Lecture Notes 1: Matrix Algebra Part B: Determinants and Inverses

Transcription

1 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 1 of 75 Lecture Notes 1: Matrix Algebra Part B: Determinants and Inverses Peter J. Hammond revised 2016 September 18th

2 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 2 of 75 Lecture Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

3 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 3 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

4 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 4 of 75 Permutations and Their Signs Definition Given N n = {1,..., n} where n 2, a permutation of N n is a bijective mapping π : N n N n. The family Π n of all permutations of N n includes: the identity mapping ι defined by ι(h) = h for all h N n ; for each π Π n, a unique inverse π 1 Π n for which π 1 π = π π 1 = ι Definition 1. Given any permutation π Π n, an inversion of π is a pair (i, j) N n such that i > j and π(i) < π(j). 2. A permutation π : N n N n is either even or odd according as it has an even or odd number of inversions. 3. The sign or signature of a permutation π, denoted by sgn(π), is defined as: (i) +1 if π is even; (ii) 1 if π is odd.

5 A Product Rule for the Signs of Permutations Theorem For any two permutations π, ρ Π n, one has sgn(π ρ) = sgn(π) sgn(ρ) The proof on the next slides uses the following: Definition First, define the signum or sign function { +1 if x > 0 R \ {0} x s(x) := 1 if x < 0 Next, for each permutation π Π n, let S(π) denote the matrix { 1 if i > j and π(i) < π(j) whose elements satisfy s ij (π) = +1 otherwise Finally, let S(π) := n i=1 n j=1 s ij(π) { 1, +1}. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 5 of 75

6 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 6 of 75 A Product Formula for the Sign of a Permutation Lemma For all π Π n one has sgn(π) = i>j s(π(i) π(j)) = S(π). Proof. Let p := #{(i, j) N n N n i > j & π(i) < π(j)} denote the number of inversions of π. By definition, sgn(π) = ( 1) p = ±1 according as p is even or odd. But the definitions on the previous slide imply that p = #{(i, j) N n N n i > j & s(π(i) π(j)) = 1} = #{(i, j) N n N n s ij (π) = 1} Therefore sgn(π) = ( 1) p = i>j s(π(i) π(j)) = n n i=1 j=1 s ij(π) = S(π)

7 Proving the Product Rule Suppose the three permutations π, ρ, σ Π n satisfy σ = π ρ. Then sgn(σ) S(σ) sgn(ρ) = = n n s ij (σ) S(ρ) i=1 j=1 s ij (ρ) and also s(σ(i) σ(j)) = s(π(ρ(i)) π(ρ(j))). The definition of the two matrices S(σ) and S(ρ) implies that their elements satisfy s ij (σ)/s ij (ρ) = 1 whenever i j. In particular, s ij (σ)/s ij (ρ) = 1 unless both i > j and also s(σ(i) σ(j)) = s(ρ(i) ρ(j)). Hence, given any i, j N n with i > j, one has s ij (σ)/s ij (ρ) = 1 if and only if s(π(ρ(i)) π(ρ(j))) = s(ρ(i) ρ(j)), or equivalently, if and only if: either ρ(i) > ρ(j) and (ρ(i), ρ(j)) is an inversion of π; or ρ(i) < ρ(j) and (ρ(j), ρ(i)) is an inversion of π. Let p denote the number of inversions of the permutation π. Then sgn(σ)/ sgn(ρ) = ( 1) p = sgn(π), implying that sgn(σ) = sgn(π) sgn(ρ). University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 7 of 75

8 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 8 of 75 Transpositions For each disjoint pair k, l {1, 2,..., n}, the transposition mapping i τ kl (i) on {1, 2,..., n} is the permutation defined by l if i = k; τ kl (i) := k if i = l; i otherwise; That is, τ kl transposes the order of k and l, leaving all i {k, l} unchanged. Evidently τ kl = τ lk and τ kl τ lk = ι, the identity permutation, and so τ τ = ι for every transposition τ. It is also evident that τ has exactly one inversion, so sgn(τ) = 1.

9 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 9 of 75 Transposition is Not Commutative Exercise Show that transpositions defined on a set containing more than two elements may not commute because, for example, τ 12 τ 23 = π 231 τ 23 τ 12 = π 312

10 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 10 of 75 Permutations are Products of Transpositions Theorem Any permutation π Π n on N n := {1, 2,..., n} is the product of at most n 1 transpositions. We will prove the result by induction on n. As the induction hypothesis, suppose the result holds for permutations on N n 1. Any permutation π on N 2 := {1, 2} is either the identity, or the transposition τ 12, so the result holds for n = 2.

