A Little Multilinear Algebra 1. Let M and N Be Positive Integers with M

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