Determinants: Introduction and existence

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Ernest Cody Allen
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1 Math 5327 Spring 2018 Determinants: Introduction and existence In this set of notes I try to give the general theory of determinants in a fairly abstract setting. I will start with the statement of the theorem, and then define the terms and notation before actually proving the theorem. Theorem 1 The determinant of and n n matrix A is the unique n-linear, alternating function from F n n to F that takes the identity to 1. Along the way to proving this theorem, we will prove Theorem 2 If D : F n n A, D(A) = det(a)d(i). F is n-linear and alternating, then for all n n matrices Determinants exist more generally than over fields. If you know the theory, we could replace F with any commutative ring with identity. We need to define some terms for Theorems 1 and 2 to make sense. Definition 1 Let V be a vector space over a field F. A function f : V V V F, where there are n copies of V in the product is called n-linear if for each index i, f(v 1, v 2,..., v i,..., v n ) is linear in v i when all the other arguments of f are held fixed. For example, a 2-linear function from V V to F, f(u, v), has the property that f(cu 1 + u 2, v) = cf(u 1, v) + f(u 2, v) and f(u, cv 1 + v 2 ) = cf(u, v 1 ) + f(u, v 2 ). That is, f is linear in u if v is held fixed, and also linear in v when u is fixed. We view the determinant as being a function ( on ) the n rows of A, and insist that it be linear in each row. For example, the function f = ac is 2-linear in the rows of the matrix. Here is a check for linearity in the second row of the matrix. Let (c, d) = k(w, x) + (y, z) = (kw + y, kx + z). We have ( ) ( ) f = f = a(kw + y) = k aw + ay. kw + y kx + z On the other hand, ( ) ( ) kf + f = k aw + ay. w x y z Since these agree, f is linear in the second row of A. You should be able to prove the following. Lemma 1 Any linear combination of n-linear functions is again n-linear. ( ) Thus, if I told you a second 2-linear function was g = bc then it would follow that ( ) h = 5f 7g, the function defined by h = 5ac 7bc is also 2-linear.

2 The other term we need to define is the term alternating. Definition 2 Let V be a vector space over a field F. A function f : V V V F, where there are n copies of V in the product is called alternating if (a) and f(v 1, v 2,..., v n ) = 0 whenever two of the vectors are equal, (b) f(v 1,..., v i,..., v j,... v n ) = f(v 1,..., v j,..., v i,... v n ). That is, if any v i = v j with i j then f = 0, and if we were to interchange v i and v j then f changes sign. If we define f on matrices, then as before, the v-vectors in f( are the ) rows of the matrix. An example of an alternating function on 2 2 matrices is f = b d. ( ) Another example, of course, is the usual determinant, g = ad bc. To show how the linearity and alternating conditions interact, let s prove theorems 1 and 2 when n = 2. ( ) Lemma 2 The determinant, det = ad bc is the unique 2-linear, alternating function satisfying det(i) = 1. Moreover, if D is any 2-linear, alternating function on 2 2 matrices, then D(A) = det(a)d(i) for all 2 2 matrices A. Proof: We do both parts at the same time. I will let you check that the usual determinant is, in fact, 2-linear and alternating. Suppose D is 2-linear and alternating. We use linearity in the first row, and then the second, using, for example, (a, b) = a(1, 0) + b(0, 1). We have ( ) ( ) ( ) 1 0 D = ad + bd c d = acd ( c ) + add d ( 1 0 ) + bcd ( ) + bdd 1 0 ( ). By alternating property (a), the first and last of the four terms are 0. By property (b), we can write ( ) ( ) ( ) 1 0 D = add + bcd 1 0 ( ) ( ) 1 0 = add bcd ( ) 1 0 = (ad bc)d = (ad bc)d(i), as desired. Page 2

3 With regard to alternating functions, some simplifications are possible. First, if we are in a field of characteristic other than 2, then the second condition implies the first. That is, in the two-argument case, suppose that for all u and v, f(u, v) = f(v, u). Then when u = v we have f(u, u) = f(u, u), or 2f(u, u) = 0. A field of characteristic 2 has the property that 2x = 0 for all x in the field. These fields can be annoying. We won t worry about such things, and say all our fields will not be like this. Certainly R, Q, C don t have this property. Since alternating property (b) implies (a) in characteristic 2 we can dispense with property (a). However, property (a) is easier to check than property (b). Here is the nicest result about alternating functions. Lemma 3 Suppose that f : V V V F in an n-linear function and for any i < n, f(v 1, v 2,..., v n ) = 0 if v i = v i+1. Then f is an alternating function. Proof: We proceed in a few steps. First, f changes sign if we interchange two adjacent vectors. This is the key part, and the part where n-linearity is needed. I will only show the case of interchanging v 1 and v 2. It goes like this: Consider f(v 1 + v 2, v 1 + v 2, v 3,..., v n ), which I will abbreviate f(v 1 + v 2, v 1 + v 2 ), ignoring positions 3 to n. By linearity in the first coordinate, f(v 1 + v 2, v 1 + v 2 ) = f(v 1, v 1 + v 2 ) + f(v 2, v 1 + v 2 ). Using linearity in the second coordinate, f(v 1 + v 2, v 1 + v 2 ) = f(v 1, v 1 ) + f(v 1, v 2 ) + f(v 2, v 1 ) + f(v 2, v 2 ). However, by hypothesis, f(v 1, v 1 ) = 0 and f(v 2, v 2 ) = 0. Moreover, f(v 1 + v 2, v 1 + v 2 ) is also 0 since the first vector equals the second. Thus, f(v 1, v 2 ) + f(v 2, v 1 ) = 0 or f(v 2, v 1 ) = f(v 1, v 2 ). Next, knowing f changes sign when we interchange adjacent arguments, we show that f changes sign when interchanging any two arguments. The trick is that any two arguments can be interchanged via a sequence of adjacent interchanges. For example, suppose we want to interchange b and e in f(a, b, c, d, e, f). Here is how that might look: f(a, b, c, d, e, f) = f(a, c, b, d, e, f) = ( 1) 2 f(a, c, d, b, e, f) = ( 1) 3 f(a, c, d, e, b, f) = ( 1) 4 f(a, c, e, d, b, f) = ( 1) 5 f(a, e, c, d, b, f). More generally, if we want to interchange v i and v j, it takes j i 1 adjacent interchanges to move v i into position j 1, one more and v i is in position j while v j is in position j 1 and j i 1 additional interchanges to move v j into the i th position, for a total of 2j 2i 1 transpositions. Each of these carries a sign of -1, and 2j 2i 1 is odd, so the net result is to change the sign of f. Finally, assuming the field does not have characteristic 2, as mentioned above, the sign changing property shows that if any two arguments in f are the same, then f = 0, completing the proof. Page 3

