A Little Multilinear Algebra n 1. Let m and n be positive integers with m ≤ n. Let a1, . , am be column vectors in R , and A = [a1 ··· am]. n Suppose that the ai generate a parallelpiped in R of m-dimesnional volume V . Then V 2 = det (ATA). (1) In this project, you’ll find a formula for det (CD), where Let C be an m × n matrix and D an n × m matrix, m ≤ n. As usual, vectors denoted by lower case letters are columns, and those by upper case are n rows. For a1, . , am in R , A = [a1 ··· am] m is the n × m matrix with columns aj. And For A1,...,An in R , A1 . A = . An n is the n × m matrix with rows Ai. Let f be a function of m column vectors a1, . , am in R , such that (i) f(a1, . , baj, . , am) = bf(a1, . , aj, . , am) for any scalar b. 0 0 (ii) f(a1, . , aj + aj, . , am) = f(a1, . , aj, . , am) + f(a1, . , aj, . , am). (iii) f(a1, . , aj, . , ak, . , am) = −f(a1, . , ak, . , aj, . , am). Since f is a function of m vectors in Rn, linear in each, it is called an m-multilinear function on n n R . Functions satisfying (iii) are called alternating. Denote by Am(R ) the set of all alternating, m-multilinear functions on Rn. *a. Set n = m. Let A be an m × m matrix with columns a1, . , am. Show that g(a1, . , am) = det A lies m m in Am(R ), i.e. that g is an alternating, m-multilinear function on R . 2. Let k1, . , km be distinct integers such that 1 ≤ k1 < k2 < ··· < km ≤ n. For column vectors a1, . , am in Rn, set ak1,1 ··· ak1,m Φ (a , . , a ) = . .. (2) k1···km 1 m . akm,1 ··· akm,m If you think of a1, . , am as the columns of the n × m matrix A with rows A1,...,An, then Ak1 Φ (a , . , a ) = det . (3) k1···km 1 m . Akm *a. Set 1 2 0 3 0 −2 a1 = 0 , a2 = 1 and a3 = 3 . −1 1 2 2 −3 1 Compute Φ1,3,4(a1, a2, a3), Φ2,4,5(a1, a2, a3) and Φ1,2,5(a1, a2, a3). n *b. Show that Φk1···km belongs to Am(R ). (Hint: Use (2) and familiar properties of determinants. You may want to think about a concrete example first, like Φ1,3,4 from the previous problem.) * Do for credit. 3. As you can see, multilinear algebra is prodigal in its use of indices. In order to control the subscript clutter, let’s use the multi-index notation, k = (k1, . , km), (4) where the ki are integers such that 1 ≤ k1 < k2 < ··· < km ≤ n. (5) With this notation, we can write Φk instead of Φk1···km Denote by M(m, n) the set of all multi-indices k defined by (3) and (4). Thus M(2, 3) = {(1, 2), (1, 3), (2, 3)}. Just for practice, write out M(3, 5). You should have 10 elements. n 4. Functions in Am(R ) are vectors in the sense that you can add them and multiply them by real scalars. n If f and g are elements of Am(R ) and c is any scalar, you define f + g and cf in the obvious way: (f + g)(a1, . , am) = f(a1, . , am) + g(a1, . , am). (6) And, (cf)(a1, . , am) = cf(a1, . , am). (7) n It turns out that the set of functions {Φk}k∈M(m,n) is a basis of Am(R ). Thus every alternating m-multilinear function f on Rn has a unique representation X f = bkΦk, (8) k where the bk are scalars and the sum is taken over all multi-indices k in M(m, n). 3 *a. Let e1, 2 and e3 be the standard basis vectors in R . Show that Φ1,2(e1, e2) = Φ1,3(e1, e3) = Φ2,3(e2, e3) = 1, and that Φ1,2(e1, e3) = Φ1,2(e2, e3) = Φ2,3(e1, e2) = 0. Is there a pattern? Try Φ1,3(e2, e3) and Φ2,3(e1, e3). *b. Let i = (i1, . , im) ∈ M(m, n). Show that 1 if k = i, Φ (e , . , e ) = (9) k i1 im 0 if k 6= i. Note that this generalizes the result of part (a). *c. Use (8) and (9) to show that bk = f(ek1 , . , ekm ). (10) 5. Let C be an m × n matrix and D, n × m, where m ≤ n. Think of C as fixed and det (CD) as a function f(d1, . , dm) = det (C[d1 ··· dm]) = det ([Cd1 ··· Cdm]), (11) of m column vectors in Rn. *a. Show that f is an alternating m-multilinear function on Rn. *b. Show that f(ek1 , . , ekm ) = det ([ck1 ··· ckm ]). (12) *c. Use (8), (9) and (10) to show that Dk1 X det (CD) = det ([c ··· c ]) det . (13) k1 km . k Dkm *d. Let A be n × m, with n ≤ m. Use (13) to show 2 Ak1 T X . det (A A) = det . . (14) k Akm Try the formula out on 2 0 1 −1 1 3 A = . 0 2 2 3 1 0.