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Linear Algebra Chapter 2 Matrices and Linear Transformations 大葉大學 資訊工程系 黃鈴玲 2.1 Addition, Scalar Multiplication, and Multiplication of Matrices • aij: the element of matrix A in row i and column j. Definition Two matrices are equal if they are of the same size and if their corresponding elements are equal. Thus A = B if aij = bij i, j. ( 代表 for every, for all) Ch2_2 Addition of Matrices Definition Let A and B be matrices of the same size. Their sum A + B is the matrix obtained by adding together the corresponding elements of A and B. The matrix A + B will be of the same size as A and B. If A and B are not of the same size, they cannot be added, and we say that the sum does not exist. Thus if C A B, then cij aij bij i,j. Ch2_3 Example 4 7, B 2 5 6, and C 5 4. Let A 1 8 0 2 3 3 1 2 7 Determine A + B and A + C, if the sum exist. Solution (自行練習) 4 7 2 5 6 (1) A B 1 8 0 2 3 3 1 4 5 7 6 1 2 0 3 2 1 3 8 3 9 1. 3 1 11 (2) A and C are not of the same size, A + C does not exist. Ch2_4 Scalar Multiplication of matrices Definition Let A be a matrix and c be a scalar. The scalar multiple of A by c, denoted cA, is the matrix obtained by multiplying every element of A by c. The matrix cA will be the same size as A. Thus if B cA, then bij caij i, j. Example Let A 1 2 4. 7 2 0 3 1 3 (2) 3 4 3 6 12 3A . 3 7 3 (3) 3 0 21 9 0 Observe that A and 3A are both 2 3 matrices. Ch2_5 Negation and Subtraction Definition The matrix (1)C is written –C and is called the negative of C. Example 0 7 1 0 7 1 C , then C 3 6 2 3 6 2 We now define subtraction in terms of addition and scalar multiplication. Let A – B = A + (–1)B Example Thus if C A B, then cij aij bij , i, j. Suppose A 5 0 2 and B 2 8 1. 3 6 5 0 4 6 5 2 0 8 2 (1) 3 8 1 A B . 2 11 3 0 6 4 5 6 3 Ch2_6 Matrix Multiplication Examples 2 1 2 1 2 2 5 (1 2) (2 5) 12 3 4 5 3 4 2 (3 2) (4 5) 26 5 5 1 3 1 3 5 0 1 3 2 0 3 2 6 2 0 5 3 0 1 3 2 0 2 0 2 1 1 3 14 6 19 6 1 10 0 2 2 0 6 Ch2_7 Example 1 3 1 2 7 2 Let A ,B . 4 1 5 6 3 Show that the product AB does not exist. Sol. 3 1 2 7 2 AB 6 3 4 1 5 3 1 4 1 7 2 6 7 5 6 3 1 4 1 2 2 3 2 5 3 無法相乘 AB does not exist. Note. In general, ABBA. 如此題之BA存在 Ch2_8 Definition Let the number of columns in a matrix A be the same as the number of rows in a matrix B. The product AB then exists. Let A: mn matrix, B: nk matrix, The product matrix C=AB has elements cij ai1 ai 2 b1 j b 2j ain ai1b1 j ai 2b2 j ainbnj bnj C is a mk matrix. If the number of columns in A does not equal the number of row B, we say that the product does not exist. Ch2_9 Example 2 0 1 1 3 5 Let A ,B . 2 0 3 2 6 Determine AB and BA, if the products exist. Sol. 0 1 14 6 19 AB 1 3 5 2 0 3 2 6 10 0 2. BA does not exist. Note. In general, ABBA. 隨堂作業:5(f) Example 3 Let C = AB, A 2 1 and B 7 3 2 Determine c23. 3 4 5 0 1 2 c23 3 4 1 (3 2) (4 1) 2 隨堂作業:11(d) Ch2_10 Size of a Product Matrix If A is an m r matrix and B is an r n matrix, then AB will be an m n matrix. A B = AB mr rn mn Example If A is a 5 6 matrix and B is an 6 7 matrix. Because A has six columns and B has six rows. Thus AB exits. And AB will be a 5 7 matrix. 