3.3 Linear Independence

Transcription

1 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) Linear Independence In this section we look more closely at the structure of vector spaces. To begin with, we restrict ourselves to vector spaces that can be generated from a finite set of elements. Each vector in the vector space can be built up from the elements in this generating set using only the operations of addition and scalar multiplication. This generating set is usually referred to as a spanning set. In particular, it is desirable to find a minimal spanning set. By minimal we mean a spanning set with no unnecessary elements (i.e. all the elements in the set are needed in order to span the vector space). To see how to find a minimal spanning set, it is necessary to consider how the vectors in the collection depend on each other. Consequently, we introduce the concepts of linear dependence and linear independence. These simple concepts provide the key to understanding the structure of vector spaces. Consider the following vectors in R 3 : x =, x = 3, x 3 = 3 8 Let S be the subspace of R 3 spanned by x, x, x 3. Actually, S can be represented in terms of two vectors x and x, since the vector x 3 is already in the span of x and x. x 3 = 3x +x (3.) Any linear combination of x, x, x 3 can be reduced to a linear combination of x and x : Thus α x +α x +α 3 x 3 = α x +α x +α 3 (3x +x ) = (α +3α 3 )x +(α +α 3 )x S = Span{x,x,x 3 } = Span{x,x } Equation (3.) can be rewritten in the form 3x +x x 3 = (3.) Since the three coefficients in (3.) are nonzero, we could solve for any vector in terms of the other two. It follows that x = 3 x + 3 x 3, x = 3 x + x 3, x 3 = 3x +x Span{x,x,x 3 } = Span{x,x 3 } = Span{x,x 3 } = Span{x,x } Because of the dependency relation (3.), the subspace S can be represented as the span of any two of the given vectors.

2 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 85 On the other hand, no such dependency relationship exists between x and x. Indeed, if there were scalars k and k, not both, such that k x +k x = (3.3) then we could solve for one of the vectors in terms of the other. x = k k x (k ) or x = k k x (k ) However, neither of the two vectors in question is a multiple of the other. Therefore, Span{x } and Span{x } are both proper subspaces of Span{x,x }, and the only way that (3.3) can hold is if k = k =. We can generalize this example by making the following observations. (I) If {v,v,...,v n } span a vector space V and one of these vectors can be written as a linear combination of the other n vectors, then those n vectors span V. (II) Given n vectors v,v,...,v n, it is possible to write one of the vectors as a linear combination of the other n vectors if and only if there exist scalars k,k,...,k n not all zero such that k v +k v + +k n v n = Definition 3.5 If S = {v,v,...,v r } is a nonempty set of vectors, then the vector equation k v +k v + +k r v r = has at least one solution, namely k =, k =,..., k r = If this is the only solution, then S is called a linearly independent set. If there are other solutions, then S is called a linearly dependent set. It follows from (I) and (II) that, if {v,v,...,v r } is a minimal spanning set, then v,v,...,v r are linearly independent. Conversely, if v,v,...,v r are linearly independent and span V, then {v,v,...,v r } is a minimal spanning set for V. Example 3. If v = (,,,3), v = (,,5, ), and v 3 = (7,,5,8), then the set of vectors S = {v,v,v 3 } is linearly dependent, since 3v +v v 3 =. Example 3. The polynomial p = x, p = 5+3x x, p 3 = +3x x form a linear dependent set in P since 3p p +p 3 =.

3 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 86 Example 3.3 Consider the vectors i = (,,), j = (,,), and k = (,,) in R 3. In terms of components, the vector equation becomes or, equivalently, k i+k j+k 3 k = k (,,)+k (,,)+k 3 (,,) = (,,) (k,k,k ) = (,,) This impliesthat k =, k =, and k 3 =, so the set S = {i,j,k} is linear independent. A similar argument can be used to show that the vectors e = (,,,...,), e = (,,,...,),..., e n = (,,,...,) form a linear independent set in R n. Example 3.4 Show that, x, x,...,x n form a linearly independent set of vectors in P n. Solution Assume that some linear combination of polynomials is zero, say We must show that a +a x+a x + +a n x n = for all x (, ) (3.4) a = a = a = = a n = To see that this is so, recall from algebra that a nonzero polynomial of degree n has at most n distinct roots. But this is implies that a = a = a = = a n = ; otherwise, it would follow from (3.4) that a +a x+a x + +a n x n is a nonzero polynomial with infinitely many roots. Example 3.5 Determine whether the vectors v = (,,3), v = (5,6, ), v 3 = (3,,) form a linear dependent set or a linear independent set. Solution... Theorem 3.6 Let v,v,...,v n be n vectors in R n and let X = [ v v v n ]. The vectors v, v,..., v n will be linearly dependent if and only if X is singular. We can use Theorem 3.6 to test whether n vectors are linearly independent in R n. Simply form a matrix X whose columns are the vectors being tested. To determine whether X is singular, calculate the value of det(x). If det(x) =, the vectors are linearly dependent. If det(x), the vectors are linearly independent.

4 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 87 Example 3.6 Determine wether the vectors are linearly dependent. Solution... v = (4,,3), v = (,3,), v 3 = (, 5,3) To determine wether r vectors v, v,..., v r in R n are linearly independent, we can rewrite the equation k v +k v + +k r v r = as a linear system Xk =, where X = [ v v v r ]. If r n, then the matrix X is not square, so we cannot use determinants to decide wether the vectors are linearly independent. The system is homogeneous, so it has the trivial solution k =. It will have nontrivial solutions if and only if the row echelon forms of X involve free variables. If there are nontrivial solutions, then the vectors are linearly dependent. If there are no free variables, then k = is the only solution, and hence the vectors must be linearly independent. Example 3.7 Given v = (,,,3), v = (,3,, ), v 3 = (,,7,7) To determine if the vectors are linearly independent, we reduce the system Xk = to row echelon form Since the echelon form involves a free variables k 3, there are nontrivial solutions and hence the vectors must be linearly dependent. To determine whether a set of vectors is linear independent in R n, we must solve a homogeneous linear system of equations. A similar situation holds for the vector space P n. Example 3.8 Determine whether the vectors p = x x+3, p = x +x+8, p 3 = x +8x+7 form a linear dependent set or a linear independent set. Solution... Next we consider a very important property of linearly independent vectors. Linear combinations of linearly independent vectors are unique. More precisely, we have the following theorem.