11 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 11 of 75 Proof of Induction Step For general n, let j := π 1 (n) denote the element that π moves to the end. By construction, the permutation π τ jn must satisfy π τ jn (n) = π(τ jn (n)) = π(j) = n. So the restriction π of π τ jn to N n 1 is a permutation on N n 1. By the induction hypothesis, for all k N n 1, there exist transpositions τ 1, τ 2,..., τ q such that π(k) = π τ jn (k) = τ 1 τ 2... τ q (k) where q n 2 is the number of transpositions in the product. For p = 1,..., q, because τ p interchanges only elements of N n 1, one can extend its domain to include n by letting τ p (n) = n. Then π τ jn (k) = τ 1 τ 2... τ q (k) for k = n as well, so π = (π τ jn ) τ 1 jn = τ 1 τ 2... τ q τ 1 jn. Hence π is the product of at most q + 1 n 1 transpositions. This completes the proof by induction on n.

12 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 12 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

13 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 13 of 75 Permutation Matrices: Definition Definition For each permutation π Π on {1, 2,..., n}, let P π denote the unique associated n-dimensional permutation matrix which is derived by applying π to the rows of the identity matrix I n. That is, for each i = 1, 2,..., n, the ith row vector of the identity matrix I n is moved to become row π(i) of P π. This definition implies that the only nonzero element in row i of P π occurs not in column j = i, as it would in the identity matrix, but in column j = π 1 (i) where i = π(j). Hence the matrix elements p π ij of P π are given by p π ij = δ i,π(j) for i, j = 1, 2,..., n.

14 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 14 of 75 Permutation Matrices: Examples Example There are two 2 2 permutation matrices, which are given by: ( ) P 12 = I 2 ; P =. 1 0 Their signs are respectively +1 and 1. There are 3! = 6 permutation matrices in 3 dimensions given by: P 123 = I 3 ; P 132 = ; P 213 = ; P 231 = ; P 312 = ; P 321 = Their signs are respectively +1, 1, 1, +1, +1 and 1.

15 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 15 of 75 Permutation Matrices: Exercise Exercise Suppose that π, ρ are permutations in Π, whose composition is the function π ρ defined by Show that: {1, 2,..., n} i (π ρ)(i) = π(ρ(i)) {1, 2,..., n} 1. the mapping i π(ρ(i)) is a permutation on {1, 2,..., n}; 2. the associated permutation matrices satisfy P π ρ = P π P ρ. Then use result 2 to prove by induction on q that, if the permutation π equals the q-fold composition π 1 π 2 π q, then P π equals the q-fold matrix product P π1 P π2 P πq.

16 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 16 of 75 Transposition Matrices A special case of a permutation matrix is a transposition T h,i of rows h and i. As the matrix I with rows h and i transposed, it satisfies δ rs if r {h, i} (T h,i ) rs = δ is if r = h if r = i Exercise Prove that: δ hs 1. any transposition matrix T = T h,i is symmetric; 2. T h,i = T i,h ; 3. T h,i T i,h = T i,h T h,i = I.

17 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 17 of 75 More on Permutation Matrices Theorem Any permutation matrix P = P π satisfies: 1. P = q s=1 Ts for the product of some collection of q n 1 transposition matrices. 2. PP = P P = I Proof. The permutation π is the composition τ 1 τ 2 τ q of q n 1 transpositions τ s (for s S := {1, 2,..., q}). As shown previously, P π = q s=1 Ts where T s = P τ s for s S. Then, because each T s is symmetric, the transpose (P π ) equals the reversed product T q T 2 T 1. But each transposition T s also satisfies T s T s = I, so PP = T 1 T 2 T q T q T 2 T 1 = T 1 T 2 T q 1 T q 1 T 2 T 1 which equals I by induction on q, and similarly P P = I.

18 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 18 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

19 A First Elementary Row Operation Suppose that row r of the m m identity matrix I m is multiplied by a scalar α R, leaving all other rows unchanged. This gives the m m diagonal matrix S r (α) = diag(1, 1,..., 1, α, 1,..., 1) whose diagonal elements are 1, except the rth element which is α. Hence the elements of S r (α) satisfy { δ ij (S r (α)) ij = αδ ij if i r if i = r Exercise For the particular m m matrix S r (α) and the general m n matrix A, show that the transformed m n matrix S r (α)a is the result of multiplying row r of A by the scalar α, leaving all other rows unchanged. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 19 of 75

20 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 20 of 75 A Second Elementary Row Operation Suppose a multiple of α times row q of the identity matrix I m is added to its rth row, leaving all the other m 1 rows unchanged. Provided that q r, the resulting m m matrix E r+αq equals I m, but with an extra non-zero element equal to α in the (r, q) position. Its elements therefore satisfy (E r+αq ) ij = δ ij + αδ ir δ jq. Exercise For the particular m m matrix E r+αq and the general m n matrix A, show that the transformed m n matrix E r+αq A is the result of adding the multiple of α times its row q to the rth row of matrix A, leaving all other rows unchanged.

