MATRIX THEORY CHAPER 0 FANG WANG 1.BRIEF REVIEW OF MATRICES AND VECTOR SPACES. or entries of A.We usually write 021a22 A= ami am2 ..amn (1)Operations on matrices: Matrices: :c-6 二fww9g9eEc4aL」 ☒业8一OBc权C 2x1 +3r3=3, 1-x1+4a2+5z3=1. 一[a:副图-] 1= )a(g)() =6++=a,, 3has a another basi m-()-()-() L3-x2」 Date:September 25.019
MATRIX THEORY - CHAPER 0 FANG WANG 1. Brief Review of Matrices and Vector Spaces. Definition 1. A matrix is a rectangular array of numbers or symbols arranged in rows and columns. The individual items in an m × n matrix A, often denoted by aij where 1 ≤ i ≤ m, 1 ≤ j ≤ n, are called elements or entries of A. We usually write A = a11 a12 · · · a1n a21 a22 · · · a2n . . . . . . . . . am1 am2 · · · amn . The space of all m × n matrices with entries in field F is denoted by Mm,n(F). In this course, we usually take F = R or C. (1) Operations on matrices: – Matrices: A = 2 0 3 −1 4 5 , B = 1 3 2 2 3 1 , C = i −1 2 0 −i 1 + i . – Linear operation: Mm,n(C) −→ Mm,n(C). Ex: aA + bC. – Product: Mm,n(C) × Mn,p(C) −→ Mm,p(C). Ex: AB, BA. – Transpose and Hermitian adjoint: Mm,n(C) −→ Mn,m(C). Ex: AT , C∗ . (2) Matrix and linear equation: ( 2x1 + 3x3 = 3, −x1 + 4x2 + 5x3 = 1. ⇐⇒ 2 0 3 −1 4 5 x1 x2 x3 = 3 1 ⇐⇒ Ax = b. (3) Matrix and changing basis: – R 3 has a standard basis e1 = 1 0 0 , e2 = 0 1 0 , e3 = 0 0 1 x = x1 x2 x3 = x1e1 + x2e2 + x3e3 = [e1, e2, e3] x1 x2 x3 – R 3 has a another basis α1 = 1 1 1 , α2 = 0 1 1 , α3 = 0 0 1 x1 x2 x3 = [α1, α2, α3] x1 x2 − x1 x3 − x2 = [α1, α2, α3] y1 y2 y3 Date: September 25, 2019. 1
2 PANG WANG Relations: 「1001 a1,a2,agl=[e1,e2,e]110 111 [- 111g- (4)Matrix and linear transformation: -Given AM(C),then it defines a linear map A:Cm-→Cm IAr pm子m片o管2长rod n=∑ae,aieE If the basis contains only n vectors,then n is called the dimension of Ve.In this course,we only consider finite dimensional vector spaces. 2.DETERMINANT Theorem 1.(Laplace erpansion)Given AE Mn,then det A= C（-lr+ade(4）=∑-i)det(Al where Aj is a(n-1)x(n-1)matrir defined by A deleting the i-th row and j-th column Example 1.Find the determinant of following matrices: 3894 Theorem 2.Given any A,BE Mn,then det(AB)=det(A)det(B). Thec rem3.fAeL.(图stisfies det(≠0.hem (1)A-1 is well-defined A-hw=-1)det(4 det(A) (d):
2 FANG WANG – Relations: [α1, α2, α3] = [e1, e2, e3] 1 0 0 1 1 0 1 1 1 , 1 0 0 1 1 0 1 1 1 x1 x2 − x1 x3 − x2 = x1 x2 x3 (4) Matrix and linear transformation: – Given A ∈ Mm,n(C), then it defines a linear map A : C n −→ C m x 7−→ Ax Conversely, a linear map A : C n −→ C m must be represented by a matrix A. – Product of matrices ∼ Composition of linear transformations. Ex: AB and BA. Definition 2. A vector space VF is a collection of objects called vectors, which may be added together and multiplied by scalar numbers in field F. In this course, we usually take F = R or C. Definition 3. Given a vector space VF. A basis of VF is a set of vectors {ei : i ∈ I} ⊂ VF such that for any vectors v ∈ V , there exists a unique expression: v = X i∈I aiei , ai ∈ F. If the basis contains only n vectors, then n is called the dimension of VF. In this course, we only consider finite dimensional vector spaces. 2. Determinant Theorem 1. (Laplace expansion) Given A ∈ Mn, then det A = Xn j=1 (−1)i+j aij det(Aij ) = Xn i=1 (−1)i+j aij det(Aij ), where Aij is a (n − 1) × (n − 1) matrix defined by A deleting the i-th row and j-th column. Example 1. Find the determinant of following matrices: A = 2 3 4 5 , B = 1 2 3 4 5 6 7 8 9 , C = 1 3 0 1 3 0 1 4 1 1 2 1 0 1 1 0 Theorem 2. Given any A, B ∈ Mn, then det (AB) = det(A) det(B). Theorem 3. If A ∈ Mn(F) satisfies det(A) 6= 0. Then (1) A−1 is well-defined: [A −1 ]ij = (−1)i+j det (Aji) det(A) ; (2) Ax = b has a unique solution x = A−1 b; (3) if e1, e2, · · · , en is a basis of V , then [α1, α2, · · · , αn] = [e1, e2, · · · , en]A is also a basis of V ; (4) A : F n −→ F n is one-by-one