5 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 88 Theorem 3.7 Let v, v,..., v n be vectors in a vector space V. A vector v in Span{v,v,...,v n } can be written uniquely as a linear combination of v, v,..., v n if and only if v, v,..., v n are linearly independent. The following theorem gives two simple facts about linear independence that are important to know. Theorem 3.8 (a) A finite set of vectors that contain the zero vector is linearly dependent. (b) A set with exactly two vectors is linear independent if and only if neither vector is a scalar multiple of the other. Geometric Interpretation of Linear Independence Linear independence has some useful geometric interpretations in R and R 3. If x and y are linearly dependent in R, then k x+k y = where k and k are not both. If, say, k, we can write x = k k y If two vectors in R are linearly dependent, one of the vectors can be written as a scalar multiple of the other. Thus, if both vectors are placed at the origin, they will lie along the same line. (x,x ) (y,y ) (x,x ) x y (y,y ) x and y linearly dependent x and y linearly independent If x y x = x and y = y x 3 y 3 are linearly independent in R 3, then the two points (x,x,x 3 ) and (y,y,y 3 ) will not lie on the same line through the origin in 3-space. Since (,,), (x,x,x 3 ), and (y,y,y 3 ) are not collinear, they determine the plane. If (z,z,z 3 ) lies on this plane, the vector z = (z,z,z 3 ) can be written as a linear combination of x and y, and hence x, y, and z are linearly dependent. If (z,z,z 3 ) does not lie on the plane, the three vectors will be linearly independent.

6 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 89 y z z y x x x, y, and z linearly dependent x, y, and z linearly independent The next theorem shows that a linearly independent set in R n can contain at most n vectors. Theorem 3.9 Let S = {v,v,...,v n } be a set of vectors in R n. If r > n, then S is linearly dependent. Remark Theorem 3.9 tells us that a set in R with more than two vectors is linearly dependent and a set in R 3 with more than three vectors is linearly dependent. Linear Independence of Functions In Example 3.6 a determinant was used to test wether three vectors were linearly independent in R 3. Determinants can also be used to help to decide if a set of n vectors is linearly independent in C (n ) [a,b]. Indeed, let f, f,..., f n be elements of C (n ) [a,b]. If these vectors are linearly dependent, then there exist scalars c, c,..., c n not all zero such that c f (x)+c f (x)+ +c n f n (x) = (3.5) for each x [a,b]. Taking the derivative with respect to x of both sides of yields c f (x)+c f (x)+ +c n f n (x) = If we continue taking derivatives of both sides, we end up with the system c f (x) +c f (x) + +c n f n (x) = c f (x) +c f (x) + +c nf n (x) =.... c f (n ) (x)+c f (n ) (x)+ +c n f n (n ) (x) = For each fixed x [a,b], the matrix equation f (x) f (x) f n (x) c f (x) f (x) f n (x) c.... =. f (n ) (x) f (n ) (x) f n (n ) (x) c n (3.6) will have the same nontrivial solution (c,c,...,c n ). Thus, if f, f,..., f n are linearly dependent in C (n ) [a,b], then, for fixed x [a,b], the coefficient matrix of system (3.6) is singular, its determinant is zero.

7 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 9 Definition 3.6 Let f, f,..., f n be functions in C (n ) [a,b] and define the function W(x) on [a,b] by f (x) f (x) f n (x) f (x) f (x) f n(x) W(x) =... f (n ) (x) f (n ) (x) f n (n ) (x) The function W(x) is called the Wronskian of f, f,..., f n. Theorem 3. Let f, f,..., f n be elements of C (n ) [a,b]. If there exists a point x in [a,b] such that W(x ), then f, f,..., f n are linearly independent. Example 3.9 Show that, e x, and e x are linearly independent in C(, ). Solution... Example 3.3 Show that the vectors, x, x, and x 3 are linearly independent in P 3. Solution... Example 3.3 Determine whether the functions x and x x are linearly independent in C[,]. Solution... Exercise 3.3. Explain why the following are linearly dependent sets of vectors. (a) v = (,,4) and v = (5,, ) in R 3 (b) v = (3, ), v = (4,5), v 3 = ( 4,7) in R (c) p = 3 x+x and p = 6 4x+x in P (d) A = and B = in M. Determine whether the following vectors are linearly independent in R. (a) (,),(3,) (b) (,3),(4,6) (c) (,),(,3),(,4) (d) (,),(, ),(, 4) (e) (,),(,) 3. Determine whether the following vectors are linearly independent in R 3. (a) (,,),(,,),(,,) (b) (,,),(,,),(,,),(,,3) (c) (,, ),(3,, ),(,,) (d) (,, ),(,,),(4,, 4) (e) (,,3),(,,)

8 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 9 4. Determine whether the following vectors are linearly independent in R 4. (a) (3,8,7, 3),(,5,3, ),(,,,6),(,4,,3) (b) (,,,),(3,3,,),(,,, ) (c) (,3, 3, 6),(,,, 6),(, 4,, ),(, 8,4, 4) (d) (3,, 3,6),(,,3,),(,,,),(,,,) 5. Determine whether the following vectors are linearly independent in M. (a) (c),,, 3 (b),, 6. Determine whether the following vectors are linearly independent in P 3. (a),x,x (c) x+,x+,x (b),x,x,x+3 (d) x+,x 7. (a) Show that the vectors v = (,,3,, ),v = (6,,5,), and v 3 = (4, 7,,3) form a linearly dependent set in R 4 (b) Express each vector as a linear combination of the other two. 8. For which real values of λ do the following vectors form a linear dependent set in R 3? v = (λ,, ), v = (,λ, ), v 3 = (,,λ) 9. For what values of α will the two vectors cos(x+α) and sinx be linearly dependent in C[ π,π]?. Under what conditions is a set with one vector linearly independent?. Indicate whether each statement is always true or sometimes false. Justify your answer by giving a logical argument or a counterexample. (a) If {v,v } is a linearly dependent set, then each vector is a scalar multiple of the other. (b) If {v,v,v 3 } is a linearly independent set, then so is the set {kv,kv,kv 3 } for every nonzero scalar k. (c) The converse of Theorem 3.8(a) is also true. (d) If v,v,...,v n span R n, then they are linearly independent. (e) If v,v,...,v n span a vector space V, then there are linearly independent. (f) If v,v,...,v r are vectors in a vector space V and Span{v,v,...,v r } = Span{v,v,...,v r } then v,v,...,v r are linearly dependent.

