11 Adjoint of a Matrix If A is any nxn matrix and Cij is the cofactor of ij a then the matrix n1 n2 nn 21 22 2n 11 12 1n C C C C C C C C C is called the matrix of cofactors of A. The transpose of this matrix is called the adjoint of A and is denoted by adj(A). Example 11 Find the minors, cofactors and the adjoint of A if − − = 2 4 0 1 6 3 3 2 1 A
12 Theorem If A is invertible, then A 1 A -1 = adj(A) Example 12 Find the inverse of the matrix given in Example 11.
13 2.5 Cramer’s Rule If Ax = b is a system of linear equations with n equations and n unknowns and A 0 then the system of linear equations has a unique solution. This solution is: A A , , x A A , x A A x n n 2 2 1 1 = = = where Ai is the matrix obtained from A by replacing column-i in A by b . Example 13 Use Cramer’s Rule to solve the following system of linear equations. 2 3 8 3 4 6 30 2 6 1 2 3 1 2 3 1 3 − − + = − + + = + = x x x x x x x x
14 Exercise 2 1. Find the determinants of the following matrices using the cofactor expansion of the first column and the first row respectively: a. − − 0 1 2 2 0 2 3 4 4 6 4 12 5 3 0 6 b. − 1 5 2 10 4 1 1 2 2 6 11 12 3 0 7 0 2. A is a 3x3 matrix with A = −2 . Find the following determinants: a. 3 A b. ( ) -1 2A c. -3 A d. T 3A e. T AA f. - 7A 3. A and B are both 4x4 matrices with A = −2 and B = 7 . Find the following determinants: a. AB b. A B 3 c. -1 ABA d. -1 BAB e. T AB f. 3AB 4. Let = u v w p q r a b c A where A = 3 . If − − − = w c r v b q u a p 4 2 4 2 4 2 B find -1 3B . 5. Find each of the following determinants by reducing each matrix to a triangular matrix: a. 2 1 0 1 2 0 1 1 0 4 8 4 1 2 3 1 − − b. 4 5 3 2 3 6 1 1 3 4 5 5 1 2 7 9 − − − 6. If = −4 g h i d e f a b c , evaluate: a. a b c g h i d e f b. g h i d e f a b c 4 4 4 3 3 3 − − −
15 c. g d h e i f d e f a b c 4 4 4 3 3 3 − − − − − − d. i g h f d e c a b 2 2 2 7. If 2 2 2 1 3 3 3 = − − g − h − i d e f a b c , evaluate: . g h i d e f a b c 8. Use properties of determinants to evaluate the following: a. 93 94 95 91 92 93 89 90 91 b. c b c b a c a b c + + + 1 1 1 c. x x x 1 1 1 1 1 1 1 1 1 1 1 1 1 d. 1 2 1 3 1 4 3 1 9 2 4 4 4 4 4 3 3 3 3 3 0 0 2 2 2 − − 9. Determine whether or not each of the following matrices has an inverse: a. − − − 8 9 1 9 1 4 1 0 1 b. − − 3 1 6 2 1 4 4 2 8 10. Determine the value/s of k so that each of the following matrices is non-singular: a. - 2 k - 2 k - 3 - 2 b. k 3 2 3 1 6 1 2 4 11. Solve each of the following systems of equations by first finding the inverse of the coefficient matrix. a. 2 0 4 − + = − + = x y x y b. 2 0 0 − + = − + = x y x y c. 4 4 3 2 2 2 2 5 3 2 2 0 − + + = + + = + + = x y z x y z x y z d. 4 8 4 2 3 2 2 1 + + = − + = + = − x y z x y z x z
16 12. Find the inverse of each of the following matrices using the adjoint formula: a. 0 4 -1 0 b. − − − − 0 1 1 2 4 3 3 5 7 c. − 2 2 2 0 1 1 1 2 3 d. 0 − 4 −12 0 2 6 1 0 0 13. Use Cramer’s Rule to solve each of the following systems of linear equations: a. 1 2 5 − + = + = x y x y b. 5 2 2 1 2 2 3 10 4 1 − − = − + + = − − = x y z x y z x y z c. 6 6 3 4 4 6 11 3 4 4 11 − = − + = + + = x y x y z x y z d. 5 9 17 4 3 5 9 2 3 3 5 1 + + = + + = + + = x y z x y z x y z 14. P and Q are 3x3 matrices and PQ 2 1 R = . If P 2 -1 = and P Q 1 T = , find the determinants of P, Q and R. 15. Consider the following matrix: = m 5 0 0 1 2 1 2 3 A a. Determine the value of m if A = −6 by using cofactor expansion. b. Obtain the cofactor matrix of A. c. Find -1 A using the adjoint method. 16. Consider the following system of equations given in matrix form, Ax = b : = − − 4 - 2 m x x x 0 3 2 4 1 0 1 3 2 3 2 1 a. Evaluate A by cofactor expansion along column 1. b. If the value of x1 is 3, find the value of m using Cramer’s Rule.
