a in C jn A adj A (det A) I n A 1 1 adj A det A { det A if i j, 0 if i j Proof Applying the cofactor expansion formula for A along the ith row, we have a i1 C i1 + a i2 C i2 + + a in C in det A Let M be the matrix obtained from A by replacing the jth row of A with the ith row of A Then det M 0 Applying the cofactor expansion formula for M along the jth row, we have a i1 C j1 + a i2 C j2 + + a in C jn 0, i j Corollary 53 A square matrix A is invertible if and only if det A 0 Example 52 For the matrix its adjoint matrix is given by Then Thus A 1 1 1 1 1 1 adj A A 1 6 Other properties of determinant Theorem 61 For any n n matrices A and B, 1 1 1 1 1 1 2 2 0 0 2 2 2 0 2 2 2 0 0 2 2 2 0 2, 1/2 1/2 0 0 1/2 1/2 1/2 0 1/2 det(ab) det A det B 4 0 0 0 4 0 0 0 4 Proof The proof is divided into three steps Step 1: det(ab) det A det B if A is not invertible We claim that AB is not invertible Suppose AB is invertible Then there is a matrix C such that (AB)C I, ie, A(BC) I This means that A is invertible, a contradiction Since det(ab) det B 0, we thus have det(ab) det A det B Step 2: det(eb) det E det B for elementary matrix E This follows from the interpretation for left multiplication of elementary matrices, and the properties of determinant Step 3: det(ab) det A det B if A is invertible 6
Since A is invertible, it can be written as product of elementary matrices, ie, A E 1 E 2 E k Applying Sept 2, we see that det(ab) det(e 1 E 2 E k B) det E 1 det(e 2 E k B) det E 1 det E 2 det E k det B Applying the same property, Hence det(ab) det A det B det A det E 1 det E 2 det E k Theorem 62 For any square matrix A, det A det A T 7 Cramer s rule Theorem 71 Let A [a ij ] be an invertible n n matrix Then the solution of the linear system Ax b is given by x i det A i, i 1, 2,, n, det A where A i is matrix obtained from A by replacing the ith column with b Proof Let X i be the matrix obtained from the identity matrix I n by replacing the ith column with x Then AX i A [ e 1,, e i 1, x, e i+1,, e n ] [ Ae 1,, Ae i 1, Ax, Ae i+1,, Ae n ] Then det(ax i ) det A i Note that It follows that x i det A i / det A [ a 1,, a i 1, b, a i+1,, a n ] Ai det(ax i ) det A det X i (det A)x i 8 Algebraic formula of determinant A permutation of {1, 2,, n} is a bijection σ : {1, 2,, n} {1, 2,, n} The set of all such permutations are denoted by S n ; the number of such permutations is n! A permutation σ can be described as ( ) 1 2 n σ or σ j j 1 j 2 j 1 j 2 j n n An inversion of σ is an ordered pair (j i, j k ) of integers such that j i > j k We denote by inv(σ) the set of all inversions of σ A permutation σ is called even (odd) if the number of inversions of σ is an even (odd) number We define { sgn σ ( 1) #inv(σ) 1 if σ is even, 1 if σ is odd For instance, the permutation 3241 of {1, 2, 3, 4} has inversions (3, 2), (3, 1), (2, 1), (4, 1); so 3241 is an even permutation The permutation 2341 has inversions (2, 1), (3, 1), (4, 1); so 2341 is an odd permutation 7
Theorem 81 For permutations σ, τ S n, we have Moreover, sgn σ 1 sgn σ sgn (τ σ) sgn (τ) sgn (σ) Proof Consider the polynomial ( xi x j ) g g(x 1,, x n ) i<j and the 1-dimensional vector space {cg c R} For any cg, we define the polynomial σ(cg) : c i<j ( xσ(i) x σ(j) ) Since σ is a permutation of {1, 2,, n}, we have σ ( {1, 2,, n} ) {1, 2,, n} Thus σ(cg) ±c i<j ( xi x j ) (sgn σ)cg Thus, on the one hand, On the other hand, It follows that sgn (τ σ) sgn (τ)sgn (σ) (τ σ)(g) ( sgn (τ σ) ) g (τ σ)(g) τ ( σ(g) ) τ ( (sgn σ)g ) (sgn τ)sgn (σ)g Theorem 82 For any n n matrix A [a ij ], det A sgn (σ)a i1,1a i2,2 a in,n σi 1 i 2 i n S n (sgn σ)a σ(1),1 a σ(2),2 a σ(n),n σ S n Proof Let A [a 1, a 2,, a n ] Then for each j, a j a 1j e 1 + a 2j e 2 + + a nj e n n a ij e i i1 By linearity, we have det A [ det ai1,1e i1, a i2,1e i2,, ] a in,ne in a i1,1a i2,2 a in,n det[e i1, e i2,, e in ] i 1,i 2,,i n If i 1 i 2 i n is not a permutation of {1, 2,, n}, then two of the vectors e i1, e i2,, e in are the same, then det[e i1, e i2,, e in ] 0 If σ i 1 i 2 i n is a permutation of {1, 2,, n}, then det[e i1, e i2,, e in ] ( 1) #inv(σ) sgn σ Therefore, det A σi 1 i 2 i n S n sgn (σ)a i1,1a i2,2 a in,n 8
The above algebraic formula is sometimes used to define determinant Theorem 83 For any n n matrix A [a ij ], det A T det A Proof Let B A T [b ij ] Then det B σ (sgn σ)b σ(1),1 b σ(2),2 b σ(n),n σ (sgn σ)a 1σ(1) a 2σ(2) a nσ(n) σ (sgn σ)a σ 1 (1),1a σ 1 (2),2 a σ 1 (n),n τ (sgn τ 1 )a τ(1),1 a τ(2),2 a τ(n),n det A (as sgn τ 1 sgn τ) Corollary 84 For any n n matrix A [a ij ], det A sgn (σ)a 1j1 a 2j2 a njn σj 1 j 2 j n S n (sgn σ)a 1σ(1) a 2σ(2) a nσ(n) σ S n For a 3 3 matrix A [a ij ], since we then have sgn (123) sgn (231) sgn (312) 1, sgn (132) sgn (213) sgn (321) 1, det A a 11 a 22 a 33 + a 12 a 23 a 31 + a 13 a 21 a 32 a 11 a 23 a 32 a 12 a 21 a 33 a 13 a 22 a 31 Theorem 85 Let T : R n R n be a linear transformation with the standard matrix A Let P be a parallelotope spanned by the vector v 1,, v n Then vol ( T (P ) ) det(a) vol (P ) Proof Let v i [v 1i, v 2i,, v ni ] T, ie, v i v 1i e 1 + v 2i e 2 + + v ni e n (1 i n) Then Thus T (v i ) v 1i T (e 1 ) + v 2i T (e 2 ) + + v ni T (e n ) [ T (v1 ), T (v 2 ),, T (v n ) ] [ T (e 1 ), T (e 2 ),, T (e n ) ] Therefore v 11 v 12 v 1n v 21 v 22 v 2n v n1 v n2 v nn vol ( T (P ) ) det [ T (v1 ), T (v 2 ),, T (v n ) ] det(a) det [ v1, v 2,, v n ] det(a) vol (P ) A[ ] v 1, v 2,, v n 9