9 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 9. (a) v is a scalar multiple of v. Answer to Exercise 3.3 (b) The vectors are linear dependent by Theorem 3.9. (c) p is a scalar multiple of p. (d) B is a scalar multiple of A.. (a), (e) 3. (a), (e) 4. (a), (b), (c), (d) 5. (a), (b) 6. (c), (d) 7. (b) v = 7 v 3 7 v 3, v = 7 v + 3 v 3, v 3 = 7 3 v + 3 v 8. λ =, λ = 9. When α is an odd multiple of π/.. If and only if the vector is not zero.. (a) False (b) True (c) False (d) True (e) False (f) True 3.4 Basis and Dimension In Section 3.3 we showed that a spanning set for a vector space is minimal if its elements are linearly independent. The elements of a minimal spanning set form the basic building blocks for the whole vector space and, consequently, we say that they form a basis for the vector space. Definition 3.7 If V is any vector space and S = {v,v,...,v n } is a set of vectors in V, then S is called a basis for V if the following two conditions hold: (a) S is linear independent. (b) S spans V. Theorem 3. (Uniqueness of Basis Representation) If S = {v,v,...,v n } is a basis for a vector space V, then every vector v in V can be expressed in the form v = c v +c v +...+c n v n in exactly one way. Coordinates Relative to a Basis If S = {v,v,...,v n } is a basis for a vector space V, and v = c v +c v +...+c n v n is the expression for a vector v in terms of the basis S, then the scalars c,c,...,c n are called coordinates of V relative to the basis S. The vectors (c,c,...,c n ) in R n constructed from these coordinates is called the coordinate vector of V relative to S; it is denoted by (v) S = (c,c,...,c n ) Remark It should be noted that coordinate vectors depend not only on the basis S by also on the order in which the basis vectors are written; a change in the order of the basis vectors results in a corresponding change of order for the entries in the coordinate vectors.

10 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 93 Example 3.3 (Standard Basis for R 3 ) In Example 3.3, we showed that if i = (,,), j = (,,), k = (,,) then S = {i,j,k} is a linearly independent set in R 3. This set also spans R 3 since any vector v = (a,b,c) in R 3 can be written as v = (a,b,c) = a(,,)+b(,,)+c(,,) = ai+bj+ck (3.7) Thus S is a basis for R 3 ; it is called that standard basis for R 3. Looking at the coefficients of i, j, and k in (3.7), it follows that the coordinates of v relative to the standard basis are a, b, and c, so Comparing this result to (3.7), we see that (v) S = (a,b,c) v = (v) S Example 3.33 (Standard Basis for R n ) In Example 3.3, we showed that if then e = (,,,...,), e = (,,,...,),..., e n = (,,,...,) S = {e,e,...,e n } is a linearly independent set in R n. Moreover, this set also spans R n since any vector v = (v,v,...,v n ) in R n can be written as v = v (,,,...,)+v (,,,...,)+ +v n (,,,...,) (3.8) Thus S is a basis for R n ; it is called the standard basis for R n It follows from (3.8) that the coordinates of v = (v,v,...,v n ) R n relative to the standard basis are v,v,...,v n, so (v) S = (v,v,...,v n ) As is Example 3.3, we have v = (v) S, so a vector v and its coordinate vector relative to the standard basis for R n are the same. RemarkIngeneral, avector anditscoordinatevector neednotbethesame; theequality that we observed in the two preceding examples is a special situation the occurs only with that standard basis for R n. Remark In R andr 3, the standard vectors are commonly denoted by i, j, and k, rather than e, e, and e 3. We shall use both notations, depending on the particular situation. Example 3.34 Let v = (,,3), v = (,, ), and v 3 = (3,,). Show that the set S = {v,v,v 3 } is a basis for R 3. Solution...

11 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 94 Example 3.35 Let S = {v,v,v 3 } be the basis for R 3 in the preceding example. (a) Find the coordinate vector of v = (,,) with respect to S. (b) Find the vector v in R 3 whose coordinate vector with respect to the basis S is (v) S = (,3,). Solution... Example 3.36 (Standard Basis for P n ) (a) Show that S = {,x,x,...,x n } is a basis for the vector space P n of polynomials of the form a +a x+a x + +a n x n. (b) Fine the coordinate vector of the polynomial p = a + a x + a x relative to the basis S = {,x,x } for P. Solution... Example 3.37 (Standard Basis for M ) Let M =, M =, M 3 =, M 4 = The set S = {M,M,M 3,M 4 } is a basis for the vector space M of matrices. To see that S spans M, note that an arbitrary vector (matrix) a b c d can be written as a b c d = a +b +c +d = am +bm +cm 3 +dm 4 To see that S is linearly independent, assume that That is, It follow that a +b am +bm +cm 3 +dm 4 = +c a b = c d +d = Thus a = b = c = d =, so S is linear independent. The basis in this example is called the standard basis for M. More generally, the standard basis for M m n consists of the mn different matrices with a single and zeros for the remaining entries.

