Section 5.5. Matrices and Vectors A matrix is a rectangular array of objects arranged in rows and columns. The objects are called the entries. A matrix with m rows and n columns is called an m n matrix. m n is called the size of the matrix, and the numbers m and n are its dimensions. 1
A matrix in which the number of rows equals the number of columns, m = n, is called a square matrix of order n. 2
If A is an m n matrix, then a ij denotes the element in the i th row and j th column of A: A = a 11 a 12 a 13 a 21 a 22 a 23 a 31 a 32 a 33.... a m1 a m2 a m3 a 1n a 2n a 3n a mn. The notation A = (a ij ) also represents this display. 3
Special Cases: Vectors A 1 n matrix v = (a 1 a 2... a n ) also written as v = (a 1, a 2,..., a n ) is called an row vector. 4
An m 1 matrix v = a 1 a 2. a m is called a column vector. The entries of a row or column vector are called the components of the vector. 5
Arithmetic of Matrices Let A = (a ij ) be an m n matrix and let B = (b ij ) be a p q matrix. Equality: A = B if and only if 1. m = p and n = q; 2. a ij = b ij for all i and j. That is, A = B if and only if A and B are identical. 6
Example: a b 3 2 c 0 = 7 1 x 2 4 0 if and only if a = 7, b = 1, c = 4, x = 3. 7
Addition: Let A = (a ij ) and B = (b ij ) be m n matrices. A+B is the m n matrix C = (c ij ) where c ij = a ij + b ij for all i and j. That is, A + B = (a ij + b ij ). Addition of matrices is not defined for matrices of different sizes. 8
Examples: (a) 2 4 3 2 5 0 + 4 0 6 1 2 0 = 2 4 3 1 7 0 (b) 2 4 3 2 5 0 + 1 3 5 3 0 6 is not defined. 9
Properties: Let A, B, and C be matrices of the same size. Then: 1. A+B = B +A (Commutative) 2. (A + B) + C = A + (B + C) (Associative) A matrix with all entries equal to 0 is called a zero matrix. E.g., 0 0 0 0 0 0 and 0 0 0 0 0 0 10
The symbol 0 will be used to denote the zero matrix of arbitrary size. 3. A + 0 = 0 + A = A for all A. The zero matrix is called the additive identity. 11
4. To each m n matrix A there is a unique matrix m n matrix A such that A + ( A) = 0 The entries of A are the negatives of the entries of A. A is called the additive inverse of A 12
Example: A = 1 7 2 2 0 6 4 1 5 A = 1 7 2 2 0 6 4 1 5 1 7 2 2 0 6 4 1 5 + 1 7 2 2 0 6 4 1 5 = 0 0 0 0 0 0 0 0 0 13
Subtraction: Let A = (a ij ) and B = (b ij ) be m n matrices. Then A B = A + ( B). Example: 2 4 3 2 5 0 4 0 6 1 2 0 = 2 4 3 2 5 0 + 4 0 6 1 2 0 = 14
Properties: Let A, B, and C be matrices of the same size. Then: 1. A+B = B+A (commutative) 2. (A + B) + C = A + (B + C) (associative) 3. A + 0 = 0 + A = A (additive identity) 4. A + ( A) = ( A) + A = 0 (additive inverse) 15
Multiplication of a Matrix by a Number The product of a number k and a matrix A, denoted ka, is given by ka = (ka ij ). This product is also called multiplication by a scalar. 16
Example: 3 2 1 4 1 5 2 4 0 3 = 17
Properties: Let A, B be m n matrices and let α, β be real numbers. Then 1. 1 A = A 2. 0 A = 0 3. α(a + B) = α A + α B 4. (α + β)a = α A + β A 18