21 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 21 of 75 Levi-Civita Lists For any n N, define the set N n := {1, 2,..., n} of the first n natural numbers. Definition A Levi-Civita list j = (j 1, j 2,..., j n ) (N n ) n is a representation of a mapping N n i j i N n := {1, 2,..., n}. This list can be regarded as a mapping from the row numbers of an n n matrix to the column numbers of the same matrix.

22 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 22 of 75 Levi-Civita Symbols Definition The Levi-Civita function (N n ) n j ɛ j { 1, 0, 1} of order n maps each Levi-Civita list into an associated value called its Levi-Civita symbol. This value depends on whether the mapping N n i j i N n is an even or an odd permutation of the ordered list (1, 2,..., n), or is not a permutation at all. Specifically, +1 if i j i is an even permutation ɛ j = ɛ j1 j 2...j n := 1 if i j i is an odd permutation 0 if i j i is not a permutation

23 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 23 of 75 The Levi-Civita Matrix Definition Given the Levi-Civita mapping N n i j i N n := {1, 2,..., n}, the associated n n Levi-Civita matrix L j has elements defined by (L j ) rs = (L j1 j 2...j n ) rs := δ jr,s This implies that the rth row of L j equals row j r of the matrix I n. That is, L j = (e j1, e j2,..., e jn ) is the n n matrix produced by stacking the n row vectors e j r (r = 1, 2,..., n) of the canonical basis on top of each other, with repetitions allowed. For a general n n matrix A, the matrix L j A = L j1 j 2...j n A is the result of stacking the n row vectors a j r (r = 1, 2,..., n) of A on top of each other, with repetitions allowed. Specifically, (L j A) rs = a jr s for all r, s {1, 2,..., n}.

24 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 24 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

25 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 25 of 75 Determinants of Order 2: Definition Consider again the pair of linear equations a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 12 x 2 = b 2 with its associated coefficient matrix ( ) a11 a A = 12 a 21 a 22 Let us define the number D := a 11 a 22 a 21 a 12. Provided that D 0, the equations have a unique solution given by x 1 = 1 D (b 1a 22 b 2 a 12 ), x 2 = 1 D (b 2a 11 b 1 a 21 ) The number D is called the determinant of the matrix A. It is denoted by either det(a), or more concisely, by A.

26 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 26 of 75 Determinants of Order 2: Simple Rule Thus, for any 2 2 matrix A, its determinant D is A = a 11 a 12 a 21 a 22 = a 11a 22 a 21 a 12 For this special case of order 2 determinants, a simple rule is: 1. multiply the diagonal elements together; 2. multiply the off-diagonal elements together; 3. subtract the product of the off-diagonal elements from the product of the diagonal elements. Exercise Show that the determinant satisfies 1 0 A = a 11 a a a

27 Cramer s Rule in the 2 2 Case Using determinant notation, the solution to the equations a 11 x 1 + a 12 x 2 = b 1 a 21 x 1 + a 12 x 2 = b 2 can be written in the alternative form x 1 = 1 D b 1 a 12 b 2 a 22, x 2 = 1 D a 11 b 1 a 21 b 2 This accords with Cramer s rule for the solution to Ax = b, which is the vector x = (x i ) n i=1 each of whose components x i is the fraction with: 1. denominator equal to the determinant D of the coefficient matrix A (provided, of course, that D 0); 2. numerator equal to the determinant of the matrix (A i, b) formed from A by replacing its ith column with the b vector of right-hand side elements, while keeping all the columns in their original order. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 27 of 75

28 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 28 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

29 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 29 of 75 Determinants of Order 3: Definition Determinants of order 3 can be calculated from those of order 2 according to the formula a A = a 22 a a 32 a 33 a 12 a 21 a 23 a 31 a 33 + a 13 a 21 a 22 a 31 a 32 = 3 j=1 ( 1)1+j a 1j C 1j where, for j = 1, 2, 3, the 2 2 matrix C 1j is the (1, j)-cofactor obtained by removing both row 1 and column j from A. The result is the following sum A = a 11 a 22 a 33 a 11 a 23 a 32 + a 12 a 23 a 31 a 12 a 21 a 33 + a 13 a 21 a 32 a 13 a 22 a 31 of 3! = 6 terms, each the product of 3 elements chosen so that each row and each column is represented just once.