12 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 95 Example 3.38 ( Basis for the Subspace Span(S) ) If S = {v,v,...,v r } is a linearly independent set in a vector space V, then S is a basis for the subspace Span(S) since the set S spans Span(S) by definition of Span(S). Definition 3.8 A nonzero vector space V is called finite-dimensional if it contains a finite set of vectors {v,v,...,v n } that forms a basis. If no such set exists, V is called infinite-dimensional. In addition, we shall regard the zero vector space to be finite-dimensional. Example 3.39 By Example 3.33, 3.36, and 3.37, the vector spaces R n, P n, and M m n are finite-dimensional. The vector spaces F(, ), C(, ), C n (, ), and C (, ) are infinitely-dimensional. Theorem 3. Let V be a finite-dimensional vector space, and let {v,v,...,v n } be any basis. (a) If a set has more than n vectors, then it is linearly dependent. (b) If a set has fewer than n vectors, then it does not span V. It follows from the preceding theorem that if S = {v,v,...,v n } is any basis for a vector space V, then all sets in V that simultaneously span V and are linearly independent must have precisely n vectors. Thus, all bases for V must have the same number of vectors as the arbitrary basis S. This yield the following result, which is one of the most important in linear algebra. Theorem 3.3 All bases for a finite-dimensional vector space have the same number of vectors. To see how this theorem is related to the concept of dimension, recall that the standard basis for R n has n vectors (Example 3.33). Thus Theorem 3.3 implies that all bases for R n have n vectors. In particular, every basis for R 3 has three vectors, every basis forr has two vectors, andevery basis for R (= R)has onevector. Intuitively, R 3 is three-dimensional, R (a plane) is two-dimensional, and R (a line) is one-dimensional. Thus, for familiar vector spaces, the number of vectors in a basis is the same as the dimension. This suggests the following definition. Definition 3.9 The dimension of a finite-dimensional vector space V, denoted by dim(v), is defined to be the number of vectors in a basis for V. In addition, we define the zero vector space to have dimension zero. Remark From here on we shall follow a common convention of regarding the empty set to be a basis for the zero vector space. Example 3.4 (Dimensions of Some Vector Spaces) dim(r n ) = n dim(p n ) = n+ dim(m m n ) = mn The standard basis has n vectors. The standard basis has n+ vectors. The standard basis has mn vectors.

13 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 96 Example 3.4 Determine a basis for and the dimension of the solution space of the homogeneous system Solution... Some Fundamental Theorems x x +x 3 + x 4 = x + x 3 x 4 = x 3x +5x 4 = The following theorem reveal the subtle interrelationships among the concepts of spanning, linear independence, basis, and dimension. These theorems are not idle exercises in mathematical theory they are essential to the understanding of vector spaces, and many practical applications of linear algebra build on them. Theorem 3.4 Let S be a nonempty set of vectors in a vector space V. (a) If S is a linearly independent set, and if v is a vector in V that is outside of span(s), then the set S {v} that result by inserting v into S is still linearly independent. (b) If v is a vector in S that is expressible as a linear combination of other vectors in S, and if S {v} denotes the set obtained by removing v from S, then S and S {v} span the same space; that is, span(s) = span(s {v}) In general, to show that a set of vectors {v,v,...,v n } is a basis for a vector space V, we must show that the vectors are linearly independent and span V. However, if we happen to know that V has dimension n (so that {v,v,...,v n } contains the right number of vectors for a basis), then it suffices to check either linear independence or spanning the remaining condition will hold automatically. This is the content of the following theorem. Theorem 3.5 If V is an n-dimensional vector space, and if S is a set in V with exactly n vectors, then S is a basis for V if either S spans V or S is linearly independent. Example 3.4 Show that v = ( 3,7) and v = (5,5) form a basis for R by inspection. Solution... 3 Example 3.43 Show that S =,, is a basis for R3. Solution...

14 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 97 Theorem 3.6 Let S be a finite set of vectors in a finite-dimensional vector space V. (a) If S spans V but is not a basis for V, then S can be reduced to a basis for V by removing appropriate vector form S. (b) If S is a linearly independent set that is not already a basis for V, then S can be enlarged to a basis for V by inserting appropriate vectors into S. Theorem3.7 If W is asubspace ofafinite-dimensionalvector spacev, thendim(w) dim(v); moreover, if dim(w) = dim(v), then W = V. Exercise 3.4. Which of the following sets of vectors are bases for R. (a) (,),(3,) (b) (,3),(4,6) (c) (,),(,3),(,4) (d) (,),(, ),(, 4) (e) (,),(,) (f) (,),(,3),(,7). Which of the following sets of vectors are bases for R 3. (a) (,,3),(,,) (b) (,,),(,,),(,,) (c) (,3,),(6,,) (e) (,,),(,,),(3,3,3) (g) (,,),(,,),(,,),(,,3) (d) (,, ),(3,, ),(,,) (f) (3,, 4),(,5,6),(,4,8) 3. Which of the following sets of vectors are bases for P. (a) {,+x,+x+x } (c) {x,x +,x } (b) {,x,x+} (d) {x,x x,x+} 4. Which of the following sets of vectors are bases for M. (a),, (b),,, (c),,, Find the coordinate vector of w relative to the basis S = {u,u } for R. (a) u = (,), u = (,), w = (3, 7) (b) u = (, 4), u = (3,8), w = (,)

15 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 98 (c) u = (,), u = (,), w = (a,b) (d) u = (3,), u = (,), w = (7,4) (e) u = (, ), u = (,3), w = (,) 6. Find the coordinate vector of v relative to the basis S = {v,v,v 3 }. (a) v = (,, ), v = (,,), v = (,,), v 3 = (,,) (b) v = (4,6,), v = (,,), v = (,,), v 3 = (,,) (c) v = (3,3, ), v = (,3, ), v = (4,,3), v 3 = (,,) (d) v = (,,3), v = (,,), v = (,,), v 3 = (3,3,3) (e) v = (5,,3), v = (,,3), v = ( 4,5,6), v 3 = (7, 8,9) 7. Find the coordinate vector of p relative to the basis S = {p,p,p 3 }. (a) p = +x 3x, p =, p = x, p 3 = x (b) p = +x+x, p =, p = x, p 3 = (x ) 8. Find the coordinate vector of A relative to the basis S = {A,A,A 3,A 4 } A =,A 3 =,A =,A 3 =,A 4 = 9. Determine the dimension of and a basis for the solution space of the system. (a) 3x +x +x 3 +x 4 = 5x x +x 3 x 4 = (b) x 3x + x 3 = x 6x +x 3 = 3x 9x +3x 3 =. Determine the dimensions of the following subspaces of R 4. (a) all vectors of the form (a,b,c,) (b) all vectors of the form (a,b,c,d), where d = a+b and c = a b (c) all vectors of the form (a,b,c,d), where a = b = c = d. Determine the dimensions of the subspaces of P 3 consisting of all polynomials a +a x+a x +a 3 x 3 for which a =.. Find standard basis vector that can be added to the set {v,v } to produce a basis for R 3. (a) v = (,,3), v = (,, ) (b) v = (,,), v = (3,, ) Answer to Exercise 3.4. (a), (e). (b), (e) (f) 3. (a), (c) 4. (b), (c)