17 ANSWER 1. a. 0 b. 0 2. a. -8 b. − 1 16 c. − 1 8 d. -54 e. 4 f. 686 3. a. -14 b. -56 c. 7 d. -2 e. -14 f. -1134 4. − 9 8 5. a. -24 b. 136 6. a. -4 b. 48 c. 12 d. -8 7. 1 6 8. a. 0 b. 2 − + − c. 3 − 3 2 + 3 − 1 d. 0 9. a. inverse b. no inverse 10. a ≠ 5±√17 2 b. ≠ −1 11.a. ( ) = ( 4 8 ) b. ( ) = ( 0 0 ) c. ( ) = ( −5 − 81 2 48 ) d. ( ) = ( 23 8 −12 ) 12. a. ( −1 0 0 1 4 ) b. ( 7 3 4 − 13 3 − 2 3 −1 5 3 − 2 3 −1 2 3 ) c. ( − 2 3 − 1 3 5 6 1 3 2 3 − 1 6 1 3 − 1 3 − 1 6) d. no inverse
18 13.a. ( ) = ( 1 2 ) b. ( ) = ( 1 1 2 ) c. ( ) = ( 1 1 2 3 2 ) d. ( ) = ( 0 − 1 2 1 2 ) 14. || = 1 2 , || = 2, || = 1 8 15. a. 4 b. ( 10 8 −4 15 −12 3 1 −2 1 ) c. ( − 5 3 − 5 2 − 1 6 − 4 3 2 1 3 2 3 − 1 2 − 1 6) 16. a. -2 b. 7
1 CHAPTER 3 VECTORS AND VECTOR SPACES 3.1 Introduction to Vectors Vectors are characterised by 2 quantities – length (magnitude)and direction, for example, velocity, force, etc. Scalars have magnitude but no direction for example, temperature, length,etc. Vectors in 2- and 3- space are repressented by arrows. B (terminal point) (Initial point) A Lowercase letters are used to denote vectors, → v = AB . Vectors u and v are said to be equivalent (or equal) if they have the same length and direction. Sum of Vectors If u and v are 2 vectors, u + v is obtained by positioning the initial point of v at the terminal point of u and joining the initial point of u to the terminal point of v. v Clearly, u + v = v + u u The zero vector has length zero and any direction. -v is the vector that has the same magnitude as v but is oppositely directed. v -v Hence, v + (-v)= 0
2 Subtraction of Vectors v – w = v + (-w) Position v and w so that their initial points coincide. The vector from the terminal point of w to the terminal point of v is the vector v – w. v v – w w If k is a scalar, then kv is the vector whose length is k times the length of v and whose direction is the same as that of v if k > 0 and opposite to that of v if k< 0. kv is called a scalar multiple of v. Vectors that are scalar multiples of each other are parallel to each other. Vectors in Coordinate Systems Let the initial point of vector v be positioned at the origin. The coordinates ( ) 1 v2 v , of the terminal point of v are called the components of v and we write ( ) 1 v2 v = v , . ( ) 1 v2 v , v If ( ) 1 v2 v = v , , ( ) 1 w2 w = w , , 1. ( ) ( ) ( ) 1 2 1 2 1 1 v2 w2 v w = v ,v w ,w = v w , 2. ( ) ( ) 1 2 1 kv2 kv = k v ,v = kv , The vector with initial point ( ) 1 y1 P x , and terminal point ( ) 2 y2 Q x , is: ( ) 2 1 y2 y1 PQ = x − x , − → Example 1 Find the components of the vectors → PQ with initial point P(2,-1) and the terminal point Q(7,-5).
3 Theorem Let u, v, w be vectors and m, n be scalars. 1. u + v = v + u 2. (u + v) + w = u + (v + w) 3. u + 0 = 0 + u = u 4. u + (-u) = 0 5. m (nu) = (mn) u 6. m (u + v) = mu + mv 7. (m + n) u = mu + nu 8. 1.u = u Vector Norms • The length of vector u is called the norm of u and is denoted by u . If ( ) 1 u2 u = u , , then 2 2 2 u = u1 + u . • A vector of norm 1 is called a unit vector. • The unit vector that has the same direction as uis u u . • If ( ) 1 y1 P x , and ( ) 2 y2 Q x , are two points in 2-space then, the distance, d between them is the norm of vector → PQ and is given by: ( ) ( ) 2 2 1 2 d = PQ = x2 − x1 + y − y → • ku = k u Example 2 If u = (3,-4) and v = (- 2,1) , find a. -u + v b. 2u - v c. v d. u + v e. Find vector w given that w = 10 and w has the same direction as u.
4 Vectors in 3-Space Vectors in 3-space can be described by 3-components i.e. the x-, y- and zcoordinates. Thus, if ( ) 1 2 v3 v = v , v , and ( ) 1 2 w3 w = w ,w , then • ( ) 1 1 2 2 v3 w3 v w = v w , v w , • ( ) 1 2 kv3 kv = kv ,kv , • The components of the vector → PQ with initial point ( ) 1 1 1 x , y ,z and terminal point ( ) 2 2 2 x , y ,z are given by: ( ) ( ) ( ) 2 1 2 1 2 1 2 2 2 1 1 1 x - x , y y , z z x , y , z x , y , z PQ OQ OP = − − = − = − → → → • If ( ) 1 2 u3 u = u ,u , , then 2 3 2 2 2 u = u1 + u + u . • If ( ) 1 1 1 P x , y ,z and ( ) 2 2 2 Q x , y ,z are two points in 3-space then, the distance, d between them is the norm of vector → PQ and is given by: ( ) ( ) ( ) 2 2 1 2 2 1 2 2 1 d = PQ = x − x + y − y + z − z → . • ku = k u . Example 3 Sketch the following vectors in 3-space. a. (0,-2,0) b. (1,1,3) c. (-1,2,-1) d. (2,0,-2)
5 Example 4 If u = (1,-2,4), v = (-3,2,-3),w = (3,-7,0), find the following: a. v + w b. -3u +2v c. -2(w - 3u) d. w e. 3w f. v − w g. The unit vector that is in the opposite direction to vector u. Example 5 Find the distance between the points P (2,-1,-5) and Q (4,-3,1).