Matrix Multiplication 1. The Product of a Row Vector and a Column Vector: The product of a 1 n row vector and an n 1 column vector is the number given by (a 1, a 2, a 3,..., a n ) b 1 b 2 b 3. b n = a 1 b 1 + a 2 b 2 + a 3 b 3 + + a n b n. 19
Also called scalar product (because the result is a number (scalar)), dot product, and inner product. The product of a row vector and a column vector (of the same dimension and in that order!) is a number. The product of a row vector and a column vector of different dimensions is not defined. 20
Examples: (3, 2, 5) 1 4 1 ( 2, 3, 1, 4) 2 4 3 5 21
2. The Product of Two Matrices: Let A = (a ij ) be an m p matrix and let B = (b ij ) be a p n matrix. The matrix product of A and B (in that order), denoted AB, is the m n matrix C = (c ij ), where c ij is the product of the i th row of A and the j th column of B. 22
AB=C 23
Let A and B be given matrices. The product AB, in that order, is defined if and only if the number of columns of A equals the number of rows of B. If the product AB is defined, then the size of the product is: (no. of rows of A) (no. of columns of B): A m p B = C p n m n 24
Examples: 1. A = 1 4 2 3 1 5, B = 3 0 1 2 1 2 AB = 1 4 2 3 1 5 3 0 1 2 1 2 = BA = 3 0 1 2 1 2 1 4 2 3 1 5 = 25
2. A = 1 2 3 4, B = 1 1 2 3 0 5 AB = 5 1 12 9 3 26 BA does not exist. 26
3. A = 2 1 4 2, B = 1 1 2 2 AB = 2 1 4 2 1 1 2 2 BA = 1 1 2 2 2 1 4 2 Conclusion: Matrix multiplication is not commutative; AB BA in general. 27
Properties of Matrix Multiplication: Let A, B, and C be matrices. 1. AB BA in general; NOT COMMUTATIVE. 2. (AB)C = A(BC) matrix multiplication is associative. 28
Identity Matrices: Let A be a square matrix of order n. The entries a 11, a 22, a 33,..., a nn form the main diagonal of A. For each positive integer n > 1, let I n denote the square matrix of order n whose entries on the main diagonal are all 1, and all other entries are 0. The matrix I n is called the n n identity matrix. 29
I 2 = 1 0 0 1, I 3 = 1 0 0 0 1 0 0 0 1, I 4 = 1 0 0 0 0 1 0 0 0 0 1 0 0 0 0 1 and so on. 30
3. If A is an m n matrix, then I m A = A and AI n = A. If A is a square matrix of order n, then AI n = I n A = A, 4. Inverse????? 31
Distributive Laws: 1. A(B + C) = AB + AC. This is called the left distributive law. 2. (A+B)C = AC + BC. This is called the right distributive law. 3. k(ab) = (ka)b = A(kB) 32
Other ways to look at systems of linear equations. A system of m linear equations in n unknowns: a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a 31 x 1 + a 32 x 3 + + a 3n x n = b 3... a m1 x 1 + a m2 x 2 + + a mn x n = b m 33
Because of the way we defined multiplication, a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a 31 x 1 + a 32 x 3 + + a 3n x n = b 3... a m1 x 1 + a m2 x 2 + + a mn x n = b m is the same as a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n a 31 a 32 a 33.... a 3n. a m1 a m2 a m3 a mn x 1 x 2 x 3. x n = b 1 b 2 b 3. b m 34