16 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) (a) (w) S = (3, 7) (b) (w) S = ( 5 8, 3 4 ) (c) (w) S = (d) (w) S = (3, ) (e) (w) S = (7,3) ( a, b a ) 6. (a) (v) S = ( 4,3,) (b) (v) S = (,6, 4) (c) (v) S = (,,3) (d) (v) S = (3,,) (e) (v) S = (,,) 7. (a) (p) S = (,, 3) (b) (p) S = (,5,) 8. (A) S = (,,,3) 9. (a) Basis: ( 4,,,), (,,,); dimension = 4 (b) Basis: (3,,), (,,); dimension =. (a) 3-dimensional (b) -dimensional (c) -dimensional. 3-dimensional. (a) {v,v,e } or {v,v,e } (b) {v,v,e } or {v,v,e } or {v,v,e 3 } 3.5 Row Space, Column Space, and Nullspace In this section we shall study three important vector spaces that are associated with matrices. Our work here will provide us with a deeper understanding of the relationships between the solutions of a linear system of equations and properties of its coefficient matrix. We begin with some definition. Definition 3. For an m n matrix a a a n a a a n A =... a m a m a mn the vectors r = [ a a a n ] r = [ a a a n ]. r m = [ a m a m a mn ] in R n formed from the rows of A are called the row vectors of A, and the vectors c = a a. a m, c = a a. a m,..., c n = a n a n in R m formed from the columns of A are called the column vectors of A.. a mn

17 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) Example 3.44 Let The row vectors of A are A = 3 4 r = [ ] and r = [ 3 4 ] and the column vectors of A are c =, c 3 =, c 3 = 4 The following definition defines three important vector spaces associate with a matrix. Definition 3. If A is an m n matrix, then the subspace of R n spanned by the row vectors of A is called the row space of A, and the subspace of R m spanned by the column vectors of A is called the column space of A. The solution space of the homogeneous system of equations Ax =, which is a subspace of R n, is called the nullspace of A. In this section and the next we shall be concerned with the following two general questions: What relationships exist between the solutions of a linear system Ax = b and the row space, column space, and nullspace of the coefficient matrix A? What relationships exist among the row space, column space, and nullspace of a matrix? To investigate the first of these questions, suppose that a a a n x a a a n A =... and x = x. a m a m a mn It follows from Formula (.) of Section.3 that if c,c,...,c n denote the column vectors of A, then the product Ax can be expressed as a linear combination of these column vectors with coefficients from x; that is, Ax = x c +x c + +x n c n (3.9) Thus a linear system, Ax = b, of m equations in n unknowns can be written as x c +x c + +x n c n = b (3.) from which we conclude that Ax = b is consistent if and only if b is expressible as a linear combination of the column vectors of A or, equivalently, if and only if b is in the column space of A. This yields the following theorem. x n

18 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) Theorem 3.8 A system of linear equations Ax = b is consistent if and only if b is in the column space of A. Example 3.45 Let Ax = b be the linear system x x = 3 x 3 4 Show that b is in the column space of A, and express b as a linear combination of the column vectors of A. Solution... The next theorem establishes a fundamental relationship between the solutions of a nonhomogeneous linear system Ax = b and those of the corresponding homogeneous linear system Ax = with the same coefficient matrix. Theorem 3.9 If x denotes any single solution of a consistent linear system Ax = b, and if v,v,...,v k form a basis for the nullspace of A that is, the solution space of the homogeneous system Ax = then every solution of Ax = b can be expressed in the form x = x +c v +c v + +c k v k (3.) and, conversely, for all choices of scalars c,c,...,c k, the vector x in this formula is a solution of Ax = b. General and Particular Solutions There is some terminology associate with Formula (3.). The vector x is called a particular solution of Ax = b. The expression x +c v +c v + +c k v k is called the general solution of Ax = b, and the expression c v +c v + +c k v k is called the general solution of Ax =. With this terminology, Formula (3.) state that the general solution of Ax = b is the sum of any particular solution of Ax = b and the general solution of Ax =. Example 3.46 In Example.4 of Section. we solved the nonhomogeneous linear system x +3x x 3 +x 5 = and obtained x +6x 5x 3 x 4 +4x 5 3x 6 = 5x 3 +x 4 +5x 6 = 5 x +6x +8x 4 +4x 5 +8x 6 = 6 (3.) x = 3r 4s t, x = r, x 3 = s, x 4 = s, x 5 = t, x 6 = 3

19 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) This result can be written in vector form as x 3r 4s t 3 4 x r x 3 x 4 = s s = + r +s +t x 5 t x }{{} } {{ } x x (3.3) which is the general solution of (3.). The vector x in (3.3) is a particular solution of (3.); the linear combination x is the general solution of the homogeneous system x +3x x 3 +x 5 = x +6x 5x 3 x 4 +4x 5 3x 6 = 5x 3 +x 4 +5x 6 = x +6x + 8x 4 +4x 5 +8x 6 = Bases for Row Spaces, Column Spaces, and Nullspaces We first developed elementary row operations for the purpose of solving linear system, and we know from that work that performing an elementary row operation on an augmented matrix does not change the solution set of the corresponding linear system. It followings that applying an elementary row operation to a matrix A does not change the solution set of the corresponding linear system Ax =, or, stated another way, it does not change the nullspace of A. Thus we have the following theorem. Theorem 3. Elementary row operations do not change the nullspace of a matrix. Example 3.47 Find a basis for the nullspace of A =. 3 5 Solution... The following theorem is a companion to Theorem 3.. Theorem 3. Elementary row operations do not change the row space of a matrix. In light of Theorem 3. and 3., one might anticipate that elementary row operations should not change the column space of a matrix. However, this is not so elementary row operations can change the column space. Theorem 3. If A and B are row equivalent matrices, then (a) A given set of column vectors of A is linearly independent if and only if the corresponding column vectors of B are linearly independent.