隨堂作業:8(f) Ch2_11 Special Matrices Definition A zero matrix is a matrix in which all the elements are zeros. A diagonal matrix is a square matrix in which all the elements not on the main diagonal are zeros. An identity matrix is a diagonal matrix in which every diagonal element is 1. 0mn 0 0 0 0 0 0 zero matrix 0 0 0 a11 0 0 a 22 A 0 0 0 0 ann 1 0 0 I n 0 1 0 0 0 1 identity matrix diaginal matrix A Ch2_12 Theorem 2.1 Let A be m n matrix and Omn be the zero m n matrix. Let B be an n n square matrix. On and In be the zero and identity n n matrices. Then A + Omn = Omn + A = A BOn = OnB = On BIn = InB = B Example 4 Let A 2 1 3 and B 2 1. 8 4 5 3 4 2 A O23 4 2 BO2 3 2 BI 2 3 1 3 0 0 0 2 1 3 A 5 8 0 0 0 4 5 8 1 0 3 0 1 1 4 0 0 0 0 0 0 2 1 3 0 O2 0 1 B 4 Ch2_13 Matrix multiplication in terms of columns (a) A: mn, B: nr Let the columns of B be the matrices B1, B2, …, Br. Write B=[B1 B2 … Br]. Thus AB=A[B1 B2 … Br]=[AB1 AB2 … ABr]. Example 2 0 4 1 3 A and B 1 5 0 2 1 8 2 6 AB 4 11 2 4 1 3 B1 , B2 , B3 0 2 1 8 2 6 and AB1 , AB2 , AB3 4 11 2 Ch2_14 Matrix multiplication in terms of columns (b) b1 A: mn, B: n1, where A=[A1 A2 … An] and B . bn b1 We get AB A1 A2 An b1 A1 b2 A2 bn An . bn Example 3 2 3 1 2 A and B 4 8 5 5 2 3 1 A1 , A2 , A3 4 8 5 2 3 1 5 AB 3 2 5 4 8 5 3 Ch2_15 Partitioning of Matrices A matrix can be subdivided into a number of submatrices. Example 0 1 2 P Q where P 0, Q 1 2, R 2 and S 5 1 A 3 1 4 3 1 4 R S 2 5 1 Example 1 2 1 1 0 P Q M A 3 0 2 and B 2 1 R S N 4 3 2 5 4 P Q M PM QN AB R S N RM SN Ch2_16 Example 5 1 1 1 2 1 Let A 3 0 and B . 1 3 1 2 4 Consider the following partition of A. 1 1 P Q A 3 0 R S 2 4 Under this partition A is interpreted as a 22 matrix. For the product AB to be exist, B must be partitioned into a matrix having two rows. Let B 1 2 1 H . 1 3 1 J AB P R 1 1 Q H PH QJ 1 2 1 1 3 3 0 S J RH SJ 21 2 1 41 3 1 隨堂作業:21(a) 0 1 2 1 3 6 3 6 16 2 Ch2_17 Homework Exercise 2.1: 5, 8, 11, 17, 21 Exercise 17 Let A be a matrix whose third row is all zeros. Let B be any matrix such that the product AB exists. Prove that the third row of AB is all zeros. Solution b1i b1i b b ( AB)3i [a31 a32 a3n ] 2i [0 0 0] 2i 0, i. b ni bni Ch2_18 2.2 Algebraic Properties of Matrix Operations Theorem 2.2 -1 Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed. Properties of Matrix Addition and scalar Multiplication 1. A + B = B + A Commutative property of addition 2. A + (B + C) = (A + B) + C Associative property of addition 3. A + O = O + A = A (where O is the appropriate zero matrix) 4. c(A + B) = cA + cB Distributive property of addition 5. (a + b)C = aC + bC Distributive property of addition 6. (ab)C = a(bC) Ch2_19 Theorem 2.2 -2 Let A, B, and C be matrices and a, b, and c be scalars. Assume that the size of the matrices are such that the operations can be performed. Properties of Matrix Multiplication 1. A(BC) = (AB)C Associative property of multiplication 2. A(B + C) = AB + AC Distributive property of multiplication 3. (A + B)C = AC + BC Distributive property of multiplication 4. AIn = InA = A (where In is the appropriate zero matrix) 5. c(AB) = (cA)B = A(cB) Note: AB BA in general. Multiplication of matrices is not commutative. Ch2_20 Proof of Thm 2.2 (A+B=B+A) Consider the (i,j)th elements of matrices A+B and B+A: ( A B)ij aij bij bij aij ( B A)ij . A+B=B+A Example 1 1 3 3 7 0 2 Let A ,B , and C . 1 4 5 2 3 1 1 3 3 7 0 2 3 5 2 A 3B 5C 2 4 5 2 1 3 1 2 9 0 6 21 10 11 25 . 