6 Dot Products u u v v u v The angle between vectors u and v satisfies: 0 . Definition of Dot Product If u and v are vectors in 2- or 3-space and is the angle between them, then the dot product or Euclidean inner product, u • v is defined by: = = • = 0 if 0 and 0 cos if 0 and 0 u v u v u v u v Example 6 If the angle between vector u = (0,0,1) and vector v = (0,2,2) is 45 0 , find u • v .
7 Components Formula for the Dot Product Let ( ) 1 u2 u = u , and ( ) 1 v2 v = v , . P u Q O v From the Law of Cosines, The above formula can be used to determine the angle between 2 vectors. Example 7 Find the angle between the following pairs of vectors: a. u = (3,-4), v = (-1,−5). b. u = (2,-1,1), v = (1,1,2). u v u • v cos =
8 Theorem Let u and v be vectors in 2- or 3-space. a. 2 v • v = v . b. If is the angle between u and v, then is acute u • v 0 is obtuse u • v 0 = 2 u • v = 0 Example 8 If u = (1,-2,3), v = (-3,4,2) and w = (3,6,3) , determine the type of angle between each pair of vectors. Orthogonal Vectors Perpendicular vectors are called orthogonal vectors. Thus, if is the angle between vectors u and v, u and v are orthogonal u • v = 0 . Theorem If u , v and w are vectors in 2- or 3-space and k is a scalar, then, a. u • v = v • u. b. u• (v + w) = u• v +u•w c. k(u• v) = (ku)• v = u• (kv) d. v • v 0if v 0 and v • v = 0if v = 0
9 3.2 Real Vector Spaces Let V be a set containing vectors u, v, w and m, n be scalars. V is called a vector space if the following axioms are satisfied by all vectors u, v, w in V. 1. If u, v V, then u + v V 2. u + v= v + u 3. u + (v+ w) = (u + v) + w 4. There exists a zero vector, 0 V such that 0 + u = u + 0 = u 5. For each u V, there exists -u V, such that u + (- u) = (- u) + u = 0 6. If m is any scalar and u V ,then m u V 7. m (u + v) = mu + mv 8. (m +n)u = mu +nu 9. m (nu) = (mn) u 10. 1.u = u Examples of vector spaces are , , , ,M ,M , ,M . 22 23 mn 2 3 n Exercise V is a set with the stated addition and scalar multiplication operators. Determine whether the given axioms are satisfied. 1. *(a,b) + (c,d) = (a+c, b+d) * k(a,b) = (ka,b) Axiom A: u + 0 = 0 + u = u Axiom B: k (u + v) = ku +kv 0,u,v 2 ,k scalar 2. *(a,b) + (c,d) = (a,b) * m(a,b) = (ma,mb) Axiom A: u + v = v + u Axiom B: (m + n) u = mu +nu u,v 2 ,m,n scalar 3. *(a+bx) + (c+dx) = (a+c) + (b+2d)x * k(p+q) = kp + kq Axiom A: p + (q + r) = (p + q) + r Axiom B: k (p + q) = kp +kq p, q, r P1 ,k scalar
10 Theorem Let V be a vector space, ua vector in V and k a scalar. Then, a. 0u = 0 b. k0 = 0 c. (-1) u = -u d. If ku = 0, then k = 0 or u = 0
11 3.3 Subspaces Definition A subset W of a vector space V is called a subspace of V if W is itself a vector space under the addition and scalar multiplication defined for V. Theorem If W is a set of one or more vectors from a vector space V, then W is a subspace of V if and only if the following conditions hold: a. If u, v W, then u + v W b. If k is any scalar and u W, then ku W Example 9 Let W be the set of all points (x,y) 2 such that x,y 0. Show that W is not a subspace of 2 .
12 Example 10 1. Determine whether the set W ={(a, b, a+b); a, b } is a subspace of 3 . 2. Determine whether the set W = ;a, b,c c 1 a b is a subspace of M22. 3. Determine whether the set W ={a+bx+abx2 ; a, b } is a subspace of P2.
13 Example The subspace of 2 and 3 are as follows: • {0} • {0} • Lines through the origin • Lines through the origin • 2 • Planes through the origin • 3 Every vector space, V has at least 2 subspaces: V itself is a subspace and the set {0} (zero subspace) 2 3
14 3.4 Linear Combinations of Vectors Definition A vector w is called a linear combination of the vectors v1 v2 vr , , , , if w can be expressed as: w = k1v1 + k2 v2 ++ krvr where 1 2 kr k ,k , , are all scalars. Example Every vector ( ) 3 v = a,b,c is expressible as a linear combination of the standard basis vectors: i = (1,0,0),j = (0,1,0),k = (0,0,1) since ( ) i j k v a b c a, b,c a(1,0,0) b(0,1,0) c(0,0,1) = + + = = + + Example 11 Consider the vectors u = (1,2,-1), v = (6,4,2) 3 . Show that w = (9,2,7) is a linear combination of u and v and x = (4,-1,8) is not a linear combination of u and v.