or in the vector-matrix form Ax = b c.f. ax = b. (1) Solution: x = A 1 b????? 35
Section 5.6. Square matrices 1. Inverse Let A be an n n matrix. An n n matrix B with the property that AB = BA = I n is called the multiplicative inverse of A or, more simply, the inverse of A. 36
Uniqueness: If A has an inverse, then it is unique. That is, there is one and only one matrix B such that AB = BA = I. B is denoted by A 1. 37
Procedure for finding the inverse: A = 1 2 3 4 We want 1 2 3 4 x y z w = 1 0 0 1 38
Examples: 1. A = 1 2 1 0 3 4 0 1 1 2 3 4 1 0 2 1 0 1 3 2 1 2 A 1 = 2 1 3 2 1 2 39
2. B = 2 1 4 2 2 1 1 0 4 2 0 1 B does not have an inverse. 40
NOTE: Not every nonzero n n matrix A has an inverse! 41
3. C = 1 0 2 2 1 3 4 1 8 Form the augmented matrix: 1 0 2 1 0 0 2 1 3 0 1 0 4 1 8 0 0 1 42
1 0 2 1 0 0 2 1 3 0 1 0 4 1 8 0 0 1 1 0 0 11 2 2 0 1 0 4 0 1 0 0 1 6 1 1 C 1 = 11 2 2 4 0 1 6 1 1 43
Finding the inverse of A. Let A be an n n matrix. a. Form the augmented matrix (A I n ). b. Reduce (A I n ) to reduced row echelon form. If the reduced row echelon form is (I n B), then B = A 1 If the reduced row echelon form is not (I n B), then A does not have an inverse. 44
That is, if the reduced row echelon form of A is not the identity, then A does not have an inverse. Signal: The row reduction process produces a row of 0 s 45
Application: Solve the system x +2z = 3 2x y +3z = 5 4x +y +8z = 2 Written in matrix form: 1 0 2 2 1 3 4 1 8 x y z = 3 5 2 Solution: Cx = b; x = C 1 b x y z = 11 2 2 4 0 1 6 1 1 3 5 2 = 29 14 15 46
Note: The n n system of equations Ax = b has a unique solution if and only if the matrix of coefficients, A, has an inverse. 47
2. Determinants Find the solution set: ax + by = α cx + dy = β a b c d x y = α β, Ax = b (ad bc)x = (αd βb) (ad bc)y = (aβ cα) The number ad bc is called the determinant of A, denoted det A. 48
A. Calculations 1. 1 1 A = (a) det A = a 2. 2 2 A = det A = a b c d a b c d = ad bc. 49
2. 3 3 A = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 deta = a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 b 2 c 3 a 1 b 3 c 2 a 2 b 1 c 3 +a 2 b 3 c 1 +a 3 b 1 c 2 a 3 b 2 c 1 OMG 50
a 1 b 2 c 3 a 1 b 3 c 2 a 2 b 1 c 3 +a 2 b 3 c 1 +a 3 b 1 c 2 a 3 b 2 c 1 a 1 (b 2 c 3 b 3 c 2 ) a 2 (b 1 c 3 b 3 c 1 )+a 3 (b 1 c 2 b 2 c 1 ) = a 1 b 2 b 3 c 2 c 3 a 2 b 1 b 3 c 1 c 3 + a 3 b 1 b 2 c 1 c 2 This is called the expansion across the first row. 51
Example: A = 2 3 3 0 4 2 2 1 3 2 3 3 0 4 2 2 1 3 = 2 4 2 1 3 ( 3) 0 2 2 3 +3 0 4 2 1 = 2(12 2) ( 3)(0+4)+3(0+8) = 2(10) ( 3)(4) + 3(8) = 56 52
a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 (b 2 c 3 b 3 c 2 ) a 2 (b 1 c 3 b 3 c 1 )+a 3 (b 1 c 2 b 2 c 1 ) = a 2 (b 1 c 3 b 3 c 1 )+b 2 (a 1 c 3 a 3 c 1 ) c 2 (a 1 b 3 a 3 b 1 ) = a 2 b 1 b 3 c 1 c 3 + b 2 a 1 a 3 c 1 c 3 c 2 a 1 a 3 b 1 b 3 Expansion down the second column. 53