30 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 30 of 75 Determinants of Order 3: Cofactor Expansion The determinant expansion A = a 11 a 22 a 33 a 11 a 23 a 32 + a 12 a 23 a 31 a 12 a 21 a 33 + a 13 a 21 a 32 a 13 a 22 a 31 is very symmetric, suggesting (correctly) that the cofactor expansion along the first row (a 11, a 12, a 13 ) A = 3 j=1 ( 1)1+j a 1j C 1j gives the same answer as the other cofactor expansions A = 3 along, respectively: j=1 ( 1)r+j a rj C rj = 3 the rth row (a r1, a r2, a r3 ) the sth column (a 1s, a 2s, a 3s ) i=1 ( 1)i+s a is C is

31 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 31 of 75 Determinants of Order 3: Alternative Expressions The same result A = a 11 a 22 a 33 a 11 a 23 a 32 + a 12 a 23 a 31 a 12 a 21 a 33 + a 13 a 21 a 32 a 13 a 22 a 31 can be obtained as either of the two expansions 3 3 A = 3 ɛ j j 1 =1 j 2 =1 j 3 =1 1 j 2 j 3 a 1j1 a 2j2 a 3j3 = sgn(π) 3 π Π i=1 a iπ(i) Here ɛ j = ɛ j1 j 2 j 3 { 1, 0, 1} denotes the Levi-Civita symbol associated with the mapping i j i from {1, 2, 3} into itself. Also, Π denotes the set of all 3! = 6 possible permutations on {1, 2, 3}, with typical member π, whose sign is denoted by sgn(π).

32 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 32 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

33 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 33 of 75 The Determinant Function Let D n denote the domain R n n of n n matrices. When n = 1, 2, 3, the determinant mapping D n A A R specifies the determinant A of each n n matrix A as a function of its n row vectors (a i ) n i=1. For a general natural number n N, consider any mapping defined on the domain D n. D n A D(A) = D ((a i ) n i=1) R Notation: Let D(A/b r ) denote the new value D(a 1,..., a r 1, b r, a r+1,..., a n ) of the function D after the rth row a r of the matrix A has been replaced by the new row vector b r.

34 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 34 of 75 Row Multilinearity Definition The function D n A D(A) of A s n rows (a i ) n i=1 is (row) multilinear just in case, for each row number i {1, 2,..., n}, each pair b i, c i R n of new versions of row i, and each pair of scalars λ, µ R, one has D(A/λb i + µc i ) = λd(a/b i ) + µd(a/c i ) Formally, the mapping R n a i D(A/a i ) R is required to be linear, for fixed each row i N n. That is, D is a linear function of the ith row vector a i on its own, when all the other rows a h (h i) are fixed.

35 The Three Characterizing Properties Definition The function D n A D(A) is alternating just in case for every transposition matrix T, one has D(TA) = D(A) i.e., interchanging any two rows reverses its sign. Definition The mapping D n A D(A) is of the determinant type just in case: 1. D is multilinear in its rows; 2. D is alternating; 3. D(I n ) = 1 for the identity matrix I n. Exercise Use the previous definitions of the determinant for n 3 to show that the mapping D n A A R is of the determinant type provided that n 3. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 35 of 75

36 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 36 of 75 First Implication of Multilinearity in the n n Case Lemma Suppose that D n A D(A) is multilinear in its rows. For any fixed B D n, the value of D(AB) can be expressed as the linear combination D(AB) = n n... j 1 =1 j 2 =1 n a 1j1 a 2j2 a njn D(L j1 j 2...j n B) j n=1 of its values at all possible matrices L j B = L j1 j 2...j n B := (b jr ) n r=1 whose rth row, for each r = 1, 2,..., n, equals the j r th row b jr of the matrix B.

37 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 37 of 75 Characterizing 2 2 Determinants 1. In the case of 2 2 matrices, the lemma tells us that multilinearity implies D(AB) = a 11 a 21 D(b 1, b 1 ) + a 11 a 22 D(b 1, b 2 ) + a 12 a 21 D(b 2, b 1 ) + a 12 a 22 D(b 2, b 2 ) where b 1 = (b 11, b 21 ) and b 2 = (b 12, b 22 ) are the rows of B. 2. If D is also alternating, then D(b 1, b 1 ) = D(b 2, b 2 ) = 0 and D(B) = D(b 1, b 2 ) = D(b 2, b 1 ), implying that D(AB) = a 11 a 22 D(b 1, b 2 ) + a 12 a 21 D(b 2, b 1 ) = (a 11 a 22 a 12 a 21 )D(B) 3. Imposing the additional restriction D(B) = 1 when B = I 2, we obtain the ordinary determinant D(A) = a 11 a 22 a 12 a Then, too, one derives the product rule D(AB) = D(A)D(B).