20 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 3 (b) A given set of column vectors of A forms a basis for the column space of A if and only if the corresponding column vectors of B form a basis for the column space of B. The following theorem makes it possible to find bases for the row and column spaces of a matrix in row-echelon form by inspection. Theorem 3.3 If a matrix R is in row-echelon form, then the row vectors with the leading s (the nonzero row vectors) form a basis for the row space of R, and the column vectors with the leading s of the row vectors form a basis for the column space of R. Example 3.48 The matrix 5 3 R = 3 is in row-echelon form. From Theorem 3.3, the vectors r = [ 5 3 ] r = [ 3 ] r 3 = [ ] form a basis for the row space of R, and the vectors c =, c =, c 3 = form a basis for the column space of R. Example 3.49 Find bases for the row and column space of A = Solution Since elementary row operations do not change the row space of a matrix, we can find a basis for the row space of A by finding a basis for the row space of any row-echelon form of A. Reducing A to row-echelon form, we obtain (verify) R = 3 6 5

21 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 4 By Theorem 3.3, the nonzero row vectors of R form a basis for the row space of R and hence form a basis for the row space of A. These basis vectors are r = [ ] r = [ 3 6 ] r 3 = [ 5 ] Keeping in mind that A and R may have different column spaces, we cannot find a basis for the column space of A directly from the column vectors of R. However, it follows from Theorem 3.(b) that if we can find a set of column vectors of R that forms a basis for the column space of R, then the corresponding column vectors of A will form a basis for the column space of A. The first, third, and the fifth columns of R contain the leading s of the row vectors, so 4 5 c =, c 3 =, c 5 = form a basis for the column space of R; thus the corresponding column vector of A namely, 4 5 c =, c 3 = 9 9, c 5 = form a basis for the column space of A. Example 3.5 Find bases for the column space of A = Solution... Example 3.5 Find a basis for the space spanned by the vectors Solution... v = (,,,,3) v = (, 5, 3,,6) v 3 = (,5,5,,) v 4 = (,6,8,8,6) Observe that in Example 3.49 the basis vectors obtained for the column space of A consisted of column vector of A, but basis vectors obtained for the row space of A were not all row vectors of A. The following example illustrate a procedure for finding a basis for the row space of a matrix A that consists entirely of row vectors of A.

22 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 5 Example 3.5 Find a basis for the row space of consisting entirely of row vectors from A. Solution... Example 3.53 (a) Find a subset of the vector v = (,,,3), v = (, 5, 3,6) v 3 = (,,3,), v 4 = (,,4, 7), v 5 = (5, 8,,) that forms a basis for the space spanned by these vectors. (b) Express each vector not in the basis as a linear combination of the basis vectors. Solution... The procedure illustrated in the Example 3.53 is sufficiently important that we shall summarize the steps: Given a set of vectors S = {v,v,...,v k } in R n, the following procedure produces a subset of these vectors that forms a basis for Span(S) and expresses those vector vectors of S that are not in the basis as linear combinations of the basis vectors. Step. Form the matrix A having v,v,...,v k as its column vectors. Step. Reduce the matrix A to its reduced row-echelon form R, and let w,w,...,w k be the column vectors of R. Step 3. Identify the columns that contain the leading s in R. The corresponding column vectors of A are the basis vectors for Span(S). Step 4. Express each column vector of R that does not contain a leading as a linear combination of preceding column vectors that do contain leading s. This yields a set of dependency equations involving the column vectors of R. The corresponding equations for the column vectors of A express the vectors that are not in the basis as linear combinations of the basis vectors. Exercise 3.5. List the row vectors and column vectors of the matrix

23 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 6. Express the product Ax as a linear combination of the column vectors of A. (a) (b) (c) (d) 3 3. Determine whether b is in the column space of A, and if so, express b as a linear combination of the column vectors of A. 3 (a) A = ; b = 4 6 (b) A = ; b = 3 5 (c) A = 9 3 ; b = 4 (d) A = 3 ; b = Suppose that x =, x = 3, x 3 = 5, x 4 = is a solution of a nonhomogeneous linear system Ax = b and that the solution set of the homogeneous system Ax = is given by the formulas 4 x = r s, x = r +s, x 3 = r, x 4 = s (a) Find the vector form of the general solution of Ax =. (b) Find the vector form of the general solution of Ax = b. 5. Find the vector form of the general solution of the given linear system Ax = b; then use that result to find the vector form of the general solution of Ax =. (a) x 3x = x 6x = (b) x +x +x 3 = 5 x + x 3 = x +x +3x 3 = 3 (c) x x + x 3 +x 4 = x 4x +x 3 +4x 4 = x +x x 3 x 4 = 3x 6x +3x 3 +6x 4 = 3 (d) x +x 3x 3 + x 4 = 4 x + x +x 3 + x 4 = x +3x x 3 +x 4 = 3 4x 7x 5x 4 = 5

24 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 7 6. Find a basis for the nullspace of A. 3 (a) A = (b) A = (c) A = (d) A = In each part, a matrix in row-echelon from is given. By inspection, find bases for the row and column spaces of A. (a) (c) 3 3 (b) 5 (d) Find a basis for the row space of A by reducing the matrix to row-echelon form. 3 (a) A = (b) A = (c) A = 3 (d) A = (e) A = For the matrices in Exercise 8, find a basis for the column space of A.. For the matrices inexercise 8, findabasis forthe row space of Aconsisting entirely of row vectors of A.. Find a basis for the subspace of R 4 spanned by the given vectors. (a) (,, 4, 3), (,,, ), (,,3,) (b) (,,,), (3,3,6,), (9,,,3)

25 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 8 (c) (,,,), (,,,), (,,,), (, 3,,3). Find a subset of the vectors thatforms a basis for thespace spanned by the vectors; then express each vector that is not in the basis as a linear combination of the basis vectors. (a) v = (,,,), v = ( 3,3,7,), v 3 = (,3,9,3), v 4 = ( 5,3,5, ) (b) v = (,,,3), v = (, 4,,6), v 3 = (,,,), v 4 = (,,,3) (c) v = (,,5,), v = (,3,,), v 3 = (4, 5,9,4), v 4 = (,4,, 3), v 5 = ( 7,8,, 8) Answer to Exercise 3.5. r = [ ], r = [ ], r 3 = [ 4 6 ] c = 5, c = 3, c 3 =, c 4 = (a) +3 (b) (c) (d) (a) = (b) b is not in the column space of A (c) = (d) = +(t ) +t 4 (e) 3 5 = (a) r +s (b) 3 5 +r +s (a) +t ; t (b) 7 +t ; t

