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Linear Algebra II Chapter 2 Vector Spaces 2.1 2.2 2.3 2.4 2.5 2.6 2.7 Vectors in Rn Vector Spaces Subspaces of Vector Spaces Spanning Sets and Linear Independence Basis and Dimension Rank of a Matrix and Systems of Linear Equations Coordinates and Change of Basis n 2.1 Vectors in R An ordered n-tuple: a sequence of n real number ( x1 , x2 ,, xn ) n n-space: R the set of all ordered n-tuple Ex: 1 n=1 R = 1-space = set of all real number n=2 R = 2-space = set of all ordered pair of real numbers ( x1 , x2 ) n=3 2 3 R = 3-space = set of all ordered triple of real numbers ( x1 , x2 , x3 ) n=4 4 R = 4-space = set of all ordered quadruple of real numbers ( x1 , x2 , x3 , x4 ) Notes: n (1) An n-tuple ( x1 , x2 ,, xn ) can be viewed as a point in R with the xi’s as its coordinates. (2) An n-tuple ( x1 , x2 ,, xn ) can be viewed as a vector x ( x1 , x2 ,, xn ) in Rn with the xi’s as its components. Ex: x1 , x2 x1 , x2 a point 0,0 a vector u u1 , u2 ,, un , v v1 , v2 ,, vn Equal: u v if and only if (two vectors in Rn) u1 v1 , u2 v2 , , un vn Vector addition (the sum of u and v): u v u1 v1 , u2 v2 , , un vn Scalar multiplication (the scalar multiple of u by c): cu cu1 , cu2 ,, cun Notes: The sum of two vectors and the scalar multiple of a vector n in R are called the standard operations in Rn. Negative: u (u1 ,u2 ,u3 ,...,un ) Difference: u v (u1 v1 , u2 v2 , u3 v3 ,..., un vn ) Zero vector: 0 (0, 0, ..., 0) Notes: (1) The zero vector 0 in Rn is called the additive identity in Rn. (2) The vector –v is called the additive inverse of v. Thm .2: (Properties of vector addition and scalar multiplication) n Let u, v, and w be vectors in R , and let c and d be scalars. (1) (2) (3) (4) (5) (6) u+v is a vector in Rn u+v = v+u (u+v)+w = u+(v+w) u+0 = u u+(–u) = 0 cu is a vector in Rn (7) c(u+v) = cu+cv (8) (c+d)u = cu+du (9) c(du) = (cd)u (10) 1(u) = u Ex 5: (Vector operations in R4) Let u=(2, – 1, 5, 0), v=(4, 3, 1, – 1), and w=(– 6, 2, 0, 3) be 4 vectors in R . Solve x for x in each of the following. (a) x = 2u – (v + 3w) (b) 3(x+w) = 2u – v+x Sol: (a) x 2u ( v 3w ) 2u v 3w (4, 2, 10, 0) (4, 3, 1, 1) (18, 6, 0, 9) (4 4 18, 2 3 6, 10 1 0, 0 1 9) (18, 11, 9, 8). (b) 3(x w ) 2u v x 3x 3w 2u v x 3x x 2u v 3w 2x 2u v 3w x u 12 v 32 w 2,1,5,0 2, 23 , 21 , 12 9,3,0, 29 9, 211 , 92 ,4 Thm 4.3: (Properties of additive identity and additive inverse) n Let v be a vector in R and c be a scalar. Then the following is true. (1) The additive identity is unique. That is, if u+v=v, then u = 0 (2) The additive inverse of v is unique. That is, if v+u=0, then u = –v (3) 0v=0 (4) c0=0 (5) If cv=0, then c=0 or v=0 (6) –(– v) = v Linear combination: The vector x is called a linear combination of v1 , v 2 ,..., v n , if it can be expressed in the form x c1v1 c2 v 2 cn v n c1 , c2 , , cn : scalar Ex 6: Given x = (– 1, – 2, – 2), u = (0,1,4), v = (– 1,1,2), and 3 w = (3,1,2) in R , find a, b, and c such that x = au+bv+cw. b 3c Sol: a 4a 1 b c 2 2b 2c 2 a 1, b 2, c 1 Thus x u 2 v w Notes: A vector u (u1 , u2 ,, un ) in R n can be viewed as: a 1×n row matrix (row vector): u [u1 , u2 ,, un ] or u1 u a n×1 column matrix (column vector): u 2 u n (The matrix operations of addition and scalar multiplication give the same