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Lecture 23 - Addendum
Andrei Antonenko
April 2, 2003
1 Proofs of the main results from the lecture
Let’s recall the definition from one of the previous lectures.
Definition 1.1. Function f (a 1 , a 2 ,... , am) is called multilinear if it is linear in every argu- ment, i.e. for any i
f (a 1 ,... , a′ i + a′′ i ,... , am) = f (a 1 ,... , a′ i,... , am) + f (a 1 ,... , a′′ i ,... , am) f (a 1 ,... , λai,... , am) = λf (a 1 ,... , ai,... , am).
Definition 1.2. Multilinear function f (a 1 , a 2 ,... , am) is called alternating if it changes the sign after interchanging any 2 arguments, i.e. for any i and j
f (a 1 ,... , ai,... , aj ,... , am) = −f (a 1 ,... , aj ,... , ai,... , am). If f is alternating, then it is equal to 0 if any 2 arguments are equal. It is true, because if we interchange these 2 arguments, the function will not change, but from the other hand, it should change its sign. So, it is equal to 0. Now we’re able to formulate the main result about alternating multilinear functions.
Theorem 1.3. For any c ∈ R in the vector space Rn^ there exists the unique alternating multi- linear function f , such that f (e 1 , e 2 ,... , en) = c (1)
(where ei’s are rows with 1 on i-th place, and 0’s on all other places). Moreover, this function is equal to
f (a 1 , a 2 ,... , an) = c ·
all permutations of n elements σ
sgn(σ)a 1 σ(1)a 2 σ(2) · · · anσ(n) (2)
= c ·
(k 1 ,k 2 ,...,kn)
sgn(k 1 , k 2 ,... , kn)a 1 k 1 a 2 k 2 · · · ankn , (3)
where aik is the k-th component of the row ai, and the summation is over all permutations of numbers from 1 to n.
Proof. 1. Let f be an alternating multilinear function, such that f (e 1 ,... , en) = c. Then
f (a 1 , a 2 ,... , an) = f
k 1
a 1 k 1 ek 1 ,
k 2
a 2 k 2 ek 2 ,... ,
kn
ankn ekn
k 1 ,k 2 ,...,kn
a 1 k 1 a 2 k 2 · · · ankn f (e 1 , e 2 ,... , en).
Since f is alternating, if any 2 numbers from k 1 , k 2 ,... , kn are equal, then f (ek 1 , ek 2 ,... , ekn ) =
- If all of them are different, then f (ek 1 , ek 2 ,... , ekn ) = c sgn(k 1 , k 2 ,... , kn). Let’s prove it. If this equality is true for some permutation, then it is true for any other permutation which we can get from the initial by transposition of any 2 elements (since after transposition both left-hand side and right-hand side change signs). But this equality is true for identity permutation, and since it is possible to get any permutation from identity, then this equality is true for all permutations. So, we get that f satisfies the expression (3). So, if f satisfying the given conditions exists, then it has the form (3) and so it is unique.
- Now we’ll prove that the function f given by the formula (3) is alternating multilinear, and satisfy the condition (1). Linearity by any argument is obvious, since for any i the equality (3) can be written as f (a 1 , a 2 ,... , an) =
j
aij uj ,
where uj ’s do not depend on ai. The condition (1) also holds, since in the expression for f (e 1 , e 2 ,... , en) all summands except the term, corresponding to the identity permutation are equal to 0, and the term, corresponding to the identity permutation is equal to 1. Now we should check that this function is alternating. Let we interchange arguments ai and aj. Then all permutations can be divided into pairs different only by interchanging ki and kj. Terms from these pairs are included in the expression (3) with different signs (since one of them is different from the other by one transposition). After interchanging ai and aj they change their roles, and so the whole expression changes its sign.
If c = 1 we will denote such function by det.
Definition 1.4. The determinant of the square n × n-matrix A = (aij ) is
det A = det(a 1 , a 2 ,... , an),
where a 1 , a 2 ,... , an are rows of A.
Proof. If matrices B and D are fixed, then the determinant of A is the alternating multilinear function of its last rows, and so it is the alternating multilinear function of rows of the matrix C. Thus, by the corollary 1.
det A = det
B D
0 I
· det C.
If the matrix D is fixed, then the first multiplicand is the alternating multilinear function of rows of B, and so
det
B D
0 I
= det
I D
0 I
· det B = det B
(because
I D
0 I
is triangular with 1’s on its diagonal). So,
det A = det B det C
