1 Basics of determinant

Size: px

Start display at page:

Download "1 Basics of determinant"

Brian Harrison
5 years ago
Views:

1 Linear Algebra- Review And Beyond Lecture 2 This lecture focus on function. Determinant is not a main topic in linear algebra, but it s no doubt a powerful tool. We will view this special function from a higher perspective. In the end, there is a brief introduction to trace. 1 Basics of determinant The definition of determinant is strange to beginners. In fact, we have many different ways to define determinant, and I will show you some more modern ways later. We merely use the definition to calculus a determinant. Definition(Determinant) = = a n1 a n2 a nn Fact = a n1 a n2 a nn (j 1 j 2 j n) (j 1 j 2 j n) ( 1) τ(j 1j 2 j n) a 1j1 a 2j2 a njn = ( 1) τ(j 1j 2 j n) a 1j1 a 2j2 a njn (j 1 j 2 j n) ( 1) τ(j 1j 2 j n) a j1 1a j2 2 a jnn In the following, we denote the columns in determinant by α k and write (α 1, α 2,, α n ) = α 1, α 2,, α n. {e k } is a standard orthogonal basis for R n, then (e 1, e 2,, e n ) = 1 (α 1,, α i,, α j,, α n ) = (α 1,, α j,, α i,, α n ) (α 1,, λα i + µβ i,, α n ) = λ (α 1,, α i,, α n ) + µ (α 1,, β i,, α n ) Remark These properties are easy to verify. However, these properties are essential for determinants. In other words, they uniquely determine the determinant. 1

2 = 0 The second property above is equivalent to if two columns are the same, then Prove: (α 1,, α i + µα j,, α n ) = (α 1,, α i,, α n ), j i In R n, we have different x 1, x 2,..., x m, m n. Prove {(1, x i, x 2 i,..., x m 1 i ), 1 i m} is linearly independent. We mainly use expansion theorems to calculus determinants. Expansion by column or row is a basic method, together with the second exercise above. The Laplace expansion is omitted here. Definition(Algebraic Cofactor) For a ij, deleting the row and column containing a ij, we get a determinant M ij with order n 1. A ij = ( 1) i+j M ij is called the algebraic cofactor of a ij. Fact = n a ija ij. n k=1 a ika jk = δ j i, n k=1 a kia kj = δ j i. Theorem(Cramer) We have a linear system a n1 a n2 a nn x 1 x 2. x n = b 1 b 2. b n If the coefficient determinant is non-zero, then (x 1,, x j,, x n ) = ( 1,, j,, n ) is the unique solution, where =, j = a n1 a n2 a nn a 11 b 1 a 1n a 21 b 2 a 2n a n1 b n a nn 2

3 Remark Cramer formula gives the analytic solution of a linear system. However, this method is only theoretical. For general linear system, we have to find better numerical methods to calculus the solution. Fact Suppose A is an n n matrix, the followings are equal: A 0 The column vectors of A is linearly independent The row vectors of A is linearly independent rank(a) = n Remark This fact tells us the connection between linearly independence and determinant. We often use this fact to judge whether the vectors are linearly independent. On the other hand, determinant is non-zero implies linear independence is useful. Let We can calculus det A by A = A = a b c d e f g h i a b c a b d e f d e g h i g h Then det A = aei + bfg + cdh gec hfa idb. Consider why this is true for matrix with order 3? Prove that if A, B are square matrices of the same size and AB = I, then BA = I. 2 Multiple linear function In analytical geometry, you have already known that determinant of order 2 or 3 denotes some volume. But it s not indeed a volume because the value of a determinant may be negative. So we have to introduce the concept of orientation. The main reference for this part is Peter Lax s book. 3