38 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 38 of 75 First Implication of Multilinearity: Proof Each element of the product C = AB satisfies c ik = n j=1 a ijb jk. Hence each row c i = (c ik ) n k=1 of C can be expressed as the linear combination c i = n j=1 a ijb j of B s rows. For each r = 1, 2,..., n and each arbitrary selection b j1,..., b jr 1 of r 1 rows from B, multilinearity therefore implies that D(b j1,..., b jr 1, c r, c r+1,..., c n ) = n j r =1 a ij r D(b j1,..., b jr 1, b jr, c r+1,..., c n ) This equation can be used to show, by induction on k, that D(C) = n n... j 1 =1 j 2 =1 n a 1j1 a 2j2 a kjk D(b j1,..., b jk, c k+1,..., c n ) j k =1 for k = 1, 2,..., n, including for k = n as the lemma claims.

39 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 39 of 75 Additional Implications of Alternation Lemma Suppose D n A D(A) is both row multilinear and alternating. Then for all possible n n matrices A, B, and for all possible permutation matrices P π, one has: 1. D(AB) = π Π n i=1 a iπ(i)d(p π B) 2. D(P π B) = sgn(π)d(b). 3. Under the additional assumption that D(I n ) = 1, one has: determinant formula: D(A) = π Π sgn(π) n i=1 a iπ(i); product rule: D(AB) = D(A)D(B)

40 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 40 of 75 First Additional Implication of Alternation: Proof Because D is alternating, one has D(B) = 0 whenever two rows of B are equal. It follows that for any matrix (b ji ) n i=1 = L jb whose n rows are all rows of the matrix B, one has D((b ji ) n i=1 ) = 0 unless these rows are all different. But if all the n rows of (b ji ) n i=1 = L jb are all different, there exists a permutation π Π such that L j B = P π B. Hence, after eliminating terms that are zero, one has D(AB) = n n j 1 =1 j 2 =1... n j a n=1 1j 1 a 2j2 a njn D((b jr ) n r=1 ) = n n j 1 =1 j 2 =1... n j a n=1 1j 1 a 2j2 a njn D(L j1 j 2...j n B) = n π Π i=1 a iπ(i)d(p π B) as stated in part 1 of the Lemma.

41 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 41 of 75 Second Additional Implication: Proof Because D is alternating, whenever T is a transposition matrix, one has D(TP π B) = D(P π B). Suppose that π = τ 1 τ q is one possible factorization of the permutation π as the composition of transpositions. But sgn(τ) = 1 for any transposition τ. So sgn(π) = ( 1) q by the product rule for signs of permutations. Note that P π = T 1 T 2 T q where T p denotes the permutation matrix corresponding to the transposition τ p, for each p = 1,..., q It follows that D(P π B) = D(T 1 T 2 T q B) = ( 1) q D(B) = sgn(π)d(b) as required.

42 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 42 of 75 Third Additional Implication: Proof In case D(I n ) = 1, applying parts 1 and 2 of the Lemma (which we have already proved) with B = I n gives immediately D(A) = π Π n i=1 a iπ(i)d(p π ) = π Π sgn(π) n But then, applying parts 1 and 2 of the Lemma for a general matrix B gives n i=1 a iπ(i) D(AB) = a iπ(i)d(p π B) π Π i=1 = sgn(π) n a iπ(i)d(b) = D(A)D(B) π Π i=1 as an implication of the first equality on this slide. This completes the proof of all three parts.

43 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 43 of 75 Formal Definition and Cofactor Expansion Definition The determinant A of any n n matrix A is defined so that D n A A is the unique (row) multilinear and alternating mapping that satisfies I n = 1. Definition For any n n determinant A, its rs-cofactor C rs is the (n 1) (n 1) determinant of the matrix C rs obtained by omitting row r and column s from A. The cofactor expansion of A along any row r or column s is A = n j=1 ( 1)r+j a rj C rj = n i=1 ( 1)i+s a is C is Exercise Prove that these cofactor expansions are valid, using the formula A = π Π n i=1 sgn(π)a iπ(i)

44 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 44 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

45 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 45 of 75 Eight Basic Rules (Rules A H of EMEA, Section 16.4) Let A denote the determinant of any n n matrix A. 1. A = 0 if all the elements in a row (or column) of A are A = A, where A is the transpose of A. 3. If all the elements in a single row (or column) of A are multiplied by a scalar α, so is its determinant. 4. If two rows (or two columns) of A are interchanged, the determinant changes sign, but not its absolute value. 5. If two of the rows (or columns) of A are proportional, then A = The value of the determinant of A is unchanged if any multiple of one row (or one column) is added to a different row (or column) of A. 7. The determinant of the product AB of two n n matrices equals the product A B of their determinants. 8. If α is any scalar, then αa = α n A.