18 8 6 15 10 3 5 17 Ch2_21 Example 2 4 3, and C 1 . Let A 1 2, B 0 1 Compute ABC. 3 1 1 0 2 0 Sol. A 22 B 23 C = 31 D 21 ABC = (AB)C = A(BC) 3 2 1 1. AB 1 2 0 1 3 1 1 0 2 1 3 11 4 ( AB)C 2 1 1 1 9. 1 3 11 0 1 Ch2_22 Caution In algebra we know that the following cancellation laws apply. If ab = ac and a 0 then b = c. If pq = 0 then p = 0 or q = 0. However the corresponding results are not true for matrices. AB = AC does not imply that B = C. PQ = O does not imply that P = O or Q = O. Example 1 2 1 2 3 8 (1) Consider t he matrices A ,B , and C . 2 4 2 1 3 2 3 4 Observe that AB AC , but B C. 6 8 1 2 2 6 (2) Consider t he matrices P , and Q . 4 2 1 3 Observe that PQ O, but P O and Q O. Ch2_23 Powers of Matrices Definition If A is a square matrix, then Ak AA A k times Theorem 2.3 If A is an n n square matrix and r and s are nonnegative integers, then 1. ArAs = Ar+s. 2. (Ar)s = Ars. 3. A0 = In (by definition) Ch2_24 Example 3 1 2 4 If A , compute A . 0 1 Solution 1 2 1 2 3 2 A 1 0 1 0 1 2 2 3 2 3 2 11 10 A . 2 1 2 5 6 1 4 Example 4 Simplify the following matrix expression. A( A 2B) 3B(2 A B) A2 7 B 2 5 AB Solution A( A 2 B) 3B(2 A B) A2 7 B 2 5 AB A2 2 AB 6 BA 3B 2 A2 7 B 2 5 AB 3 AB 6 BA 4 B 2 注意這兩項不能合併! Ch2_25 Systems of Linear Equations A system of m linear equations in n variables as follows a11 x1 a1n xn b1 am1 x1 amn xn bm Let a11 a1n x1 b1 A , X , and B am1 amn xn bm We can write the system of equations in the matrix form AX = B 隨堂作業:32(c) Ch2_26 Solutions to Systems of Linear Equations Consider a homogeneous system of linear equations AX=0. Let X1 and X2 be solutions. Then AX1=0 and AX2=0 A(X1 + X2) = AX1 + AX2 = 0 If c is a scalar, A(cX1) = cAX1 = 0 X1 + X2 is also a solution cX1 is also a solution Note. The set of solutions to a homogeneous system of linear equations is closed under addition and under scalar multiplication. It is a subspace. Ch2_27 Example 5 Consider the following homogeneous system of linear equations. x1 2 x2 8 x3 0 x2 3 x3 0 x1 x2 5 x3 0 It can be shown that there are many solutions, x1=2r, x2=3r, x3 = r. The solutions are vectors in R3 of the form (2r, 3r, r) or r(2, 3, 1). The solutions form a subspace of R3 of dimension 1. Note. x1=0, …, xn = 0, is a solution to every homogeneous system. The set of solutions to every homogeneous system passes through the origin. Figure 2.4 隨堂作業:36(a) Ch2_28 Solutions to Nonhomogeneous systems Exercise 41 Show that a set of solutions to a system of nonhomogeneous linear equations is not closed under addition or under scalar multiplication, and that it is thus not a subspace of Rn. Proof Let AX=Y be a nonhomogeneous system of linear equations, so Y0. Let X1 and X2 be two solutions. Then AX1=Y and AX2=Y A(X1 + X2) = AX1 + AX2 = 2Y (可省略) If c is a scalar, A(cX1) = cAX1 = cY X1 + X2 is not a solution. cX1 is not a solution. The set of solutions is not a subspace of Rn. Ch2_29 Example Consider the following nonhomogeneous system of linear equations. x1 2 x2 8 x3 8 x2 3 x3 2 x1 x2 5 x3 6 The general solution of this system is (2r+4, 3r+2, r). (2r+4, 3r+2, r) = r(2, 3, 1)+(4, 2, 0) Note that r(2, 3, 1) is the general solution of the corresponding homogeneous system. The vector (4, 2, 0) is the specific solution to the nonhomogeneous system corresponding to r=0. 