results as the corresponding vector operations) Vector addition Scalar multiplication u v (u1 , u2 , , un ) (v1 , v2 , , vn ) cu c(u1 , u2 ,, un ) (u1 v1 , u2 v2 , , un vn ) u v [u1 , u2 , , un ] [v1 , v2 , , vn ] [u1 v1 , u2 v2 , , un vn ] u1 v1 u1 v1 u v u v u v 2 2 2 2 un vn un vn (cu1 , cu2 , , cun ) cu c[u1 , u2 ,, un ] [cu1 , cu2 ,, cun ] u1 cu1 u cu cu c 2 2 un cun Keywords in Section 2.1: ordered n-tuple n-space equal vector addition scalar multiplication negative difference zero vector additive identity additive inverse 2.2 Vector Spaces Vector spaces: Let V be a set on which two operations (vector addition and scalar multiplication) are defined. If the following axioms are satisfied for every u, v, and w in V and every scalar (real number) c and d, then V is called a vector space. Addition: (1) u+v is in V (2) u+v=v+u (3) u+(v+w)=(u+v)+w (4) V has a zero vector 0 such that for every u in V, u+0=u (5) For every u in V, there is a vector in V denoted by –u such that u+(–u)=0 Scalar multiplication: (6) cu is in V. (7) c(u v) cu cv (8) (c d )u cu du (9) c(du) (cd )u (10) 1(u ) u Notes: (1) A vector space consists of four entities: a set of vectors, a set of scalars, and two operations V:nonempty set c:scalar (u, v ) u v: vector addition (c, u) cu: scalar multiplication V , (2) V 0: , is called a vector space zero vector space Examples of vector spaces: (1) n-tuple space: Rn (u1 , u2 ,, un ) (v1 , v2 ,, vn ) (u1 v1 , u2 v2 ,, un vn ) vector addition scalar multiplication k (u1 , u2 ,, un ) (ku1 , ku2 ,, kun ) (2) Matrix space: V M mn (the set of all m×n matrices with real values) Ex: :(m = n = 2) u11 u12 v11 v12 u11 v11 u12 v12 u u v v u v u v 21 22 21 22 21 21 22 22 u11 u12 ku11 ku12 k u u ku ku 22 21 22 21 vector addition scalar multiplication (3) n-th degree polynomial space: V Pn (x) (the set of all real polynomials of degree n or less) p( x) q( x) (a0 b0 ) (a1 b1 ) x (an bn ) x n kp( x) ka0 ka1 x kan x n (4) Function space: V c(, ) (the set of all real-valued continuous functions defined on the entire real line.) ( f g )( x) f ( x) g ( x) (kf )( x) kf ( x) Theorem 2.4: (Properties of scalar multiplication) Let v be any element of a vector space V, and let c be any scalar. Then the following properties are true. (1) (2) (3) (4) 0v 0 c0 0 If cv 0, then c 0 or v 0 (1) v v Notes: To show that a set is not a vector space, you need only find one axiom that is not satisfied. Ex 6: The set of all integer is not a vector space. Pf: 1V , 12 R ( 12 )(1) 12 V (it is not closed under scalar multiplication) noninteger scalar integer Ex 7: The set of all second-degree polynomials is not a vector space. Pf: Let p( x) x 2 and q( x) x 2 x 1 p ( x) q ( x) x 1 V (it is not closed under vector addition) Ex 8: V=R2=the set of all ordered pairs of real numbers vector addition: (u1 , u2 ) (v1 , v2 ) (u1 v1 , u2 v2 ) scalar multiplication: c(u1, u2 ) (cu1,0) Verify V is not a vector space. Sol: 1(1, 1) (1, 0) (1, 1) the set (together with the two given operations) is not a vector space Keywords in Section2.2: vector space n-space matrix space polynomial space function space 2.3 Subspaces of Vector Spaces Subspace: (V ,,) : a vector space W : a nonempty subset W V (W ,,) :a vector space (under the operations of addition and scalar multiplication