46 The Transpose Rule 2: Verification The transpose rule 2 is key: for any statement about how A depends on the rows of A, there is an equivalent statement about how A depends on the columns of A. Exercise Verify Rule 2 directly for 2 2 and then for 3 3 matrices. Proof of Rule 2 The expansion formula implies that A = π Π sgn(π) n i=1 a iπ(i) = π Π sgn(π) n But the product rule for signs of permutations implies that sgn(π) sgn(π 1 ) = sgn(ι) = 1, with sgn(π) = ±1. Hence sgn(π 1 ) = 1/ sgn(π) = sgn(π). So, because π π 1 is a bijection, A = π 1 Π sgn(π 1 ) n j=1 a jπ 1 (j) = A after using the expansion formula with π replaced by π 1. j=1 a π 1 (j)j University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 46 of 75

47 Verification of Rule 6 Exercise Verify Rule 6 directly for 2 2 and then for 3 3 matrices. Proof of Rule 6 Recall the notation E r+αq for the matrix resulting from adding the multiple of α times row q of I to its rth row. Recall too that E r+αq A is the matrix that results from applying the same row operation to the matrix A. Finally, recall the formula A = n j=1 a rj C rj for the cofactor expansion of A along the rth row. The corresponding cofactor expansion of E r+αq A is then E r+αq A = n j=1 (a rj + αa qj ) C rj = A + α B where B is derived from A by replacing row r with row q. Unless q = r, the matrix B will have its qth row repeated, implying that B = 0 because the determinant is alternating. So q r implies E r+αq A = A for all α, which is Rule 6. University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 47 of 75

48 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 48 of 75 Verification of the Other Rules Apart from Rules 2 and 6, note that we have already proved the product Rule 7, whereas the interchange Rule 4 just restates alternation. Now that we have proved Rule 2, note that Rules 1 and 3 follow from multilinearity, applied in the special case when one row of the matrix is multiplied by a scalar. Also, the proportionality Rule 5 follows from combining Rule 4 with multilinearity. Finally, Rule 8, concerning the effect of multiplying all elements of a matrix by the same scalar, is easily checked because the expansion of A is the sum of many terms, each of which involves the product of exactly n elements of A.

49 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 49 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

50 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 50 of 75 Expansion by Alien Cofactors Expanding along either row r or column s gives A = n j=1 a rj C rj = n when one uses matching cofactors. Expanding by alien cofactors, however, from either the wrong row i r or the wrong column j s, gives 0 = n j=1 a rj C ij = n i=1 a is C is i=1 a is C ij This is because the answer will be the determinant of an alternative matrix in which: either row i has been duplicated and put in row r; or column j has been duplicated and put in column s.

51 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 51 of 75 The Adjugate Matrix Definition The adjugate (or (classical) adjoint ) adj A of an order n square matrix A has elements given by (adj A) ij = C ji. It is therefore the transpose (C + ) of the cofactor matrix C + whose elements (C + ) ij = C ij are the respective cofactors of A.

52 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 52 of 75 Main Property of the Adjugate Matrix Theorem For every n n square matrix A one has (adj A)A = A(adj A) = A I n Proof. The (i, j) elements of the two product matrices are respectively [(adj A)A] ij = n k=1 C ki a kj and [A(adj A)] ij = n k=1 a ik C jk These are both cofactor expansions, which are expansions by: alien cofactors in case i j, implying that both equal 0; matching cofactors in case i = j, implying that they equal A. Hence for each pair (i, j) one has [(adj A)A] ij = [A(adj A)] ij = A δ ij = A (I n ) ij

53 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 53 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

54 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 54 of 75 Minors: Definition Definition Given any m n matrix A, a minor (determinant) of order k is the determinant A i1 i 2...i k,j 1 j 2...j k of a k k submatrix (a ij ), with 1 i 1 < i 2 <... < i k m and 1 j 1 < j 2 <... < j k n, that is formed by selecting all the elements that lie both: in one of the chosen rows i r (r = 1, 2,..., k); in one of the chosen columns j s (s = 1, 2,..., k).

55 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 55 of 75 Minors: Some Examples, and Rank Example 1. In case A is an n n matrix: the whole determinant A is the only minor of order n; each of the n 2 cofactors C ij is a minor of order n In case A is an m n matrix: each element of the mn elements of the matrix is a minor of order 1; the number of minors of order k is ( ) ( ) m n m! n! = k k k!(m k)! k!(n k)! Exercise Verify that the set of elements that make up the minor A i1 i 2...i k,j 1 j 2...j k of order k is completely determined by its k diagonal elements a ih,j h (h = 1, 2,..., k). (These need not be diagonal elements of A).