15 Exercise 1. Express x = (1,1,1) as a linear combination of u = (1,2,3), v = (0,1,2) and w = (-1,0,1). 2. Express x = (1,-2,2) as a linear combination of u = (1,2,3), v = (0,1,2) and w = (-1,0,1).
16 Example 12 1. Consider p = 1+3x+x2 , q = x+2x2 ,r = 1-5x2 . p is a linear combination of q and r because p = 3q+r. (Verify) 2. Consider = = = = 1 3 - 2 0 ,D 1 2 -1 3 ,C 1 0 0 2 ,B 2 1 0 8 A . A is a linear combination of B, C and D as A = B + 2C – D. (Verify)
17 Spanning Sets Definition If S = { k v1 , v2 , , v } is a set of vectors in vector space V, then the span of S is the set of all linear combinations of vectors in S. If span(S) = V, we say that V is spanned by S or S spans V. Example 1. S = {(1,0,0), (0,1,0),(0,0,1)} spans 3 since any vector u = (x,y,z) 3 can be written as: u = x(1,0,0) + y(0,1,0) + z(0,0,1) 2. S = {1,x,x2 } spans P2 since any polynomial p = a+bx+cx2 P2 can be written as: p = a(1) + b(x)+ c(x2 ) 3. S = 0 1 0 0 , 1 0 0 0 , 0 0 0 1 , 0 0 1 0 spans M22 as every matrix = c d a b A M22 can be written as: + + + = 0 1 0 0 d 1 0 0 0 c 0 0 0 1 b 0 0 1 0 A a Example 13 Show that the set S = {(1,2,3), (0,1,2),(-2,0,1)} spans 3 .
18 Example14 Determine whether the set S = {(1,2,3), (0,1,2),(-1,0,1)} spans 3 . Theorem If S = { k v1 , v2 , , v } is a set of vectors in a vector space V, then span(S) is a subspace of V. Proof
19 Linear Dependence & Linear Independence Definition A set of vector S = { k v1 , v2 , , v } in a vector space V is said to be linearly independent if the vector equation c1v1 + c2 v2 +ck vk = 0 has only the trivial solution c1 = 0, c2 = 0,….., ck = 0. If there are other non-trivial solutions, then S is said to be linearly dependent. Example 15 1. The set S = {(1,2),(2,4)} is linearly dependent because: 2. The set S = { u = (2,−1,0,3), v = (1,2,5,−1),w = (7,−1,5,8) } is linearly dependent since: 3. Consider the vector i = (1,0,0),j = (0,1,0), k = (0,0,1). The vector equation c1 i + c2 j + c3k = 0 i.e. c1 = c2 = c3 = 0 (trivial solution). Hence, S = {i,j,k} is a linearly independent set. 4. Determine whether the vectors u = (1,2,3), v = (0,1,2), w = (-2,0,1) form a linearly independent set. 5. Show that the set S = {p = 1+x-2x2 , q = 2+5x-x 2 , r = x+x2 } forms a linearly dependent set. Hence, express p as a linear combination of q and r.
20 Theorem A set S with two or more vectors is: a. Linearly dependent if and only if at least one of the vectors in S can be expressed as a linear combination of the other vectors in S. b. Linearly independent if and only if no vector in S can be expressed as a linear combination of the other vectors in S. Geometric Interpretation ▪ In 2 and 3 , two vectors are linearly independent if the vectors do not lie on the same line (initial points at the origin). ▪ In 3 , three vectors are linearly independent if the vectors do not lie in the same plane (initial points at the origin).
21 3.5 Basis & Dimension Definition A set of vectors S = { n v1 , v2 , , v } in a vector space V is called a basisfor V if the following conditions hold: 1. S spans V 2. S is linearly independent. Theorem If S is a basis for a vector space V, then every vector v V can be expressed in the form 1 2 n n v = c v1 + c v2 +c v in exactly one way. Example Consider the set S = { i = (1,0,0), j = (0,1,0), k = (0,0,1) }. We have previously shown that: 1. S is linearly independent 2. S spans 3 . Thus, S is a basis of 3 . [S is called the standard basis of 3 ] Example 16 1. Show that the set S = { u = (1,1),v = (1,-1) } is a basis for 2 . 2. Show that the set S = {p = 1+2x+x 2 , q = 2+9x, r = 3+3x+4x 2 } is a basis of P2.
22 Example 1. S = {1,x,x 2 ,…., x n } is the standard basis for Pn. 2. S = 0 1 0 0 , 1 0 0 0 , 0 0 0 1 , 0 0 1 0 is the standard basis for M22. Theorem If S = { n v1 , v2 , , v } is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Example Since 3 has a basis of four vectors, the set S = {(1,2,-1),(1,1,0),(2,3,0),(5,9,-1)} must be linearly dependent. Definition (Dimension) If a vector space V has a basis consisting of n vectors, then the dimension of V is n and we write dim(V) = n. If V consists of the zero vector alone, then dim(V) = 0. Example 17 Determine the standard basis and dimension for each of the following vector spaces: 1 2 3 n 22 mn 2 , 3 , n ,P ,P ,P ,P ,M ,M
23 Example 18 Determine a basis for and the dimension of the solution space of the following homogeneous (Refer to Section 1.5): 0 2 0 2 3 0 2 2 0 3 4 5 1 2 3 5 1 2 3 4 5 1 2 3 5 + + = + − − = − − + − + = + − + = x x x x x x x x x x x x x x x x
24 3.6 Row Space, Column Space & Null Space Definition For a mxn matrix, A = m m m n n n a a a a a a a a a 1 2 21 22 2 11 12 1 the vectors, ( ) ( ) ( ) m m m m n n n a a a a a a a a a 1 2 2 21 22 2 11 12 1 = = = r r r1 in n formed from the rows of A are called row vectors of A and the vectors = = = m n n n n m m a a a a a a a a a 2 1 2 22 12 2 1 21 11 c ,c , c 1 in m are called the column vectors of A. Example 19 If − = 3 1 4 2 1 0 A obtain the row and column vectors of A. Definition Let A be a mxn matrix. 1. The row space of A is the subspace of n spanned by the row vectors of A. 2. The column space of A is the subspace of m spanned by the column vectors of A. 3. The null space of A is the solution space of the homogeneous system, Ax = 0 . the nullspace of A is a subspace of n .