defined in V) W is a subspace of V Trivial subspace: Every vector space V has at least two subspaces. (1) Zero vector space {0} is a subspace of V. (2) V is a subspace of V. Theorem 2.5: (Test for a subspace) If W is a nonempty subset of a vector space V, then W is a subspace of V if and only if the following conditions hold. (1) If u and v are in W, then u+v is in W. (2) If u is in W and c is any scalar, then cu is in W. Ex: Subspace of R2 (1) 0 0 0, 0 (2) Lines through t he origin (3) R 2 Ex: Subspace of R3 (1) 0 0 0, 0, 0 (2) Lines through t he origin (3) Planes through t he origin (4) R 3 Ex 2: (A subspace of M2×2) Let W be the set of all 2×2 symmetric matrices. Show that W is a subspace of the vector space M2×2, with the standard operations of matrix addition and scalar multiplication. Sol: W M 22 M 22 : vector sapces Let A1, A2 W ( A1T A1, A2T A2 ) A1 W, A2 W ( A1 A2 )T A1T A2T A1 A2 ( A1 A2 W ) k R , A W (kA)T kAT kA W is a subspace of M 22 ( kA W ) Ex 3: (The set of singular matrices is not a subspace of M2×2) Let W be the set of singular matrices of order 2. Show that W is not a subspace of M2×2 with the standard operations. Sol: 1 0 0 0 A W , B W 0 0 0 1 1 0 A B W 0 1 W2 is not a subspace of M 22 Ex 4: (The set of first-quadrant vectors is not a subspace of R2) Show that W {( x1 , x2 ) : x1 0 and x2 0} , with the standard operations, is not a subspace of R2. Sol: Let u (1, 1) W 1u 11, 1 1, 1W W is not a subspace of R 2 (not closed under scalar multiplication) Ex 6: (Determining subspaces of R2) Which of the following two subsets is a subspace of R2? (a) The set of points on the line given by x+2y=0. (b) The set of points on the line given by x+2y=1. Sol: (a) W ( x, y) x 2 y 0 (2t, t ) t R Let v1 2t1 , t1 W v2 2t2 , t2 W v1 v2 2t1 t2 ,t1 t2 W (closed under addition) kv1 2kt1 ,kt1 W (closed under scalar multiplication) W is a subspace of R 2 (b) W x, y x 2 y 1 (Note: the zero vector is not on the line) Let v (1,0) W 1v 1,0W W is not a subspace of R 2 Ex 8: (Determining subspaces of R3) Which of the following subsets is a subspace of R 3? (a) W ( x1 , x2 ,1) x1 , x2 R (b) W ( x1 , x1 x3 , x3 ) x1 , x3 R Sol: (a) Let v (0,0,1) W (1) v (0,0,1) W W is not a subspace of R3 (b) Let v ( v1 , v1 v3 , v3 ) W , u (u1 , u1 u 3 , u 3 ) W v u v1 u1 , v1 u1 v3 u 3 , v3 u 3 W kv kv1 , kv1 kv3 , kv3 W W is a subspace of R3 Thm 4.6: (The intersection of two subspaces is a subspace) If V and W are both subspaces of a vector space U , then the intersecti on of V and W (denoted by V U ) is also a subspace of U . Elementary Linear Algebra: Section 4.3, p.202 Keywords in Section 3.3: Subspace trivial subspace 2.4 Spanning Sets and Linear Independence Linear combination: A vector v in a vector space V is called a linear combinatio n of the vectors u1,u 2 , ,u k in V if v can be written in the form v c1u1 c2u 2 ck u k c1,c2 , ,ck : scalars Ex 2-3: (Finding a linear combination) v1 (1,2,3) v 2 (0,1,2) v 3 (1,0,1) Prove (a) w (1,1,1) is a linear combinatio n of v1 , v 2 , v 3 (b) w (1,2,2) is not a linear combinatio n of v1 , v 2 , v 3 Sol: (a) w c1v1 c2 v 2 c3 v 3 1,1,1 c1 1,2,3 c2 0,1,2 c3 1,0,1 (c1 c3 , 2c1 c2 , 3c1 2c2 c3 ) c1 c3 2c1 c2 3c1 2c2 c3 1 1 1 1 0 1 1 Guass Jordan Elimination 2 1 0 1 3 2 1 1 1 0 1 1 0 1 2 1 0 0 0 0 c1 1 t , c2 