26 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 9 (c) +r +s +t ; r +s +t (d) s t ; s t (a), (b) 5, (c) 3 5, 8 7 (d),, 3 7. (a) r = [ ], r = [ ], c =, c = (b) r = [ 3 ], r = [ ], c =, c = 3 (c) r = [ 4 5 ], r = [ 3 ], r 3 = [ 3 ], r 4 = [ ], c =, c =, c 3 = 4 3, c 4 = 5 3 (d) r = [ 5 ], r = [ 4 3 ], r 3 = [ 7 ], r 4 = [ ], c =, c =, c 3 = 4, c 4 = (a) [ 3 ], [ 9 ] (b) [ ] (c) [ 4 5 ], [ 4 7 ] (d) [ ], [ ] (e) [ 3 ], [ ], [ ] 9. (a) 5 7, 4 6 (b) 4 (c), 4 3 (d) 3, 4 3

27 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) (e) 3, , 9 6. (a) [ 3 ], [ ] (b) [ ] (c) [ 4 5 ], [ 3 ] (d) [ ], [ 3 4 ] (e) [ 3 ], [ ], [ ]. (a) (,, 4, 3), (,, 5, ), (,,, ) (b) (,,,), (,,,), (,,, 6 ) (c) (,,,), (,,,), (,,,), (,,,) 3. (a), 3 7 ; v 3 = v +v, v 4 = v +v (b), ; v = v, v 4 = v +v 3 3 (c) 5,, Rank and Nullity ; v 3 = v v, v 5 = v +3v +v 4 In the preceding section we investigated the relationships between systems of linear equations and the row space, column space, and nullspace of the coefficient matrix. In this section we shall be concerned with relationships between the dimensions of the row space, column space, and nullspace of a matrix and its transpose. Four Fundamental Matrix Spaces If we consider a matrix A and its transpose A T together, then there are six vector spaces of interest: row space of A row space of A T column space of A nullspace of A column space of A T nullspace of A T However, transposing a matrix converts row vectors into column vectors and column vectors into row vectors, so the row space of A T is the same as the column space of A,

28 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) and the column space of A T is the same as the row space of A. This leaves four vector spaces of interest: row space of A nullspace of A column space of A nullspace of A T These are known as the fundamental matrix spaces associated with A. If A is an m n matrix, then the row space of A and the nullspace of A are subspaces of R n, and the column space of A T and the nullspace of A T are subspaces of R m. Our goal in this section is to establish relationships between the dimensions of these four vector spaces. Row and Column Spaces Have Equal Dimensions In Example 3.49 of Section 3.5, we found that the row and column spaces of the matrix A = each have three basis vectors; that is, both are three-dimensional. It is not accidental that these dimensions are the same; it is a consequence of the following general result. Theorem 3.4 If A is any matrix, then the row space and column space of A have the same dimension. Definition 3. The common dimension of the row space and column space of a matrix A is called the rank of A and is denoted by rank(a); the dimension of nullspace of A is called the nullity of A and is denoted by nullity(a). Example 3.54 Find the rank and nullity of the matrix 4 A = Solution... The following theorem states that a matrix and its transpose have the same rank. Theorem 3.5 If A is any matrix, then rank(a) = rank(a T ). The following theorem establishes an important relationship between the rank and nullity of a matrix. Theorem 3.6 (Dimension Theorem for Matrices) If A is a matrix with n columns, then rank(a) + nullity(a) = n (3.4)

29 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) Theorem 3.7 If A is an m n matrix, then (a) rank(a) = the number of leading variables in the solution of Ax =. (b) nullity(a) = the number of parameters in the general solution of Ax =. Example 3.55 The matrix 4 A = has 5 columns, so rank(a)+nullity(a) = 5 This is consistent with Example 3.54, where we showed that rank(a) = and nullity(a) = 3 Example 3.56 Find the number of parameters in the general solution of Ax = if A is a 6 9 matrix of rank 5. Solution... Suppose that A is an m n matrix of rank r; it follows from Theorem 3.5 that A T is an n m matrix of rank r. Applying Theorem 3.6 to A and A T yields nullity(a) = n r, nullity(a T ) = m r from which we deduce the following table relating the dimensions of the four fundamental spaces of an m n matrix A of rank r. Maximum Value for Rank Fundamental Spaces Dimension Row space of A r Column space of A r Nullspace of A n r Nullspace of A T m r If A is an m n matrix, then the row vectors lie in R n and the column vectors lie in R m. This implies that the row space of A is at most n-dimensional and that the column space is at most m-dimensional. Since the row and column space have the same dimension (the rank of A), we must conclude that if m n, then the rank of A is at most the smaller of the value of m and n. We denote this by writing rank(a) min(m, n) (3.5) where min(m,n) denotes the smaller of the numbers m and n if m n or denotes their common value if m = n. Example 3.57 If A is a 7 4 matrix, then the rank of A is at most 4, and consequently, the seven row vectors must be linearly dependent. If A is a 4 7 matrix, then again the rank of A is at most 4, and consequently, the seven column vectors must be linearly dependent.

30 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 3 Linear Systems of m Equations in n Unknowns The following theorem specifies conditions under which a linear system of m equations in n unknowns is guaranteed to be consistent. Theorem 3.8 If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent. (a) Ax = b is consistent. (b) b is in the column space of A. (c) The coefficient matrix A and the augmented matrix [A b] have the same rank. It is not hard to visualize why this theorem is true if one views the rank of a matrix as the number of nonzero rows in its reduced row-echelon form. For example, the augmented matrix for the system x x 3x 3 +x 4 = 4 3x +7x x 3 + x 4 = 3 x 5x +4x 3 3x 4 = 7 3x +6x +9x 3 6x 4 = which has the following reduced row-echelon form is We see from the third row in this matrix that the system is inconsistent. However, is is also because of this row that the reduced row-echelon form of the augmented matrix has fewer zero rows than the reduced row-echelon form of the coefficient matrix. This forces the coefficient matrix and the augmented matrix for the system to have different ranks. Theorem 3.9 If Ax = b is a linear system of m equations in n unknowns, then the following are equivalent. (a) Ax = b is consistent for every m matrix b. (b) The column vectors of A span R m. (c) rank(a) = m. A linear system with more equations than unknowns is called an overdetermined linear systems. If Ax = b is an overdetermined linear system of m equations in n unknowns (so that m > n), then the column vectors of A cannot span R m ; it follows from Theorem 3.9 that for a fixed m n matrix A with m > n, the overdetermined linear systems Ax = b cannot be consistent for every possible b.