4 We need one more piece of notation before beginning the proof of Theorems 1 and 2. Given an n n matrix A, let A(i j) be the (n 1) (n 1) matrix obtained by deleting the i th row and j th column of A. Recall that if A is a matrix, we denote by a i,j the element in the i th row, j th column of A. Now if D is an (n 1)-linear, alternating function on (n 1) (n 1) matrices, let D i,j (A) = D(A(i j)). The proofs of Theorems 1 and 2 will be in two steps: First, we will show that n-linear, alternating functions exist, and then show that there is at most one such function. Existence will be a recursive argument using D i,j (A). Theorem 3 If D is an (n 1)-linear, alternating function on (n 1) (n 1) matrices, then for any j with 1 j n, E j (A) = ( 1) i+j a i,j D i,j (A) is an n-linear, alternating function on n n matrices. E j (I n ) = 1. Moreover, if D(I n 1 ) = 1 then This is interesting because it makes it look like there are lots of determinants rather than just one. The formula for E j is the usual cofactor expansion down the j th column. Proof: We must show that E is n-linear and alternating, given that D is (n 1)-linear and alternating. We first consider n-linearity. We show that for any i and j that a i,j D i,j (A) is n-linear. If so, then E j is a linear combination of n-linear functions, so it will be n-linear. We look at how a i,j D i,j (A) acts on the k th row of A. There are two cases: k i and k = i to consider. For k i, we first note that if the k th row can be written u + cv where u and v are row vectors in F n (our book uses F n for row vectors, I will use this in what follows), then if u and v are the vectors in F n 1 obtained by deleting the j th element, the k th row in A(i j) will be u + cv. If we just write T k (v) = a i,j D i,j (v), the action of D i,j (A) where only v =row k can change in A then T k (u + cv) = a i,j D i,j (u + cv ) = a i,j D i,j (u ) + ca i,j D i,j (v ) = T k (c) + ct k (v), as desired. This shows T k is linear whenever k j. If k = j, then any change to row j has no effect on D i,j (A). In this case, the action takes place with a i,j. If we write u i for the i th component of row j we have T j (u + cv) = a i,j D i,j (A) = (u j + cv j )D i,j (A) = u j D i,j (A) + cv j D i,j (A) = T j (u) + ct j (v), so T j is also linear. Thus, a i,j D i,j (A) is n-linear, making E j n-linear. Next, since E j is n-linear, to show it is alternating, by Lemma 3, we need only show E j (A) = 0 whenever A has two adjacent equal rows. Suppose these rows are row k and row k + 1. If i is neither k nor k + 1 then deleting row i, column j from A leaves a matrix with Page 4

5 two identical rows. Thus, D i,j (A) = 0 in these cases. This means all but two of the terms in the sum for E j are 0, with only terms i = k and i = k + 1 contributing. We have E j (A) = ( 1) i+j a i,j D i,j (A) = ( 1) k+j a k,j D k,j (A) + ( 1) k+j+1 a k+1,j D k+1,j (A). Now D k,j (A) = D k+1,j (A) and a k,j = a k+1,j since rows k and k + 1 of A are the same. But these terms have different signs, so they cancel, and E j (A) = 0. Thus, we have shown that E j is n-linear and alternating. Finally, suppose that D(I n 1 ) = 1. If i j then removing the i th row and j th column from I n removes two 1 s from the diagonal leaving a matrix with a row of 0 s. This means D i,j (I n ) = 0 whenever i j. Consequently, E j (I n ) = ( 1) i+j a i,j D i,j (I n ) = ( 1) j+j a j,j D j,j (I n ) = D(I n 1 ) = 1, which completes the proof. Because of this, we have half of our proof of Theorem 1: Corollary 1 For every positive integer n there is at least one determinant on F n n. The proof is an induction on n, where the result is trivial for n = 1 (we just define det(a) = a), or if we prefer, we could use the case where n = 2, which we previously worked out. The key result is that the usual cofactor expansion, say across the top row always for each submatrix, gives us a determinant. Page 5

Determinants: Uniqueness and more

Math 5327 Spring 2018 Determinants: Uniqueness and more Uniqueness The main theorem we are after: Theorem 1 The determinant of and n n matrix A is the unique n-linear, alternating function from F n n to