4 A simplex in R n is a polyhedra with n + 1 vertices. Without the lost of generality, we take one of the vertices as origin and denote the others by a 1, a 2,, a n. We say 0, a 1, a 2,, a n are vertices of an ordered simplex. We say S is degenerate if S locates in an n 1 dimensional subspace of R n. If S = (0, a 1,, a n ) can be deformed continuously and nondegenerately into the standard ordered simplex (0, e 1,, e n ), we say S is positive oriented. If S is positive oriented, O(S) = 1; If S is negative oriented, O(S) = 1; If S is degenerate, O(S) = 0. Otherwise it s negative oriented. We define the signed volume by V (S) = O(s) V ol(s) = O(S) 1 n V ol n 1(Base) Height Now we study the properties of V (S). If a j = a k for some j k, then V (S) = 0 Fix all a k with k j, then V (S) is a linear function of a k Now we can define the determinant based on the discussion above. Definition(Determinant) a 1, a 2,, a n are column vectors, and S = (0, a 1,, a n ) is an ordered simplex. A column vector function D(a 1, a 2,, a n ) is defined by D(e 1, e 2,, e n ) = 1 D(a 1, a 2,, a n ) = n!v (S) If a j = a k for some j k, then D(a 1, a 2,, a n ) = 0 Fix all a k with k j, then D(a 1, a 2,, a n ) is a linear function of a k From the second property(antisymmetric), we immediately know if we change the order of a i and a j and get D, then D = D. Thus Suppose τ is a permutation on [n], then D(a 1, a 2,, a n ) = sign(τ) D(a τ(1), a τ(2),, a τ(n) ) 4

5 Note that {e i } is a basis, so any vector in R n is the linear combination of {e i }. Suppose a i = n a ije j, then D(a 1, a 2,, a n ) = D( a 1j e j, Use the property above and D(e 1, e 2,, e n ) = 1 D( a 1j e j, a 2j e j,, a nj e j ) = a 2j e j,, (f 1,f 2,,f n) a nj e j ) sign(f 1, f 2,, f n ) a 1f1 a 2f2 a nfn From the discussion above, we get the equality in the previous definition of determinant. In addition, we know that such a function is unique. Thus we can define determinant with the three properties. f(a 1, a 2,, a n ) is an n-element function on F n, and satisfying f(e 1, e 2,, e n ) = 1 f(a 1,, a i,, a j,, a n ) = f(a 1,, a j,, a i,, a n ) f(a 1,, λa i + µb i,, a n ) = λf(a 1,, a i,, a n ) + µf(a 1,, b i,, a n ) Find the form of f. 3 Jacobi determinant - trivial version Jacobi determinant is important in analysis when we calculus integration with change of variables. The proof of changing variables is difficult. We are going to discuss a trivial version here. First we prove the following theorem: Theorem If A, B are square matrices with order n, then det BA = det A det B. Suppose the columns of A are a 1, a 2,, a n, and {e i } is a standard orthogonal basis. Prove the theorem above step by step Verify that the j-th column of BA is (BA)e j = Ba j, thus det BA = D(Ba 1, Ba 2,, Ba n ). Suppose det B 0, define C(a 1, a 2,, a n ) = det BA det B = D(Ba 1, Ba 2,, Ba n ) det B 5

6 Verify that C(a 1, a 2,, a n ) satisfies the three properties of determinant, thus by u- niqueness, C(a 1, a 2,, a n ) = D(a 1, a 2,, a n ) = det A. For the condition det B = 0, let B(t) = ti + B, use the continuity of polynomial. Remark The last step above is a standard trick in linear algebra: use a small perturbation to get a invertible matrix, then use the continuity. By the connection between the volume of simplex and determinant, we know a linear mapping with matrix B maps any simplex S in R n to another simplex S, and V ol(s ) = det B V ol(s). Any open set is the union of simplexes, so the volume of the image of any open set is det B times the original volume. 4 Trace Another important function of square matrix is trace. Definition(Trace) A = (a ij ) is a square matrix of order n, then the trace tr(a) = i=1 a ii tr(ka) = k tr(a) tr(a + B) = tr(a) + tr(b) tr(ab) = tr(ba) Suppose A = (a ij ) n n, then tr(aa T ) = Remark If we regard an n-order square real matrix A as an n 2 -tuple, we can naturally define n the norm of A by A = i, a2 ij = tr(aa T ). This norm is called the Frobenius norm of matrix. Obviously, tr(aa T ) = 0 if and only if A = 0. i, a 2 ij 6

MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS

MA 1B ANALYTIC - HOMEWORK SET 7 SOLUTIONS 1. (7 pts)[apostol IV.8., 13, 14] (.) Let A be an n n matrix with characteristic polynomial f(λ). Prove (by induction) that the coefficient of λ n 1 in f(λ) is