隨堂作業:40 Figure 2.5 Ch2_30 Idempotent and Nilpotent Matrices (Exercise 24~30) Definition (1) A square matrix A is said to be idempotent (等冪) if A2=A. (2) A square matrix A is said to nilpotent (冪零) if there is a p positive integer p such that A =0. The least integer p such that Ap=0 is called the degree of nilpotency of the matrix. Example 3 6 2 3 6 (1) A ,A A. 1 2 1 2 3 9 2 0 0 (2) B ,B . The degree of nilpotency : 2 1 3 0 0 隨堂作業:28, 29(b) Ch2_31 Homework Exercise 2.2: 6, 13, 21, 25, 27, 28, 29, 32, 36, 40, 41 Exercise 21 A, B: diagonal matrix of the same size, c: scalar Prove that A+B and cA are diagonal matrices. Ch2_32 Homework Exercise 25 1 b Determine b, c, and d such that is idempotent. c d Exercise 27 Prove that if A and B are idempotent and AB=BA, then AB is idempotent. Ch2_33 2.3 Symmetric Matrices Definition The transpose (轉置) of a matrix A, denoted At, is the matrix whose columns are the rows of the given matrix A. i.e., A : m n At : n m, ( At )ij Aji i, j. Example 1 A 2 7, B 1 2 7, and C 1 3 4. 6 8 0 4 5 1 1 4 t t C 3. At 2 8 B 2 5 0 7 4 7 6 Ch2_34 Theorem 2.4 Properties of Transpose Let A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed. 1. (A + B)t = At + Bt Transpose of a sum 2. (cA)t = cAt Transpose of a scalar multiple 3. (AB)t = BtAt Transpose of a product 4. (At)t = A Ch2_35 Theorem 2.4 Properties of Transpose Proof for 3. (AB)t = BtAt ( AB)t ij ( AB) ji a j1 a j 2 a jn b1i b 2i bni a j1b1i a j 2b2i a jnbni ( B t At )ij [row i of B t ] [column j of At ] [column i of B]t [row j of A]t b1i b2i a j1 a j2 bni a j1b1i a j 2b2i a jnbni a jn Ch2_36 Symmetric Matrix Definition A symmetric matrix is a matrix that is equal to its transpose. A At , i.e., aij a ji i, j Example 5 2 5 4 match 1 0 1 4 0 8 1 7 2 4 8 3 4 0 2 4 7 3 9 3 2 3 9 3 6 match 隨堂作業:2(c) Ch2_37 Example 3 Let A and B be symmetric matrices of the same size. Let C be a linear combination of A and B. Prove that the product C is symmetric. Proof Let C = aA+bB, where a and b are scalars. Ct = (aA+bB)t = (aA)t + (bB)t by Thm 2.4 (1) = aAt + bBt by Thm 2.4 (2) = aA + bB since A and B are symmetric =C Thus C is symmetric. Ch2_38 If and only if Let p and q be statements. Suppose that p implies q (if p then q), written p q, and that also q p, we say that “p if and only if q” (若且唯若) (通常簡寫為 iff ) (也說成: p is necessary and sufficient for q ) Ch2_39 Example 4 Let A and B be symmetric matrices of the same size. Prove that the product AB is symmetric if and only if AB = BA. Proof *We have to show (a) AB is symmetric AB = BA, and the converse, (b) AB is symmetric AB = BA. () Let AB be symmetric, then AB= (AB)t by definition of symmetric matrix = BtAt by Thm 2.4 (3) = BA since A and B are symmetric () Let AB = BA, then (AB)t = (BA)t = AtBt by Thm 2.4 (3) = AB since A and B are symmetric Ch2_40 Example 3 相關題目 Let A be a symmetric matrix. Prove that A2 is symmetric. Proof ( A2 )t ( AA) t ( At At ) AA A2 Ch2_41 Trace of a matrix Definition Let A be a square matrix. The trace of A, denoted tr(A) is the sum of the diagonal elements of A. Thus if A is an n n matrix. tr(A) = a11 + a22 + … + ann Example 5 1 2 4 6. Determine the trace of the matrix A 2 5 3 0 7 Solution We get tr ( A) 4 (5) 0 1. 