1 2t , c3 t (this system has infinitely many solutions) t 1 w 2 v1 3v 2 v 3 (b) w c1 v1 c2 v 2 c3 v 3 1 0 1 1 Guass Jordan Elimination 2 1 0 2 3 2 1 2 1 0 1 1 0 1 2 4 0 0 0 7 this system has no solution ( 0 7) w c1v1 c2 v 2 c3 v 3 Elementary Linear Algebra: Section 4.4, pp.208-209 the span of a set: span (S) If S={v1, v2,…, vk} is a set of vectors in a vector space V, then the span of S is the set of all linear combinations of the vectors in S, span(S ) c1 v1 c2 v 2 ck v k ci R (the set of all linear combinatio ns of vectors in S ) a spanning set of a vector space: If every vector in a given vector space can be written as a linear combination of vectors in a given set S, then S is called a spanning set of the vector space. Elementary Linear Algebra: Section 4.4, p.209 Notes: span ( S ) V S spans (generates ) V V is spanned (generated ) by S S is a spanning set of V Notes: (1) span( ) 0 (2) S span( S ) (3) S1 , S 2 V S1 S 2 span( S1 ) span( S 2 ) Elementary Linear Algebra: Section 4.4, p.209 Ex 5: (A spanning set for R3) Show that the set S (1,2,3), (0,1,2), (2,0,1) sapns R3 Sol: We must determine whether an arbitrary vector u (u1 , u2 , u3 ) in R 3 can be as a linear combinatio n of v1 , v 2 , and v 3 . u R 3 u c1 v1 c2 v 2 c3 v 3 c1 2c1 c2 2c3 u1 u2 3c1 2c2 c3 u3 The problem thus reduces to determinin g whether this system is consistent for all values of u1 , u2 , and u3 . Elementary Linear Algebra: Section 4.4, p.210 1 0 2 A 2 1 0 0 3 2 1 Ax b has exactly one solution for every u. span( S ) R 3 Elementary Linear Algebra: Section 4.4, p.210 Thm 4.7: (Span(S) is a subspace of V) If S={v1, v2,…, vk} is a set of vectors in a vector space V, then (a) span (S) is a subspace of V. (b) span (S) is the smallest subspace of V that contains S. (Every other subspace of V that contains S must contain span (S).) Elementary Linear Algebra: Section 4.4, p.211 Linear Independent (L.I.) and Linear Dependent (L.D.): S v1 , v 2 ,, v k : a set of vectors in a vector space V c1 v1 c2 v 2 ck v k 0 (1) If the equation has only the trivial solution (c1 c2 ck 0) then S is called linearly independen t. (2) If the equation has a nontrivial solution (i.e., not all zeros), then S is called linearly dependent. Elementary Linear Algebra: Section 4.4, p.211 Notes: (1) is linearly independen t (2) 0 S S is linearly dependent. (3) v 0 v is linearly independen t (4) S1 S2 S1 is linearly dependent S2 is linearly dependent S2 is linearly independen t S1 is linearly independen t Elementary Linear Algebra: Section 4.4, p.211 Ex 8: (Testing for linearly independent) Determine whether the following set of vectors in R3 is L.I. or L.D. S 1, 2, 3, 0, 1, 2, 2, 0, 1 v1 v2 v3 c1 2c3 0 Sol: 0 c1v1 c2 v 2 c3 v 3 0 2c1 c2 3c1 2c2 c3 0 1 0 2 0 1 0 0 0 - Jordan Eliminatio n 0 1 0 0 2 1 0 0 Gauss 3 2 1 0 0 0 1 0 c1 c2 c3 0 only the trivial solution S is linearly independen t Elementary Linear Algebra: Section 4.4, p.213 Ex 9: (Testing for linearly independent) Determine whether the following set of vectors in P2 is L.I. or L.D. Sol: S = {1+x – 2x2 , 2+5x – x2 , x+x2} v1 v2 v3 c1v1+c2v2+c3v3 = 0 i.e. c1(1+x – 2x2) + c2(2+5x – x2) + c3(x+x2) = 0+0x+0x2 2 0 0 1 c1+2c2 =0 G. J. c1+5c2+c3 = 0 1 5 1 0 –2c1 – c2+c3 = 0 2 1 1 0 1 2 0 0 1 1 1 0 0 0 03 0 This system has infinitely many