56 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 56 of 75 Definition of Minor Rank Definition The (minor) rank of a matrix is the dimension of its largest non-zero minor determinant.

57 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 57 of 75 Principal and Leading Principal Minors Definition If A is an n n matrix, the minor A i1 i 2...i k,j 1 j 2...j k of order k is: a principal minor if i h = j h for h = 1, 2,..., k, implying that all its diagonal elements a ih j h are diagonal elements of A; a leading principal minor if its diagonal elements are the leading diagonal elements a hh (h = 1, 2,..., k). Exercise Explain why an n n determinant has: 1. 2 n 1 principal minors; 2. n leading principal minors.

58 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 58 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

59 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 59 of 75 Definition of Inverse Matrix Exercise Suppose that A is any invertible n n matrix for which there exist n n matrices B and C such that AB = CA = I. 1. By writing CAB in two different ways, prove that B = C. 2. Use this result to show that the equal matrices B = C, if they exist, must be unique. Definition The n n matrix X is the unique inverse of the invertible n n matrix A just in case AX = XA = I n. In this case we write X = A 1, so A 1 denotes the unique inverse. Big question: does the inverse exist?

60 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 60 of 75 Existence Conditions Theorem An n n matrix A has an inverse if and only if A 0, which holds if and only if at least one of the equations AX = I n and XA = I n has a solution. Proof. Provided that A 0, the identity (adj A)A = A(adj A) = A I n shows that the matrix X := (1/ A ) adj A is well defined and satisfies XA = AX = I n, so X is the inverse A 1. Conversely, if XA = I n has a solution, then the product rule for determinants implies that 1 = I n = XA = X A. Similarly if AX = I n has a solution. In either case one has A 0. The rest follows from the paragraph above.

61 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 61 of 75 Singularity versus Invertibility So A 1 exists if and only if A 0. Definition 1. In case A = 0, the matrix A is said to be singular; 2. In case A 0, the matrix A is said to be non-singular or invertible.

62 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 62 of 75 Example and Application to Simultaneous Equations Exercise ( ) ( Verify that A = = A 1 2 = C := by using direct multiplication to show that AC = CA = I 2. Example Suppose that a system of n simultaneous equations in n unknowns is expressed in matrix notation as Ax = b. Of course, A must be an n n matrix. Suppose A has an inverse A 1. Premultiplying both sides of the equation Ax = b by this inverse gives A 1 Ax = A 1 b, which simplifies to Ix = A 1 b. Hence the unique solution of the equation is x = A 1 b. )

63 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 63 of 75 Cramer s Rule: Statement Notation Given any m n matrix A, let [A j, b] denote the new m n matrix in which column j has been replaced by the column vector b. Evidently [A j, a j ] = A. Theorem Provided that the n n matrix A is invertible, the simultaneous equation system Ax = b has a unique solution x = A 1 b whose ith component is given by the ratio of determinants x i = [A i, b] / A. This result is known as Cramer s rule.

64 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 64 of 75 Cramer s Rule: Proof Proof. Given the equation Ax = b, each cofactor C ij of the coefficient matrix A is formed by dropping row i and column j of A. It therefore equals the (i, j) cofactor of the matrix [A j, b]. Expanding the determinant by cofactors along column j therefore gives [A j, b] = n i=1 b i C ij = n by definition of the adjugate matrix. i=1 (adj A) jib i Hence the unique solution to the equation system has components x i = (A 1 b) i = 1 n A (adj A) jib i = 1 i=1 A [A i, b] for i = 1, 2,..., n.

65 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 65 of 75 Rule for Inverting Products Theorem Suppose that A and B are two invertible n n matrices. Then the inverse of the matrix product AB exists, and is the reverse product B 1 A 1 of the inverses. Proof. Using the associative law for matrix multiplication repeatedly gives: and (B 1 A 1 )(AB) = B 1 (A 1 A)B = B 1 (I)B = B 1 (IB) = B 1 B = I (AB)(B 1 A 1 ) = A(BB 1 )A 1 = A(I)A 1 = (AI)A 1 = AA 1 = I. These equations confirm that X := B 1 A 1 is the unique matrix satisfying the double equality (AB)X = X(AB) = I.

66 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 66 of 75 Rule for Inverting Chain Products Exercise Prove that, if A, B and C are three invertible n n matrices, then (ABC) 1 = C 1 B 1 A 1. Then use mathematical induction to extend this result in order to find the inverse of the product A 1 A 2 A k of any finite chain of invertible n n matrices.

67 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 67 of 75 Matrices for Elementary Row Operations Example Consider the following two out of the three possible kinds of elementary row operation: 1. of multiplying the rth row by α R, represented by the matrix S r (α); 2. of multiplying the qth row by α R, then adding the result to row r, represented by the matrix E r+αq. Exercise Find the determinants and, when they exist, the inverses of the matrices S r (α) and E r+αq.