25 Relationship between Solutions of Ax = 0 and Ax = b Theorem If x0 denotes any single solution of a consistent non-homogeneous linear system Ax = b and if k v1 , v2 , , v form a basis for the null space of A, i.e. the solution space of the homogeneous system Ax = 0 then every solution Ax = b can be expressed in the form: 1 2 k k x = x0 + c v1 + c v2 +c v Terminology ▪ The vector 0 x is called a particular solutionof Ax = b . ▪ 1 2 k k x0 + c v1 + c v2 +c v is called the general solution of Ax = b . ▪ 1 2 k k c v1 + c v2 +c v is called the general solution of Ax = 0 . Example 20 Consider the following system of linear equations: 2 6 8 4 18 6 5 10 15 5 2 6 5 2 4 3 1 3 2 2 0 1 2 4 5 6 3 4 6 1 2 3 4 5 6 1 2 3 5 + + + + = + + = + − − + − = − + − + = x x x x x x x x x x x x x x x x x x Find the general solution for Ax = b and hence, identify the particular solution for Ax = b and the general solution for Ax = 0 .
26 Bases for Row Spaces, Column Spaces and Null Spaces Theorem Elementary row operations do not change the null space of a matrix. Theorem Elementary row operations do not change the row space of a matrix. [This means that if A and B are row equivalent, then the row space of A is the same as the row space of B.] Theorem If A is row equivalent to matrix B in row-echelon form, then the non-zero rows of B form a basis for the row space of A. Example 21 Find a basis for the row space of − − − = − − 2 0 4 2 3 4 2 1 3 0 6 1 0 1 1 0 1 3 1 3 A .
27 Example 22 Find a basis for the vector space (rowspace) spanned by the vectors u = (1,-2,0,0,3), v = (2,-5,-3,-2,6),w = (0,5,15,10,0), and x = (2,6,18,8,6).
28 Example 23 Find a basis for the space (rowspace) spanned by p = -2+4x 2 , q =x-x 2 , r = 1-2x2 and s = -1+x+x 2 .
29 Example 24 Find bases for the row space and column space of: − − − − − − − − − − = 1 3 4 2 5 4 2 6 9 1 9 7 2 6 9 1 8 2 1 3 4 2 5 4 A
30 3.7 Rank & Nullity Theorem If A is any matrix, then the row space and column space of A have the same dimension. Definition The dimension of the row/column space of a matrix A is called the rank of A and is denoted by rank(A); the dimension of the null space of A is called the nullity of A and is denoted by nullity (A). Example 25 Consider the following matrix: − − − − − − − = 4 9 2 4 4 7 2 5 2 4 6 1 3 7 2 0 1 4 1 2 0 4 5 3 A a. Find bases for the null space, row space and column space of A. b. State the rank and nullity of A.
31 Dimension Theorem for Matrices If A is a matrix with n columns, then Theorem Rank(A) = Rank(AT ) Theorem The system of linear equations Ax = b is consistent if and only if Rank(A:b) = Rank(A) Theorem If A is a mxn matrix, then a. Rank(A) = the number of leading variables in the solution of Ax = 0 . b. Nullity(A) = the number of parameters in the solution of Ax = 0 . Summary Suppose A is a mxn matrix. Determine the dimensions of the given vector spaces if the dimension of the row space of A is r. Fundamental Space Dimension Row space of A r Column space of A Null Space of A Row Space of AT Column Space of AT Null Space of AT Rank(A)+Nullity(A) = n