Example: A = 2 3 3 0 4 2 2 1 3 2 3 3 0 4 2 2 1 3 = ( 3) 0 2 2 3 +4 2 3 2 3 1 2 3 0 2 = 3(4) + 4(12) 4 = 56 54
a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = a 1 (b 2 c 3 b 3 c 2 ) a 2 (b 1 c 3 b 3 c 1 )+a 3 (b 1 c 2 b 2 c 1 ) = a 3 b 1 c 2 a 3 b 2 c 1 b 3 a 1 c 2 +b 3 a 2 c 1 +c 3 a 1 b 2 c 3 a 2 b 1 = a 3 (b 1 c 2 b 2 c 1 ) b 3 (a 1 c 2 a 2 c 1 )+c 3 (a 1 b 2 a 2 b 1 ) = a 3 b 1 b 2 c 1 c 2 b 3 a 1 a 2 c 1 c 2 + c 3 a 1 a 2 b 1 b 2 Expansion down the third column 55
and so on. You can expand across any row, or down any column. BUT: Associated with each position is an algebraic sign: + + + + + For example, across the second row: a 1 a 2 a 3 b 1 b 2 b 3 c 1 c 2 c 3 = = b 1 a 2 a 3 c 2 c 3 +b 2 a 1 a 3 c 1 c 3 b 3 a 1 a 2 c 1 c 2 56
3. 4 4 determinants a 1 a 2 a 3 a 4 b 1 b 2 b 3 b 4 c 1 c 2 c 3 c 4 d 1 d 2 d 3 d 4 Sign chart + + + + + + + + 57
Recall: Find the solution set: ax + by = α cx + dy = β a b c d x y = α β, Ax = b (ad bc)x = αd bβ (ad bc)y = aβ αc (det A) x = αd bβ = α b β d (det A) y = aβ αc = a α c β 58
(det A) x = αd βb = α b β d (det A) y = aβ cα = a α c β If det A 0, then x = y = α b β d det A a α c β det A If det A = 0, then the system either has infinitely many solutions or no solutions. 59
Cramer s Rule. Given a system of n linear equations in n unknowns: (a square system) a 11 x 1 + a 12 x 2 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + + a 2n x n = b 2 a 31 x 1 + a 32 x 2 + + a 3n x n = b 3... a n1 x 1 + a n2 x 2 + + a nn x n = b n x i = det A i, (provided det A 0) det A where A i is the matrix A with the i th b. column replaced by the vector 60
The n n system of equations Ax = b has a unique solution if and only if det A 0. If det A = 0, then the system either has infinitely many solutions or no solutions. 61
Examples: 1. Given the system x + 2y + 2z = 3 2x y + z = 5 4x + y 2z = 2 Does Cramer s rule apply? det A = 1 2 2 2 1 1 4 1 2 = 62
x + 2y + 2z = 3 2x y + z = 5 4x + y 2z = 2 Find y. y = 1 3 2 2 5 1 4 2 2 1 2 2 2 1 1 4 1 2 = 63
2. Given the system 2x + 7y + 6z = 1 5x + y 2z = 7 3x + 8y + 4z = 1 det A = 2 7 6 5 1 2 3 8 4 = 0 Cramer s Rule does not apply Does the system have infinitely many solutions or no solutions?? Compare with ax = b. 64
B. Properties of determinants: Let A be an n n matrix. 1. If A has a row or column of zeros, then det A = 0 Example: 1 0 2 2 0 3 4 0 8 Expand down second column: 0 2 3 4 8 + 0 1 2 4 8 0 1 2 3 2 = 0 65
2. If A is a diagonal matrix, A = det A = a 1 b 2 0 0 c 3 a 1 0 0 0 b 2 0 0 0 c 3 0, 0 0 0 c 3 +0 0 b 2 0 0 = a 1 b 2 c 3. In particular, det I n = 1 For example, I 3 : 1 0 0 0 1 0 0 0 1 = 1 66
3. If A is a triangular matrix, e.g., A = a 1 a 2 a 3 a 4 0 b 2 b 3 b 4 0 0 c 3 c 4 0 0 0 d 4 (upper triangular) Expand down first column!! det A = a 1 b 2 c 3 d 4 67
4. If B is obtained from by interchanging any two rows (columns), then det B = det A. 68
NOTE: If A has two identical rows (or columns), then det A = 0. 69