68 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 68 of 75 Outline More Special Matrices Permutations and Transpositions Permutation and Transposition Matrices Elementary Row Operations Determinants Determinants of Order 2 Determinants of Order 3 Characterizing the Determinant Function Rules for Determinants Expansion by Alien Cofactors and the Adjugate Matrix Minor Determinants The Inverse Matrix Definition and Existence Orthogonal Matrices

69 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 69 of 75 Inverting Orthogonal Matrices An n-dimensional square matrix Q is said to be orthogonal just in case its columns (q j ) n j=1 form an orthonormal set i.e., they must be pairwise orthogonal unit vectors (of length 1). Theorem A square matrix Q is orthogonal if and only if it satisfies Q Q = I, and so if and only if Q 1 = Q. Proof. The elements of the matrix product Q Q satisfy (Q Q) ij = n k=1 q ik q kj = n k=1 q kiq kj = q i q j where q i (resp. q j ) denotes the ith (resp. jth) column vector of Q. But the columns of Q are orthonormal iff q i q j = δ ij for all i, j = 1, 2,..., n, and so iff Q Q = I.

70 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 70 of 75 Exercises on Orthogonal Matrices Exercise Show that if the matrix Q is orthogonal, then so is Q. Use this result to show that a matrix is orthogonal if and only if its row vectors also form an orthonormal set. Exercise Show that any permutation matrix is orthogonal.

71 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 71 of 75 Rotations in R 2 Example In R 2, consider the anti-clockwise rotation through an angle θ of the unit circle S 1 = {(x 1, x 2 ) R 2 x x 2 2 = 1}. It maps: 1. the first unit vector (1, 0) of the canonical basis to the column vector (cos θ, sin θ) ; 2. the second unit vector (0, 1) of the canonical basis to the column vector ( sin θ, cos θ). So the rotation can be represented by the rotation matrix ( ) cos θ sin θ R θ := sin θ cos θ which has these vectors as its columns.

72 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 72 of 75 Rotations in R 2 Are Orthogonal Matrices Because sin( θ) = sin(θ) and cos( θ) = cos(θ), the transpose of R θ satisfies R θ = R θ, and so is the clockwise rotation through an angle θ of the unit circle S 1. Since clockwise and anti-clockwise rotations are inverse operations, it is no surprise that R θ R θ = I. We verify this algebraically by using matrix multiplication ( ) ( ) ( ) R cos θ sin θ cos θ sin θ 1 0 θ R θ = = = I sin θ cos θ sin θ cos θ 0 1 because cos 2 θ + sin 2 θ = 1, thus verifying orthogonality. Similarly ( R θ R cos θ sin θ θ = sin θ cos θ ) ( ) cos θ sin θ = sin θ cos θ ( ) 1 0 = I 0 1

73 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 73 of 75 The Unit Sphere in R 3 In R 3, the three unit vectors (1, 0, 0), (0, 1, 0) and (0, 0, 1) of the canonical basis represent respectively 1. one point (1, 0, 0) on the equatorial great circle, whose latitude is 0, and whose longitude is also 0 (the Greenwich Meridian); 2. a second point (0, 1, 0) on the equatorial great circle, whose latitude is also 0, and whose longitude is 90 East; 3. the North pole (0, 0, 1) at latitude 90 North, whose longitude is undefined. The unit sphere is the set S 2 = {(x 1, x 2, x 3 ) R 3 x x x 2 3 = 1} = {x R 3 x 2 = 1}

74 Spherical or Polar Coordinates in R 3 x = r sin ϕ cos θ, y = r sin ϕ sin θ, z = r cos ϕ r = ( ) x 2 + y 2 + z 2, θ = arctan(y/x), ϕ = arctan x 2 + y 2 /z Spherical coordinates (r, θ, φ) as often used in ϕ is the polarmathematics: angle radial (η = 1 distance r, azimuthal angle θ, and 2π ϕ is latitude ); More details Dmcq Own work θ is the azimuthal angle ( longitude ). 1/2 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 74 of 75

75 University of Warwick, EC9A0 Maths for Economists Peter J. Hammond 75 of 75 Orthogonal Matrices: Rotations in R 3 Example A rotation in R 3 is described by two angles, denoted by θ and η. Representing a rotation through the angle θ about the z-axis: cos θ sin θ 0 R θ := sin θ cos θ Representing a rotation through the angle η about the y-axis: cos η 0 sin η S η := sin η 0 cos η Exercise Calculate the two matrix products R θ S η and S η R θ. Show that they are equal iff either θ = 0 or η = 0.