隨堂作業:15(b) Ch2_42 Theorem 2.5 Properties of Trace Let A and B be matrices and c be a scalar. Assume that the sizes of the matrices are such that the operations can be performed. 1. tr(A + B) = tr(A) + tr(B) 2. tr(AB) = tr(BA) 3. tr(cA) = c tr (A) 4. tr(At) = tr(A) Proof of (1) Since the diagonal element of A + B are (a11+b11), (a22+b22), …, (ann+bnn), we get tr(A + B) = (a11 + b11) + (a22 + b22) + …+ (ann + bnn) = (a11 + a22 + … + ann) + (b11 + b22 + … + bnn) = tr(A) + tr(B). Ch2_43 Example of (2) tr(AB)=tr(BA) 3 1 3 2 1 A 2 0 , B 1 0 1 1 2 2 2 0 8 11 AB 6 4 2, BA 2 5 1 2 1 tr( AB) 3 tr( BA) Ch2_44 Matrices with Complex Elements The element of a matrix may be complex numbers. A complex number is of the form z = a + bi Where a and b are real numbers and i 1. a is called the real part and b the imaginary part (虛部) of z. The conjugate (共軛複數) of a complex number z = a + bi is defined and written z = a bi. Ch2_45 Example 7 2i . Let A 2 i 3 2i and B 3 5i 4 1 i 2 3i Compute A + B, 2A, and AB. Solution 2i 2 i 3 3 2i 2i 5 i 3 2 i 3 2i 3 A B 4 5 i 1 i 2 3 i 4 1 i 5 i 2 3 i 5 i 2 8 i 2 i 3 2i 4 2i 6 4i 2 A 2 4 5 i 8 10 i 2i 2 i 3 2i 3 AB 1 i 2 3i 4 5 i 10 9i (2 i )3 (3 2i)(1 i ) (2 i)( 2i) (3 2i )( 2 3i) 11 4i ( 4 )( 3 ) ( 5 i )( 1 i ) 4 ( 2 i ) ( 5 i )( 2 3 i ) 7 5 i 15 18 i Ch2_46 Definition (i) The conjugate of a matrix A is denoted A and is obtained by taking the conjugate of each element of the matrix. (ii) The conjugate transpose of A is written and defined by A*=A t. (iii) A square matrix C is said to be hermitian if C=C*. Example (i), (ii) 2 3i 1 4i t 2 3i 1 4i * 2 3i A A A A 7i 7i 6 6 1 4i Example (iii) 3 4i 2 * C C 3 4 i 6 隨堂作業:23 6 7i Ch2_47 Homework Exercise 2.3: 2, 5, 11, 15, 23 Exercise 11 A matrix A is said to be antisymmetric if A = At. (a) give an example of an antisymmetric matrix. (b) Prove that an antisymmetric matrix is a square matrix having diagonal elements zero. (c) Prove that the sum of two antisymmetric matrices of the same size is an antisymmetric matrix. Ch2_48 2.4 The Inverse of a Matrix Definition Let A be an n n matrix. If a matrix B can be found such that AB = BA = In, then A is said to be invertible (可逆) and B is called the inverse (反矩陣) of A. If such a matrix B does not exist, then A has no inverse. (denote B = A1, and Ak=(A1)k ) Example 1 2 1 Prove that the matrix A 1 2 has inverse B 3 1 . 3 4 2 2 Proof 1 2 AB 1 2 3 1 1 0 I 2 3 4 2 0 1 2 1 1 2 1 0 2 BA 3 1 I2 3 4 0 1 2 2 Thus AB = BA = I2, proving that the matrix A has inverse B. Ch2_49 Theorem 2.7 If a matrix has an inverse, that inverse is unique. Proof Let B and C be inverses of A. Thus AB = BA = In, and AC = CA = In. Multiply both sides of the equation AB = In by C. C(AB) = CIn Thm2.2 乘法結合律 (CA)B = C InB = C B=C Thus an invertible matrix has only one inverse. Ch2_50 Let A be an invertible nn matrix How to find A-1? Let A1 X 1 X 2 X n , I n C1 C2 Cn . We shall find A1 by finding X1, X2, …, Xn. Since AA1 =In, then AX 1 X 2 X n C1 C2 Cn . i.e., AX1 C1 , AX 2 C2 ,, AX n Cn . Solve these systems by using Gauss-Jordan elimination: augmented matrix : A : C1 C2 Cn I n : X 1 X 2 X n . A : I n I n : A1 . Note. 若是最後沒有變成 [In:B] 的形式,則 A1不存在。 Ch2_51 Gauss-Jordan Elimination for finding the Inverse of a Matrix Let A be an n n matrix. 1. Adjoin the identity n n matrix In to A to form the matrix [A : In]. 