5. Multiply a row (column) of A by a nonzero number k to obtain a matrix B. Then det B = kdet A. 70
6. Multiply a row (column) of A by a number k and add it to another row (column) to obtain a matrix B. Then det B = det A. 71
7. Let A and B be n n matrices. Then det[ab] = det Adet B. 72
Example: Calculate 3 3 1 5 2 2 0 2 4 1 3 2 2 10 3 2 73
= 6 1 1 0 1 0 1 1 2 0 0 1 8 0 0 0 60 = 360 74
Equivalences: 1. The system of equations: Ax = b has a unique solution. 2. The reduced row echelon form of A is I n. 3. The rank of A is n. 4. A has an inverse. 5. det A 0. 75
A is nonsingular if det A 0; A is singular if det A = 0. 76
Section 5.7. Vectors; Linear Dependence/Independence in R n R 2 = {(a, b) : a, b R} the plane R 3 = {(a, b, c) : a, b, c R} 3- space R 4 = {(x 1, x 2, x 3, x 4 )} 4-space and so on 77
Let S = {v 1,v 2,,v k } be a set of vectors in R n. The set S is linearly dependent if there exist k numbers c 1, c 2,, c k NOT ALL ZERO such that c 1 v 1 + c 2 v 2 + + c k v k = 0. (c 1 v 1 + c 2 v 2 + +c k v k is a linear combination of v 1, v 2,...) 78
S is linearly independent if it is not linearly dependent. That is, S is linearly independent if c 1 v 1 + c 2 v 2 + + c k v k = 0 holds ONLY WHEN c 1 = c 2 = = c k = 0. 79
The set S is linearly dependent if there exist k numbers c 1, c 2,, c k NOT ALL ZERO such that c 1 v 1 + c 2 v 2 + + c k v k = 0. Another way to say this: The set S is linearly dependent if one of the vectors can be written as a linear combination of the others. 80
1. Two vectors v 1, v 2. Linearly dependent iff one vector is a multiple of the other. Examples: v 1 = (1, 2, 4), v 2 = ( 1, 1, 2) 2 linearly dependent: v 1 = 2v 2 81
v 1 = (2, 4, 5), v 2 = (0, 0, 0) linearly dependent: v 2 = 0v 1 v 1 = (5, 2, 0), v 2 = ( 3, 1, 9) linearly independent 82
2. v 1 = (1, 1), v 2 = (2, 3) v 3 = (3, 5) {v 1, v 2 } Dependent or independent?? {v 1, v 2, v 3 } Dependent or independent?? 83
Does there exist three numbers, c 1, c 2, c 3, not all zero such that c 1 v 1 + c 2 v 2 + c 3 v 3 = (0,0) That is, does the system of equations c 1 + 2c 2 + 3c 3 = 0 c 1 3c 2 5c 3 = 0 have nontrivial solutions? 84
A homogeneous system with more unknowns than equations always has infinitely many nontrivial solutions. Let v 1,v 2,,v k be a set of k vectors in R n. If k > n, then the set of vectors is (automatically) linearly dependent. 85
3. v 1 = (1, 1, 2), v 2 = (2, 3, 0), v 3 = ( 1, 2, 2), v 4 = (0, 4, 3) Dependent or independent?? a. {v 1, v 2, v 3, v 4 } b. {v 1, v 2, v 3 } c. {v 1, v 2 } 86
v 1 = (1, 1, 2), v 2 = (2, 3, 0), v 3 = ( 1, 2, 2) Does c 1 v 1 + c 2 v 2 + c 3 v 3 = 0 have non-trivial solutions? That is, does c 1 + 2c 2 c 3 = 0 c 1 3c 2 2c 3 = 0 2c 1 + 2c 3 = 0 have non-trivial solutions 87
Augmented matrix and row reduce: 1 2 1 0 1 3 2 0 2 0 2 0 88
NOTE: It s enough to row reduce: 1 2 1 1 3 2 2 0 2 89
Calculate the determinant 1 2 1 1 3 2 2 0 2 90