31 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 4 Example 3.58 The linear system x x = b x x = b x +x = b 3 x +x = b 4 x +3x = b 5 is overdetermined, so it cannot be consistent for all possible values of b, b, b 3, b 4, and b 5. Exact conditions under which the system is consistent can be obtained by solving the linear system by Gauss-Jordan elimination. We can show that the augmented matrix is row equivalent to b b b b b 3 3b +b b 4 4b +3b b 5 5b +4b Thus, the system is consistent if and only if b, b, b 3, b 4, and b 5 satisfy the conditions b 3b +b 3 = 3b 4b +b 4 = 4b 5b +b 5 = or, on solving this homogeneous linear system, b = 5r 4s, b = 4r 3s, b 3 = r s, b 4 = r, b 5 = s where r and s are arbitrary. Theorem 3.3 If Ax = b is a consistent linear system of m equations in n unknowns, and if A has rank r, then the general solution of the system contains n r parameters. Example 3.59 If A is a 6 9 matrix with rank of 5, and if Ax = b is a consistent linear system, then the general solution of the system contains 9 5 = 4 parameters. Theorem 3.3 If A is an m n matrix, then the following are equivalent. (a) Ax = has only the trivial solution. (b) The column vectors of A are linearly independent. (c) Ax = b has at most one solution (none or one) for every m matrix b. A linear system with more unknowns than equations is called an underdetermined linear systems. If Ax = b is a consistent underdetermined linear system of m equations in n unknowns (so that m < n), then it follows from Theorem 3.3 that the general solution has at least one parameter; hence a consistent underdetermined linear system have infinitely many solutions.

32 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 5 Example 3.6 If A is a 6 9 matrix, then for every 9 matrix b, the linear system Ax = b is underdetermined. Thus Ax = b must be consistent for some b, and for each such b the general solution must have 7 r parameters, where r is the rank of A. Theorem 3.3 (Equivalent Statements) If A is an n n matrix, then the following are equivalent. (a) A is invertible. (b) Ax = has only the trivial solution. (c) The reduced row-echelon form of A is I n. (d) A can be expressed as a product of elementary matrices. (e) Ax = b is consistent for every n matrix b. (f) Ax = b has exactly one solution for every n matrix b. (g) det(a). (h) The column vectors of A are linearly independent. (i) The row vectors of A are linearly independent. (j) The column vectors of A span R n. (k) The row vectors of A span R n. (l) The column vectors of A form a basis for R n. (m) The row vectors of A form a basis for R n. (n) A has rank n. (o) A has nullity. Exercise 3.6. Verify that rank(a) = rank(a T ). A = 3 5

33 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 6. Find the rank and nullity of the matrix. 3 (a) A = (b) A = (c) A = 3 (d) A = (e) A = In each part of Exercise, use the results obtained to find the number of leading variables and the number of parameters in the solution of Ax = without solving the system. 4. In each part, use the information in the table to the dimension of the row space, column space and nullspace of A, and of the nullspace of A T. (a) (b) (c) (d) (e) (f) (g) Size of A Rank (A) 3 5. In each part, find the largest possible value for the rank of A and the smallest possible value for the nullity of A. (a) A is 4 4 (b) A is 3 5 (c) A is In each part, use the information in the table to determine whether the system Ax = b is consistent. If so, state the number of parameters in its general solution. (a) (b) (c) (d) (e) (f) (g) Size of A Rank(A) 3 Rank[A b] For each of the matrices in Exercise 6, find the nullity of A, and determine the number of parameter in the general solution of the homogeneous linear system Ax =. 8. What conditions must be satisfied by b, b, b 3, b 4, and b 5 for the overdetermined

34 Prepared by Dr. Archara Pacheenburawana (MA3 Sec 75) 7 linear system to be consistent? x 3x = b x x = b x + x = b 3 x 4x = b 4 x +5x = b 5 9. Are there values of r and s for which r s r + 3 has rank or? If so, find those values.. (a) If A is 3 5 matrix, then the rank of A is at most. Why? (b) If A is 3 5 matrix, then the nullity of A is at most (c) If A is 3 5 matrix, then the rank of A T is at most (d) If A is 3 5 matrix, then the nullity of A T is at most. rank(a) = rank(a T ) = Answer to Exercise 3.6. (a) rank =, nullity = (b) rank = 3, nullity = (c) rank =, nullity = (d) rank = 3, nullity = (e) rank = 3, nullity = 3 3. (a) ; (b) 3; (c) ; (d) 3; (e) 3;3. Why?. Why?. Why? 4. (a) 3;3;; (b) ;;; (c) ;;; (d) ;;7;7 (e) ;;3;3 (f) ;;4;4 (g) ;;; 5. (a) rank = 4, nullity = (b) rank = 3, nullity = (c) rank = 3, nullity = 6. (a) Yes; (b) No (c) Yes; (d) Yes; 7 (e) No (f) Yes; 4 (g) Yes; 7. (a) nullity = ; (b) nullity = ; (c) nullity = ; (d) nullity = 7;7 (e) nullity = 3;3 (f) nullity = 4;4 (g) nullity = ; 8. b = r, b = s, b 3 = 4s 3r, b 4 = r s, b 5 = 8s 7r 9. Rank is if r = and s = ; the rank is never.. (a) 3 (b) 5 (c) 3 (d) 3

35 Chapter 4 Linear Transformations 4. Definition and Examples Definition 4. A mapping T from a vector space V into a vector space W, denoted by T : V W, is said to be a linear transformation if for all v,v V and for all scalars α and β. T(αv +βv ) = αt(v )+βt(v ) (4.) If T is a linear transformation mapping a vector space V into W, it follows from (4.) that T(v +v ) = T(v )+T(v ) (α = β = ) (4.) and T(αv) = αt(v) (v = v, β = ) (4.3) Conversely, if T satisfies (4.) and (4.3), then T(αv +βv ) = T(αv )+T(βv ) = αt(v )+βt(v ) Thus T is a linear transformation if and only if T satisfies (4.) and (4.3). In the special case when V = W, the linear transformation T : V V is called a linear operator on V. Thus a linear operator is a linear transformation that maps a vector space V into itself. Linear Operator on R Example 4. Let T be the operator defined by T(x) = 3x for each x R. Since T(αx) = 3(αx) = α(3x) = αt(x) 8