2. Compute the reduced echelon form of [A : In]. If the reduced echelon form is of the type [In : B], then B is the inverse of A. If the reduced echelon form is not of the type [In : B], in that the first n n submatrix is not In, then A has no inverse. An n n matrix A is invertible if and only if its reduced echelon form is In. Ch2_52 Example 2 1 1 2 Determine the inverse of the matrix A 2 3 5 3 5 1 Solution 1 0 0 1 1 2 1 1 2 1 0 0 [ A : I 3 ] 2 3 5 0 1 0 R2 (2) R10 1 1 2 1 0 3 1 0 1 3 5 0 0 1 R3 R1 0 2 1 1 1 2 1 0 0 3 1 0 1 0 1 (1)R2 0 1 1 2 1 0 R1 R2 0 1 1 2 1 0 3 1 0 1 R3 (2)R2 0 0 0 2 1 3 2 1 0 1 01 1 0 0 R1 R3 0 1 0 5 3 1 R2 (1)R3 0 0 1 3 2 1 1 1 0 1 Thus, A 5 3 1. 2 1 隨堂作業:4(a), 14 3 Ch2_53 Example 3 Determine the inverse of the following matrix, if it exist. 1 1 5 A 1 2 7 2 1 4 Solution 1 5 1 0 0 1 1 1 5 1 0 0 1 2 1 1 0 [ A : I 3 ] 1 2 7 0 1 0 R2 (1)R1 0 2 1 4 0 0 1 R3 (2)R1 0 3 6 2 0 1 2 1 0 1 0 3 R1 (1)R2 0 1 2 1 1 0 R3 3R2 0 0 0 5 3 1 There is no need to proceed further. The reduced echelon form cannot have a one in the (3, 3) location. The reduced echelon form cannot be of the form [In : B]. Thus A–1 does not exist. 隨堂作業:4(c) Ch2_54 Properties of Matrix Inverse Let A and B be invertible matrices and c a nonzero scalar, Then 1. ( A1 ) 1 A 1 1 1 2. (cA) A c 1 1 1 3. ( AB) B A 4. ( An ) 1 ( A1 ) n 5. ( At ) 1 ( A1 )t Proof 1. By definition, AA1=A1A=I. 2. (cA)( 1c A1 ) I ( 1c A1 )(cA) 3. ( AB)( B 1 A1 ) A( BB 1 ) A1 AA1 I ( B 1 A1 )( AB) 1 1 1 n n 4. An ( A1 ) n A A A A I ( A ) A n times n times 5. AA1 I , ( AA1 )t ( A1 )t At I , A1 A I , ( A1 A)t At ( A1 )t I , Ch2_55 Example 4 If A 4 1, then it can be shown that A1 1 1. Use this 3 1 3 4 informatio n to compute ( At ) 1. Solution t 1 1 1 3 (A ) (A ) . 3 4 1 4 t 1 1 t 隨堂作業:15,19 Ch2_56 Theorem 2.8 Let AX = Y be a system of n linear equations in n variables. If A–1 exists, the solution is unique and is given by X = A–1Y. Proof (X = A–1Y is a solution.) Substitute X = A–1Y into the matrix equation. AX = A(A–1Y) = (AA–1)Y = InY = Y. (The solution is unique.) Let X1 be any solution, thus AX1 = Y. Multiplying both sides of this equation by A–1 gives A–1AX1= A–1Y InX1 = A–1Y X1 = A–1Y. Ch2_57 Example 5 x1 x2 2 x3 1 Solve the system of equations 2 x1 3x2 5 x3 3 x1 3 x2 5 x3 2 Solution This system can be written in the following matrix form: 1 1 2 x1 1 2 3 5 x2 3 3 5 x3 2 1 If the matrix of coefficients is invertible, the unique solution is 1 x1 1 1 2 1 x2 2 3 5 3 x 1 3 5 2 3 This inverse has already been found in Example 2. We get x1 0 1 1 1 1 x2 5 3 1 3 2 x 3 2 1 2 1 3 The unique solution is x1 1, x2 2, x3 1. Ch2_58 Elementary Matrices Definition An elementary matrix is one that can be obtained from the identity matrix In through a single elementary row operation. Example 1 0 0 I 3 0 1 0 0 0 1 R2 R3 5R2 R2+ 2R1 1 0 0 E1 0 0 1 0 1 0 1 0 0 E2 0 5 0 0 0 1 1 0 0 E3 2 1 0 0 0 1 Ch2_59 Elementary Matrices 一個矩陣做 elementary row operation, 相當於在左邊乘一個對應的 elementary matrix。 