1 Exercise 3 1. Find vectors x and y such that: a. x+y = (5,0) b. 2x-3y = (10,15) 2. Find the distance between points P and Q: a. P(-2,6,-1), Q(-2,-2,4) b. P(5,3,-2,1), Q(1,2,3,-4) 3. Find the components of the vector with initial point P and terminal point Q: a. P(3,6), Q(3,7) b. P(5,-3), Q(4,-6) c. P(3,6,-4), Q(-1,3,-3) d. P(-2,0,3), Q(4,4,-4) e. P(a,b,c), Q(0,0,0) f. P(0,0,0), Q(-a,-b,-c) 4. If u =(1,-2,2), v = (3,0,-8), w = (4,2,0) and x = (1,2-1), find the following: a. u + v b. u + v c. u − v d. u − v e. v v + w w f. (1/ x )x g. (1/ x )x h. u • v i. w •u j. v •u + w•u k. u• (v + w) l. x•x m. Find the unit vector that has the opposite direction as vector v. n. Find the vector with norm 2 that has the same direction as vector u. 5. Which pairs of vectors are orthogonal? a. (1,1), (1,-1) b. (1,-1,-1), (0,1,-1) c. (4,-2,-1), (-2,3,4) d. (-7,-3,1,0), (0,2,6,4) 6. If u =(3,2,-1), v = (1,-1,2) and w = (3,4,4), find: a. u+v b. v+w c. -3(v-2w) d.-u+4w-(v+2u) 7. For each of the following find the length of vector v and the unit vector that has the same direction as v: a. v = (1,2,2) b. v = (1,0,3) c. v = (4,0,-3,5) d. v = (0,4,3,4,-4) 8. If u =(2,-2,0,2) and v = (1,3,3,-3), solve for w: a. 2w = 3u - v b. u – w= 2v 9. If v = (8,8,6), find vector u such that: a. u has the same direction as v and half its length. b. u has the opposite direction as v and is one fourth the length of v. 10. Find the distance and angle betweenu and v: a. u = (2,-3), v = (-2,3) b. u = (1,7), v = (4,3) c. u = (1,2,3,0), v = (0,1,4,-1) d. u = (1,1,2), v = (-1,3,0) 11. Determine the type of angle between the following pairs of vectors: a. (1,1), (1,-1) b. (2,-2,-2), (0,2,-2) c. (4,-2,-1),(-2,3,4) d. (-7,-3,1,0), (0,2,6,4)
2 12. If v = (-1, 2, 5), find the value of k such that kv = 4 . 13. Consider the vectors, u = (-7,-3, 1, 0) and v = (-1, 2,-2, 1). Find: a. u + 2v b. the angle between u and v. c. the vector x that is in the opposite direction from v if x = 10 . 14. If u = (6, 2, 12) and v = (-3, 0, 2), find: a. u + v . b. the angle between u and v. c. the vector x that has the same direction as v with x = 52 . 15. Consider the vectors, u = (4,1,2,3), v = (0,3,8,-2) and w = (3,1,2,2). Find: a. u • 3w b. u - v c. the angle between v and w. d.vector x that has the same direction as u with x = 4 . 16. Suppose u = (-4,2,-6), v = (2,-c,3c) and w = (1,3,4). a. Find the value of c and the scalar multiple k if u and v are parallel to each other. b. If v forms angle with w, then find the value of . (Round off your answer to one decimal place) 17. V is a set with the stated addition and scalar multiplication operators. Determine whether the given axioms are satisfied. (k, l are scalars) a. (a+bx+cx2 )+(d+ex+fx2 ) = (a+d) + (c+f)x2 k(a+bx+cx2 ) = 2ka+2kbx+2kcx2 Axiom A: k(p+q) = kp+kq Axiom B: k(lp) = (kl)p b. + + + = + c g d h a e bf g h e f c d a b = kc kd ka 1 c d a b k Axiom A: A + B = B + A Axiom B: (kl)A = kA + lA c. (a,b,c) + (d,e,f) = (d,0,0) k(a,b,c) = (a,kb,2kc) Axiom A: u + v = v + u Axiom B: (k+l) u =ku + lu
3 d. (a,b,c) + (d,e,f) = (a+d,b+e,c+f) k(a,b,c) = (ka,0,kc) Axiom A: u + v = v + u Axiom B: k(lu) = (kl)u e. (a,b,c) + (d,e,f) = (a+d,0,c+f) k(a,b,c) = (2ka,2kb,2kc) Axiom A: u + 0 = u Axiom B: k(u + v) = ku+kv f. (a+bx+cx2 )+(d+ex+fx2 ) = a + ex + (c+f)x2 k(a+bx+cx2 ) = a+kbx+k2cx2 Axiom A: p+q = q+p Axiom B: k(lp) = (kl)p g. (a,b) + (c,d) = (a-c,b+d) k(a,b) = (ka,2b) Axiom A: u + v = v + u Axiom B: k(u + v) = ku+kv h. (a,b) + (c,d) = (a+b,c-d) k(x,y) = (k 2x,2ky) Axiom A: u + 0 = 0 +u =u Axiom B: k(u + v) = ku+kv 18. V is a set of vectors with the following addition and multiplication operators. a.Determine whether the following axiom is satisfied: m(nu) = (mn)u (a,b,c) + (d,e,f) = (a+d+1,b+e+1,c+f+1) k(a,b,c) = (ka+k-1,kb+k-1,kc+k-1) b.Find the zero vector,0, using the property that for every vector vV , there exists a zero vector such that v 0 v + = .[Hint: The zero vector ,0 need not be (0,0,0) ]. 19. State whether the following sets are vector spaces. If not, state also the axioms that fail. a. The set of all 2x2 matrices of the form c 0 a b with the standard operations. b. The set of vectors with the following operations: (a,b) + (c,d) = (a+b,c+d) k(a,b) = (ka,b) 20. Determine whether each of the following sets is a subspace of 3 with the standard operations: a. W = (a,0,b): a,b b. W = (a,b,c): c = a -3b, a,b c. W = (a,b,c):b 1/a,a 0,b,c =