MATRIX THEORY-CHAPER 0 3.RANK Definition 4.Let Ve be a finite dimensional vector space.We call the vectorsv...xV linearly independent if the equation x1+22+…k,x∈ only has zero solution 1=2=...x=0. 1,e2, Definition 5.Let S be a set of vectors.The rank of S is defined by the marimum number r such that there erists r vectorsv,v2,..vS such that they are linearly independent. Example&aA-[名！g】emm到-2 Theorem 4.Let AE Mm.n(F).Then rank(A)=rank(AT)=rank(A')<min{m,n} Definition 7.The following operations are called elementary operation .Interchange of two rows; ·AMh tion of a row by a nonzero scalar; Definition 8.The RREF is the matrices satisfying 包Au er the时chm ro row has 1 as its first no Example 4.Find the RREF for the following matriz 「1 A- Then rank(A)=2.Moreover, Nul(A)={z∈C:Az=0 Ran(A)={y∈c4:y=Ax}= Theorem 6.Given any then dim(Null(A))=n-rank(A),dim(Ran(A))=rank(A)
MATRIX THEORY - CHAPER 0 3 3. Rank Definition 4. Let VF be a finite dimensional vector space. We call the vectors v1, v2, · · · , vk ∈ V linearly independent if the equation x1v1 + x2v2 + · · · xkvk, xi ∈ F. only has zero solution x1 = x2 = · · · xk = 0. Example 2. Given any vectors space V of finite dimension n. If {e1, e2, · · · , en} is a basis of V , then e1, e2, · · · , en are linearly independent. Definition 5. Let S be a set of vectors. The rank of S is defined by the maximum number r such that there exists r vectors v1, v2, · · · , vr ∈ S such that they are linearly independent. Definition 6. Let A ∈ Mn,m(F). Then the rank of A is defined by the rank of column vectors, or equivalently the rank of row vectors. Example 3. Let A = 2 0 3 −1 4 5 . Then rank(A) = 2. Theorem 4. Let A ∈ Mm,n(F). Then rank(A) = rank(A T ) = rank(A ∗ ) ≤ min{m, n} Definition 7. The following operations are called elementary operation: • Interchange of two rows; • Multiplication of a row by a nonzero scalar; • Addition of a scalar multiple of one two to another row. Theorem 5. Any matrix A ∈ Mm,n(F) corresponds to a canonical form , called row-reduced echelon form (RREF), by a sequence of elementary operations. Definition 8. The RREF is the matrices satisfying (a) Each nonzero row has 1 as its first nonzero entry; (b) All other entries in the column of such a leading 1 are 0; (c) Any rows consisting entirely of zeroes occur at the bottom of the matrix; (d) The leading 1’s occur in a stair step pattern, left to right. Example 4. Find the RREF for the following matrix: A = 1 0 2 1 1 −1 2 1 3 −1 1 2 5 5 1 2 −2 1 −2 2 −→ 1 0 2 1 1 0 2 3 4 0 0 2 3 4 0 0 −2 −3 −4 0 −→ 1 0 2 1 1 0 1 3 2 2 0 0 0 0 0 0 0 0 0 0 0 Then rank(A) = 2. Moreover, Null(A) = {x ∈ C 5 : Ax = 0} = Span −1 0 0 0 1 , −1 −2 0 1 0 , 0 − 3 2 1 0 0 , Ran(A) = {y ∈ C 4 : y = Ax} = Span 1 −1 1 2 , 0 2 2 −2 . Theorem 6. Given any A ∈ Mm,n, then dim(Null(A)) = n − rank(A), dim(Ran(A)) = rank(A)
4 FANG WANG Example 5.Find the RREF for the following matriz: 「10211b1 「10211 A-[A,= -1213-12 → 02 3 40b1+b2 0-21+b 00000 -b1+b2+b4 This tell us if there erists solution to Ar=b then vector b must satisfies the -2b1-b2+b3=0. (1) -b1+b2+b4=0. It is easy to check b satisfies (1)if and only if bE Ran(A)
4 FANG WANG Example 5. Find the RREF for the following matrix: A˜ = [A, b] = 1 0 2 1 1 b1 −1 2 1 3 −1 b2 1 2 5 5 1 b3 2 −2 1 −2 2 b4 −→ 1 0 2 1 1 b1 0 2 3 4 0 b1 + b2 0 2 3 4 0 −b1 + b3 0 −2 −3 −4 0 −2b1 + b4 −→ 1 0 2 1 1 b1 0 1 3 2 2 0 1 2 (b1 + b2) 0 0 0 0 0 −2b1 − b2 + b3 0 0 0 0 0 −b1 + b2 + b4 This tell us if there exists solution to Ax = b then vector b must satisfies the (1) ( −2b1 − b2 + b3 = 0, −b1 + b2 + b4 = 0. It is easy to check b satisfies (1) if and only if b ∈ Ran(A)