a b R2 R3 g h d e a b A d e g h c f i 5R2 a 5d g c 1 0 0 i 0 0 1 A E1 A f 0 1 0 b c 1 0 0 5e 5 f 0 5 0 A E2 A h i 0 0 1 b a d 2a e 2b R2+ 2R1 g h c 1 0 0 f 2c 2 1 0 A E3 A i 0 0 1 Ch2_60 Notes for elementary matrices Each elementary matrix is square and invertible. Example I E1 E1 I , i.e., E2 E1 I R1 2 R 2 1 0 0 I 0 1 0 0 0 1 R1 2 R 2 1 2 0 E1 0 1 0 0 0 1 1 2 0 E2 0 1 0 0 0 1 If A and B are row equivalent matrices and A is invertible, then B is invertible. Proof If A … B, then B=En … E2 E1 A for some elementary matrices En, … , E2 and E1. So B1 = (En … E2 E1A)1 =A1E11 E21 … En1. Ch2_61 Homework Exercise 2.4: 4, 7, 14, 15, 19, 21, 28 (後面的 section 2.5~2.7 都跳過) Exercise 7 d b 1 1 a b . If A , show that A (ad bc) c a c d (求22之反矩陣,此公式較快) Ch2_62 Homework Exercise 21 Prove that (AtBt)1 = (A1B1)t. Exercise 28 True or False: (a) If A is invertible A1 is invertible. (b) If A is invertible A2 is invertible. (c) If A has a zero on the main diagonal it is not invertible. (d) If A is not invertible then AB is not invertible. (e) A1 is row equivalent to In. Ch2_63 2.5 Matrix Transformations, Rotations, and Dilations A function, or transformation, is a rule that assigns to each element of a set a unique element of another set. We will be especially interested in linear transformations, which are transformations that preserve the mathematical structure of a vector space. Consider the function f(x) =3x2+4. domain (定義域) of the function: the set of all possible x f(2)=16, we say that the image of 2 is 16. Extend these ideas to functions between vector spaces We usually use the term transformation rather than function in linear algebra. Ch2_64 Consider the transformation T maps R3 into R2 defined by T(x, y, z) = (2x, yz) The domain of T is R3 and we say the codomain (對應域) is R2. T(1, 4, 2) = (2, 6) The image of (1, 4, 2) is (2, 6). Convenient representations: x T y 2 x , and the image of z y z 1 2 4 is 6. 2 Ch2_65 Definition A transformation T of Rn into Rm is a rule that assigns to each vector u in Rn a unique vector v in Rm. Rn is called the domain of T and Rm is the codomain. We write T(u) = v; v is the image of u under T. The term mapping is also used for a transformation. Ch2_66 Dilation(擴張) and Contraction(收縮) Consider the transformation T x r x , where r > 0. y y This equation can be written as the following useful matrix form. T x r 0 x y 0 r y Figure 2.11 Ch2_67 Reflection(反射) Consider the transformation T x x . T is called a reflection. y y This equation can be written as the following useful matrix form. T x 1 0 x y 0 1 y Figure 2.12 Ch2_68 Matrix transformations Every matrix defines a transformation. Let A be a matrix and x be a column vector such that Ax exists. Then A defines the matrix transformation T(x) = Ax. x For example, A 5 3 2 defines T y 50 0 4 1 z 1 3 6 1 14 , written as T 3 8 T and 4 2 2 1 , written as 3 6 4 8 隨堂作業:1 x 3 2 y 4 1 z 3 14 1 2 2 Figure 2.14 Matrix transformations. Ch2_69 Definition Let A be an mn matrix. Let x be an element of Rn written in a column matrix form. A defines a matrix transformation T(x)=Ax of Rn into Rm. The vector Ax is the image of x. The domain of the transformation is Rn and codomain is Rm. We write T: Rn Rm if T maps Rn into Rm. Geometrical Properties: Matrix transformations maps line segments (線段) into line segments (or points). If the matrix is invertible (可逆) the transformation also maps parallel lines into parallel lines. Ch2_70 Example 1 Consider the transformation T: R2 R2 defined by the matrix 4 2 . Determine the image of the unit square under this