4 21. Determine whether each of the following sets is a subspace of M22 with the standard operations: a. W = ;a b;a, b,c,d c d a b b. W = = + ;a 2c 1; a, b,c,d c d a b c. W = ;a, b,c, c 0 a b 22. Determine whether each of the following sets is a subspace of P3 with the standard operations: a. { ax+bx2 +cx3 ; ab = 1} b. {a+bx+cx2 +dx3 ; a+b+c+d = 0} c. {a+bx+cx2 +dx3 ; a,b,c,d } 23. Determine whether each of the following sets is linearly dependent or linearly independent: a. {(-2,2),(3,5)} b. {(-4,-3,4),(1,-2,3),(6,0,0)} c. {2-x,2x-x 2 ,6-5x+x2 } d. {x2 +3x+1,2x2+x-1,4x} 24. Determine the possible value/s of t so that the following sets are linearly independent: a. {(t,1,1),(1,t,1),(1,1,t)} b. {(t,1,1),(1,0,1),(1,1,3t)} 25. Express vector v as a linear combination of vectors w, x and y if possible: a. v = (10,1,4), w = (2,3,5), x = (1,2,4), y = (-2,2,3) b. v = (-1,7,2), w = (1,3,5), x = (2,-1,3), y = (-3,2,-4) 26. Express each of the following as a linear combination of p1 = 2+x+4x2 , p2 = 1-x+3x2 and p3 = 3+2x+5x2 : a.3+2x+5x2 b.6+11x+6x2 27. Determine whether each of the following sets spans of P2: a. {p = 1+x+x2 ; q = 1+2x+x2 ; r = x} b. {p = 1-x+x2 ; q = 1+x-x 2 ; r = 1 } 28. Determine whether each of the following sets spans 3 : a. {(4,7,3),(-1,2,6),(2,-3,5)} b. {(6,7,6),(3,2,-4),(1,-3,2)} 29. Show that each of the following sets is linearly dependent. Hence, express each vector as a linear combination of the other vectors. a. {(3,4), (-1,1), (2,0)} b. {(1,1,1), (1,1,0), (0,1,1), (0,0,1)} c. {(0,3,1,-1), (6,0,5,1), (4,-7,1,3)}
5 30. Determine whether each of the following sets is a basis of the given vector space: a. {(3,-2), (4,5)} for 2 b. {(1,5,3), (0,1,2), (0,0,6)} for 3 31. Determine whether each of the following sets is a basis of P3. a. {4t-t 2 , 5+t3 , 3t+5, 2t3 -3t2 } b. {t3 -2t2+1, t2 -4, t3+2t, 5t} 32. Find the value of y so that the vector (1,y,5) can be expressed as a linear combination of u = (1,-3,2) and v = (2,-1,1). 33. a. Find a basis (other than the standard basis) for the subspace of P2 that is spanned by: p = 2-3x+x2 , q = 4+x+x2 , r = -7x+x2 b. Hence, express 1+9x-x 2 as a linear combination of the basis polynomials found in (a). 34. a. Show that the set B = {u = (0,1,-1), v = (1,2,1), w = (1,1,-2)} is a basis of 3 . b. Hence, express vector x = (1,1,-6) as a linear combination of u, v and w. 35. Find a basis for the subspace of P2 spanned by: 1+5x-6x2 , 2+6x-8x2 , 3+7x-10x2 , 4+8x+12x2 36. Consider vectors u = (1,2,3,4), v = (1,0,1,2), w = (1,4,5,6) in 4 . a. Show that u, v and w for a linearly dependent set. b. Hence, express u as a linear combination of v and w. 37. Show that u = (0,3,1,-1), v = (6,0,5,1), and w = (4,-7,1,3) form a linearly dependent set in 4 . 38. a. Show that B = {p = -4+x+3x2 , q = 6+5x+2x2 , r = 8+4x+x2 } is a basis of P2. b. Express s = -10+2x+8x2as a linear combination of p, q and r. 39. Show that the vectors u = (1,3,-1,4), v = (3,8,-5,7), and w = (2,9,4,23) form a linearly dependent. 40. Show that the set S = {p = x-1, q = x2+2x, r = x2+x-2} forms a basis of P2. Hence, express x2+x-6 as a linear combination of p, q and r. 41. Show that the set B = {p = 2+2x+x2 , q = -1+2x, r = 2x+x2 } is a basis of P2. 42. Show that the set S = {u = (2,4,-3), v = (0,1,1),w = (0,1,-1) } is a basis of 3 .