4. v 1 = ( a, 1, 1), v 2 = ( 1, 2a, 3), v 3 = ( 2, a, 2), v 4 = (3a, 2, a) For what values of a are the vectors linearly dependent? 91
4. v 1 = ( a, 1, 1), v 2 = ( 1, 2a, 3), v 3 = ( 2, a, 2), v 4 = (3a, 2, a) For what values of a are the vectors v 1, v 2, v 3 linearly dependent? a c 1 c 2 2c 3 = 0 c 1 + 2a c 2 + a c 3 = 0 c 1 + 3c 2 + 2c 3 = 0 92
Augmented matrix and row reduce: a 1 2 0 1 2a a 0 1 3 2 0 93
Or row reduce: a 1 2 1 2a a 1 3 2 94
Calculate the determinant a 1 1 1 2a 3 2 a 2 95
5. v 1 = (1, 1,2,1), v 2 = (3,2,0, 1) v 3 = ( 1, 4,4,3), v 4 = (2,3, 2, 2) a. {v 1, v 2, v 3, v 4 } dependent or independent? b. What is the maximum number of independent vectors? 96
Row reduce 1 1 2 1 3 2 0 1 1 4 4 3 2 3 2 2 97
Or, calculate the determinant. 1 1 2 1 3 2 0 1 1 4 4 3 2 3 2 2 98
Tests for independence/dependence Let S = v 1,v 2,,v k be a set of vectors in R n. Case 1: k > n : S is linearly dependent. 99
Case 2: k = n : Form the n n matrix A whose rows are v 1, v 2,,v n 1. Row reduce A: if the reduced matrix has n nonzero rows independent; one or more zero rows dependent. 100
Equivalently, solve the system of equations c 1 v 1 + c 2 v 2 + + c k v k = 0. Unique solution: c 1 = c 2 = = c n = 0 independent; infinitely many solutions dependent.
2. Calculate det A: det A 0 independent; det A = 0 dependent. Note: If v 1,v 2,,v n is a linearly independent set of vectors in R n, then each vector in R n has a unique representation as a linear combination of v 1,v 2,,v n.
Case 3: k < n : Form the k n matrix A whose rows are v 1, v 2,,v k 1. Row reduce A: if the reduced matrix has k nonzero rows independent; one or more zero rows dependent. 101
Equivalently, solve the system of equations c 1 v 1 + c 2 v 2 + + c k v k = 0. Unique solution: c 1 = c 2 = = c n = 0 independent; infinitely many solutions dependent.
Linear Dependence/Independence of Functions Let S = {f 1 (x), f 2 (x), f k (x)} be a set of functions defined on an interval I. S is linearly dependent if there exist k numbers c 1, c 2,, c k NOT ALL ZERO such that c 1 f 1 (x)+c 2 f 2 (x)+ +c k f k (x) 0 on I. 102
S is linearly independent if it is not linearly dependent. That is, S is linearly independent if c 1 f 1 (x)+c 2 f 2 (x)+ +c k f k (x) 0 on I implies c 1 = c 2 = = c k = 0. Examples: 103
1. f 1 (x) = 1, f 2 (x) = x, f 3 (x) = x 2 Suppose f 1, f 2, f 3 are linearly dependent. Then there exist 3 numbers c 1, c 2, c 3, NOT ALL 0 such that c 1 1 + c 2 x + c 3 x 2 0 That is, c 1 1 + c 2 x + c 3 x 2 = 0 for all x 104
2. f 1 (x) = sin x, f 2 (x) = cos x Suppose f 1, f 2 are linearly dependent. Then there exist 2 numbers c 1, c 2, NOT BOTH 0 such that c 1 cos x + c 2 sin x 0 That is, c 1 cos x + c 2 sin x = 0 for all x 105
3. f 1 (x) = sin x, f 2 (x) = cos x, f 3 (x) = cos(x + π/3) 106