A 2 3 transformation. Sol. The unit square is the square whose vertices are the points 1 1 0 0 P , Q , R , O 0 1 1 0 Since 4 2 1 4 2 3 0 2, 4 2 1 6 2 3 1 5, 4 2 0 2 2 3 1 3 Ch2_71 The images are: P P' 1 4 0 2, Q Q' R R' O O 1 6 0 2 0 0 1 5, 1 3, 0 0 Figure 2.15 The square PQRO is mapped into the parallelogram P’Q’R’O. Ch2_72 Composition(合成) of Transformations Consider the matrix transformations T1(x)=A1x and T2(x)=A2x. The composite transformation T=T2 T1 is given by T(x) = T2(T1(x)) = T2 (A1x) =A2A1x Thus T is defined by the matrix product A2A1. T(x) = A2A1x Figure 2.16 隨堂作業:10(a) Ch2_73 Homework Exercise 2.5: 1, 10 Ch2_74 2.6 Linear Transformations A vector space has two operations: addition and scalar multiplication. Consider the matrix transformation T(u)=Au. T(u+v) = A(u+v) = Au+Av = T(u)+T(v) u+v v T(v) T u T(u+v) = T(u)+T(v) T(u) T(cu) = A(cu) = cAu = cT(u) u cu T T(cu) = cT(u) T(u) Definition Let u and v be vectors in Rn and let c be a scalar. A transformation T: Rn → Rm is said to be a linear transformation if T(u + v) = T(u) + T(v) and T(cu) = cT(u). Ch2_75 Example 1 Prove that the following transformation T: R2 → R2 is linear. T(x, y) = (x y, 3x) Solution “+”: Let (x1, y1) and (x2, y2) R2. Then T((x1, y1) + (x2, y2)) = T(x1 + x2, y1 + y2) = (x1 + x2 y1 y2, 3x1 + 3x2) = (x1 y1, 3x1) + (x2 y2, 3x2) = T(x1, y1) + T(x2, y2) “”: Let c be a scalar. T(c(x1, y1)) = T(cx1, cy1) = (cx1 cy1, 3cx1) = c(x1 y1, 3x1) = cT(x1, y1) Thus T is linear. 隨堂作業:1 Note A linear transformation T: U → U is called a linear operator. Ch2_76 Example 2 Show that the following transformation T: R3 → R2 is not linear. T(x, y, z) = (xy, z) Solution “+”: Let (x1, y1, z1) and (x2, y2, z2) R3. Then T((x1, y1, z1) + (x2, y2, z2)) = T(x1 + x2 , y1 + y2 , z1 + z2) = ((x1 + x2)(y1 y2), z1 + z2) and T(x1, y1, z1) + T(x2, y2, z2) = (x1y1, z1) + (x2y2, z2) Thus, in general T((x1, y1, z1) + (x2, y2, z2)) T(x1, y1, z1) + T(x2, y2, z2). T is not linear. 隨堂作業:4(b) Ch2_77 Example 3 Determine a matrix A that describes the linear transformation 2x y x T( ) y 3 y Solution It can be shown that T is linear. The domain of T is R2. We find the effect of T on the standard basis of R2. T ( 1) 2 0 0 and T ( 0) 1 1 3 A 2 1 0 3 Why? See the next page. T can be written as T ( x ) 2 1 x y 0 3 y Ch2_78 Matrix Representation Let T: Rn → Rm be a linear transformation. {e1, e2, …, en} be the standard basis of Rn, and u be an arbitrary vector in Rn. 1 0 a1 e1 ,, e n u 0 1 an Express u in terms of the basis, u = a1e1+a2e2+ … + anen. Since T is a linear transformation, T(u) = T(a1e1+ a2e2 + … + anen) = T(a1e1) + T(a2e2) + …+ T(anen) = a1T(e1) + a2T(e2) + …+ anT(en) a1 [T (e1 ) T (e 2 ) T (e n )] an Ch2_79 Thus the linear transformation T is defined by the matrix A = [ T(e1) … T(en) ]. A is called the standard matrix of T. Ch2_80 Example 4 The transformation T ( x ) 2 x y defines a reflection in the y 3 y line y = x. It can be shown that T is linear. Determine the standard matrix of this transformation. Find the image of 4 . 1 Solution We find the effect of T on the standard basis. T ( 1) 0 0 1 0 1 and T ( ) 1 0 1 1 0 A 0 T can be written as T ( x ) 0 1 x y 1 0 y T ( 4) 0 1 4 1 1 1 0 1 4 Ch2_81 Figure 2.22 隨堂作業:12 Ch2_82 Homework Exercise 2.6: 1, 4, 12 (2.7~2.9節跳過) Ch2_83