6 43. Find a basis for the null space, row space and column space for each of the following matrices. Find also the rank and nullity of A. a. 1 6 2 4 b. − 4 2 1 1 3 2 c. − − − − − − 0 3 0 15 8 3 8 17 3 2 3 2 2 9 2 53 d. − − − − − − − − 4 7 0 5 5 1 3 1 2 3 2 1 2 1 1 1 2 3 1 4 e. −1 − 2 −1 −1 − 5 1 2 1 2 9 2 4 3 3 20 1 2 1 1 5 f. − − − − − 0 0 1 1 2 0 0 3 0 6 1 1 2 3 2 1 1 2 0 2 44. Consider matrix A= − − − − 1 4 8 0 0 4 0 1 1 1 1 3 0 1 2 1 0 0 a. Find the rank of A by reducing A to its row-echelon form. b. Hence determine: i. the dimension of the column space of A. ii. the nullity of A. 45. Solve the following homogeneous system, Ax = 0 by Gauss-Jordan elimination. Hence, obtain a basis for the null space of A, a basis for the row space of A, a basis for the column space of A, the rank of A and the nullity of A. a. 0 3 14 0 2 7 32 0 3 10 0 + + = − + + = − + + = + + = x y z x y z x y z x y z b. 2 2 0 5 22 0 6 2 4 0 3 8 4 0 − + = + + = − + + = − + = x y z x z w y z w x y z c. 3 6 5 11 0 2 2 5 0 2 4 3 6 0 + + − = + + − = + + − = x y z w x y z w x y z w d. . 2 4 6 0 3 0 0 2 3 0 1 3 5 3 4 5 1 3 1 3 5 + − = − + + = − − = + − = x x x x x x x x x x x
7 46. Consider the following homogeneous system of linear equations: 7 0 5 7 10 3 0 2 3 4 0 3 2 0 2 3 5 1 2 3 4 5 1 2 3 4 5 1 2 3 4 5 − + + = + + + + = + − + + = + + + + = x x x x x x x x x x x x x x x x x x a. What is the coefficient matrix, A for the above system. Find the null space of A, N(A). b. Find a basis for N(A) and the nullity of A. c. Find a basis for the row space and the column space of A. d. State the rank ofA and the nullity of A. 47. Matrices A and B are row equivalent i.e. − − = 4 9 3 1 4 3 7 2 2 2 2 5 1 1 0 1 2 1 0 0 A ~ B 0 0 0 0 0 0 0 0 1 2 0 1 1 0 2 1 0 3 0 4 = − − − a. Find the rank and nullity of A. b. Find a basis for the row space, column space and null space of A. 48. Consider the matrix: − − − − − − − = 2 8 10 12 18 1 4 5 6 9 3 2 1 4 1 1 4 5 6 9 A a. Find a bases for the null space, row space and column space.the rank and nullity of A. b. State the rank and nullity of A. 49. Given the following information,determine the dimension of the row space of A,the dimension of the column space of A, the nullity of A and the dimension of the null space of AT . (i) (ii) (iii) (iv) (v) (vi) (vii) Size of A 4x4 4x4 4x4 5x7 7x5 3x3 4x2 Rank of A 4 2 1 3 3 0 2 Dim of row space Dim of column space Nullity of A Dim of null space of AT
8 Answer 1. a. b. 3. a. (0,1) b. (-1,-3) c. (-4,-3,-1) d. (6,4,-7) e. (-a,-b,-c) f. (-a,-b,-c) 5. a. orthogonal b. orthogonal c. not orthogonal d. orthogonal 7. a. 1 2 2 , , 3 3 3 b. 10 3 10 ,0, 10 10 c. 2 2 3 2 2 ,0, , 5 10 2 − d. 4 57 3 57 4 57 4 57 0, , , , 57 57 57 57 − 9. a. (4, 4,3) b. 3 2, 2, 2 − − − 11. a. right angle b. right angle c. obtuse angle d. right angle 13. a. 95 b. 92.36 c. (1, 2,2, 1 − − ) 15. a. 69 b. 9 c. 66.24 d. 8 30 2 30 4 30 6 30 ,,, 15 15 15 15 17. a. satisfied, not satisfied b. satisfied, not satisfied c. not satisfied, not satisfied d. satisfied, satisfied e. not satisfied, satisfied f. not satisfied, satisfied g. not satisfied, satisfied h. not satisfied, not satisfied 19. a. vector space b. not a vector space (Axiom 2,2,4,5,7&8 fail)
9 21. a. not a subspace b. not a subspace c. subspace 23. a. linear independent b. linear independent c. linear dependent d. linear independent 25. a. v = w + 2x -3y b. not a linear combination 27. a. does not span b. does not span 31. a. basis b. basis
1 CHAPTER 4 LINEAR TRANSFORMATIONS A function (transformation) that maps vector space V into vector space W is denoted by: T : V → W V is called the domain of T and W is called the codomain of T. If v V and w W such that T(v) = w then, w is called the image of v under T. The set of all images of vectors in V that are in W is called the range of T and v is called the pre-image of w. V W Range Domain Codomain Example 1 No Function Description Domain Codomain 1. T(x1,x2)=x1 2 +x2 2 → 2 T : 2. T(x)=x2 3. T(x,y,z)=(x+z,y+z) 4. T(v1,v2)=(v1+v2, v1-v2 ) Example 2 2 2 T : → is defined by T(x,y)=(x-y,x+2y). a. Find T(-1,2) b. Find the pre-image of w=(-1,11)
2 If the domain of a function T is n and the codomain is m then, f is called a transformation from n to m and we write n m T : → . If m=n, then T is called an operator. 4.1 Linear Transformations If the functions are all linear the transformations n m T : → is called a linear transformation. The linear transformation n m T : → can be written in matrix form as T(x)= Ax where A is a mxn matrix and is called the standard matrix of the linear transformation. Example 3 1. T(x,y)=(x-y,x+2y) can be written as: − = y x 1 2 1 1 y x T and the standard matrix, − = 1 2 1 1 A 2. 2 3 T : → is defined as follows: ( ) − − = = 2 1 x x 1 2 2 1 3 0 T x Ax a. Find T(x) if x=(2,-1) b. Find T(v) if v=(1,3) 3. The linear transformation n m T : → is defined by T(v)=Av. Find m and n for the following standard matrices. a. − = 4 2 1 2 3 0 0 1 1 A b. − − − = 0 2 5 0 2 3 A c. − = 3 1 0 0 1 0 1 2 A 4. If I is the nxn identity matrix and also the standard matrix for the transformation n m T : → then i.e. x is mapped onto itself. T is then called the identity transformation.