Test for independence/dependence Suppose that the functions f 1 (x), f 2 (x), f k (x) are (k 1)-times differentiable on I. If the determinant W(x) = f 1 (x) f 2 (x) f k (x) f 1 (x) f 2 (x) f k (x) f (k 1) 1 (x) f (k 1) 2 (x) f (k 1) (x) k 0 for at least one x I, then the functions are linearly independent on I. NOTE: If W(x) 0 in I, then?????? The functions may be dependent, they may be independent!!!! 107
W(x) is called THEEEEEE
WRONSKIAN 108
Examples: f 1 = 2x+1, f 2 = x 3, f 3 = 3x+2 Wronskian: 2x + 1 x 3 3x + 2 2 1 3 0 0 0 = 0 Independent or dependent??? Does there exist 3 numbers, c 1, c 2, c 3, not all 0, such that c 1 f 1 + c 2 f 2 + c 3 f 3 0 109
f 1 (x) = x 3, f 2 (x) = x 3, x < 0 2x 3, x 0 Wronskian: For x < 0 x 3 x 3 3x 2 3x 2 = 0 For x 0 x 3 2x 3 3x 2 6x 2 = 0 110
Suppose there exist 2 numbers c 1, c 2, not both zero, such that c 1 f 1 (x) + c 2 f 2 (x) 0 111
Interpretations of a system of linear equations Given the system of m linear equations in n unknowns: a 11 x 1 + a 12 x 2 + a 13 x 3 + + a 1n x n = b 1 a 21 x 1 + a 22 x 2 + a 23 x 3 + + a 2n x n = b 2 a 31 x 1 + a 32 x 2 + a 33 x 3 + + a 3n x n = b 3... a m1 x 1 + a m2 x 2 + a m3 x 3 + + a mn x n = b m 112
The coefficient matrix A from R n to R m is a mapping a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n a 31. a 32. a 33.. a 3n a m1 a m2 a m3 a mn x 1 x 2 x 3. x n = b 1 b 2 b 3. b m 113
A is a linear transformation! A[x + y] = Ax + Ay and A[αx] = α Ax Example: A = ( 1 2 1 3 0 2 ) is a linear transformation from R 3 to R 2. 114
Write the system as a 11 a 12 a 13 a 1n a 21 a 22 a 23 a 2n a 31. a 32. a 33.. a 3n a m1 a m2 a m3 a mn x 1 x 2 x 3. x n = b 1 b 2 b 3. b m or in the vector-matrix form Ax = b Given a vector b in R m, find a vector x in R n such that Ax = b. 115
Write the system equivalently as x 1 a 11 a 21. +x 2 a 12 a 22. + +x n a 1n a 2ṇ. = b 1 b 2. a m1 a m2 a mn b m This says, express b as a linear combination of the columns of A. 116
Section 5.8. Eigenvalues/Eigenvectors Example: Set A = 1 3 1 1 1 1 3 3 1. A is a linear transformation from R 3 to R 3. 1 3 1 1 1 1 3 3 1 1 1 1 = 1 1 1 117
1 3 1 1 1 1 3 3 1 1 3 2 = 12 2 13 1 3 1 1 1 1 3 3 1 3 2 3 = 6 4 6 = 2 3 2 3 1 3 1 1 1 1 3 3 1 1 0 1 = 2 0 2 = 2 1 0 1 118
Let A be an n n matrix. A number λ is an eigenvalue of A if there is a non-zero vector v such that Av = λv 119
To find the eigenvalues of A, find the values of λ that satisfy det(a λ I) = 0. Example: Let A = 2 3 1 4 A λi = 2 λ 3 1 4 λ det (A λi) = 2 λ 3 1 4 λ = (2 λ)(4 λ) 3 = λ 2 6λ+5 = 0 Eigenvalues: λ 1 = 5, λ 2 = 1 120
Let A = 1 3 1 1 1 1 3 3 1 det (A λi) = 1 λ 3 1 1 1 λ 1 3 3 1 λ = λ 3 + λ 2 + 4λ 4 det (A λi) = 0 implies λ 3 + λ 2 + 4λ 4 = 0 or λ 3 λ 2 4λ + 4 = 0 121
Terminology: det(a λ I) is a polynomial of degree n, called the characteristic polynomial of A. The zeros of the characteristic polynomial are the eigenvalues of A The equation det(a λi) = 0 is called the characteristic equation of A. 122
A non-zero vector v that satisfies Av = λv is called an eigenvector corresponding to the eigenvalue λ.
Examples: 1. A = 2 2 2 1 Characteristic polynomial: det(a λ I) = 2 λ 2 2 1 λ = (2 λ)( 1 λ) 4 = λ 2 λ 6 Characteristic equation: λ 2 λ 6 = (λ 3)(λ + 2) = 0 Eigenvalues: λ 1 = 3, λ 2 = 2 123
Eigenvectors: 124
2. A = 4 1 4 0 Characteristic polynomial: det(a λ I) = 4 λ 1 4 λ = λ 2 4λ + 4 Characteristic equation: λ 2 4λ + 4 = (λ 2) 2 = 0 Eigenvalues: λ 1 = λ 2 = 2 125
Eigenvectors: 126
3. A = 5 3 6 1 Characteristic polynomial: det(a λ I) = 5 λ 3 6 1 λ = λ 2 4λ + 13 Characteristic equation: λ 2 4λ + 13 = 0 Eigenvalues: λ 1 = 2 + 3i, λ 2 = 2 3i 127
Eigenvectors: 128
NOTE: If a + b i is an eigenvalue of A with eigenvector α+β i, then a b i is also an eigenvalue of A and α β i is a corresponding eigenvector. 129
4. A = 4 3 5 1 2 1 1 3 2 Characteristic polynomial: det(a λ I) = 4 λ 3 5 1 2 λ 1 1 3 2 λ = λ 3 + 4λ 2 λ 6 Characteristic equation: λ 3 4λ 2 +λ+6 = (λ 3)(λ 2)(λ+1) = 0. Eigenvalues: λ 1 = 3, λ 2 = 2, λ 3 = 1 130
Eigenvectors: 131
5. A = 4 1 1 2 5 2 1 1 2 Characteristic polynomial: det(a λ I) = 4 λ 1 1 2 5 λ 2 1 1 2 λ = λ 3 + 11λ 2 39λ + 45 Characteristic equation: λ 3 11λ 2 +39λ 45 = (λ 3) 2 (λ 5) = 0. Eigenvalues: λ 1 = λ 2 = 3, λ 3 = 5 132
Eigenvectors: 133
6. A = 3 1 1 7 5 1 6 6 2 Characteristic polynomial: det(a λ I) = 3 λ 1 1 7 5 λ 1 6 6 2 λ = λ 3 + 12λ + 16 Characteristic equation: λ 3 12λ 16 = (λ+2) 2 (λ 4) = 0. Eigenvalues: λ 1 = λ 2 = 2, λ 3 = 4 134
Eigenvectors: 135
Some Facts THEOREM If v 1, v 2,..., v k are eigenvectors of a matrix A corresponding to distinct eigenvalues λ 1, λ 2,..., λ k, then v 1, v 2,..., v k are linearly independent. 136
THEOREM Let A be a (real) n n matrix. If the complex number λ = a + bi is an eigenvalue of A with corresponding (complex) eigenvector u+iv, then λ = a bi, the conjugate of a + bi, is also an eigenvalue of A and u iv is a corresponding eigenvector. 137
THEOREM Let A be an n n matrix with eigenvalues λ 1, λ 2,..., λ n. Then det A = ( 1) n λ 1 λ 2 λ 3 λ n. That is, det A is the product of the eigenvalues of A. (The λ s are not necessarily distinct, the multiplicity of an eigenvalue may be greater than 1, and they are not necessarily real.) 138
Equivalences: (A an n n matrix) 1. Ax = b has a unique solution. 2. The reduced row echelon form of A is I n. 3. The rank of A is n. 4. A has an inverse. 5. det A 0. 6. 0 is not an eigenvalue of A 139