solutions. (i.e., This system has nontrivial solutions.) S is linearly dependent. Elementary Linear Algebra: Section 4.4, p.214 (Ex: c1=2 , c2= – 1 , c3=3) Ex 10: (Testing for linearly independent) Determine whether the following set of vectors in 2×2 matrix space is L.I. or L.D. 2 1 3 0 1 0 S , , 0 1 2 1 2 0 Sol: v1 v2 v3 c1v1+c2v2+c3v3 = 0 2 1 3 0 1 0 0 0 c1 c2 c3 0 1 2 1 2 0 0 0 Elementary Linear Algebra: Section 4.4, p.215 2c1+3c2+ c3 = 0 c1 =0 2c2+2c3 = 0 c1 + c2 =0 2 3 1 0 1 0 0 0 0 2 2 0 1 1 0 0 1 0 - Jordan Eliminatio n Gauss 0 0 0 1 0 0 0 0 1 0 0 0 0 0 c1 = c2 = c3= 0 (This system has only the trivial solution.) S is linearly independent. Elementary Linear Algebra: Section 4.4, p.215 Thm 4.8: (A property of linearly dependent sets) A set S = {v1,v2,…,vk}, k2, is linearly independent if and only if at least one of the vectors vj in S can be written as a linear combination of the other vectors in S. Pf: () c1v1+c2v2+…+ckvk = 0 S is linearly dependent ci 0 for some i ci 1 ci 1 ck c1 v i v1 v i 1 v i 1 v k ci ci ci ci Elementary Linear Algebra: Section 4.4, p.217 () Let vi = d1v1+…+di-1vi-1+di+1vi+1+…+dkvk d1v1+…+di-1vi-1-vi+di+1vi+1+…+dkvk = 0 c1=d1, …,ci-1=di-1, ci=-1,ci+1=di+1,…, ck=dk (nontrivial solution) S is linearly dependent Corollary to Theorem 4.8: Two vectors u and v in a vector space V are linearly dependent if and only if one is a scalar multiple of the other. Elementary Linear Algebra: Section 4.4, p.217 Keywords in Section 4.4: linear combination:線性組合 spanning set:生成集合 trivial solution:顯然解 linear independent:線性獨立 linear dependent:線性相依 4.5 Basis and Dimension Basis: V:a vector space S ={v1, v2, …, vn}V Generating Sets Bases Linearly Independent Sets (a ) S spans V (i.e., span(S) = V ) (b) S is linearly independent S is called a basis for V Notes: (1) Ø is a basis for {0} (2) the standard basis for R3: {i, j, k} i = (1, 0, 0), j = (0, 1, 0), k = (0, 0, 1) Elementary Linear Algebra: Section 4.5, p.221 n (3) the standard basis for R : {e1, e2, …, en} e1=(1,0,…,0), e2=(0,1,…,0), en=(0,0,…,1) Ex: R4 {(1,0,0,0), (0,1,0,0), (0,0,1,0), (0,0,0,1)} (4) the standard basis for mn matrix space: { Eij | 1im , 1jn } Ex: 2 2 matrix space: 1 0 0 1 0 0 0 0 , , , 0 0 0 0 1 0 0 1 (5) the standard basis for Pn(x): {1, x, x2, …, xn} Ex: P3(x) {1, x, x2, x3} Elementary Linear Algebra: Section 4.5, p.224 Thm 4.9: (Uniqueness of basis representation) If S v1 , v 2 ,, v n is a basis for a vector space V, then every vector in V can be written in one and only one way as a linear combination of vectors in S. Pf: 1. span(S) = V S is a basis 2. S is linearly independent span(S) = V Let v = c1v1+c2v2+…+cnvn v = b1v1+b2v2+…+bnvn 0 = (c1–b1)v1+(c2 – b2)v2+…+(cn – bn)vn S is linearly independen t c1= b1 , c2= b2 ,…, cn= bn Elementary Linear Algebra: Section 4.5, p.224 (i.e., uniqueness) Thm 4.10: (Bases and linear dependence) If S v1 , v 2 ,, v n is a basis for a vector space V, then every set containing more than n vectors in V is linearly dependent. Pf: Let S1 = {u1, u2, …, um} , m > n span ( S ) V u1 c11v1 c21v 2 cn1 v n uiV u 2 c12 v1 c22 v 2 cn 2 v n u m c1m v1 c2 m v 2 cnm v n Elementary Linear Algebra: Section 4.5, p.225 Let k1u1+k2u2+…+kmum= 0 d1v1+d2v2+…+dnvn= 0 (where di = ci1k1+ci2k2+…+cimkm) S is L.I. di=0 i i.e. c11k1 c12k 2 c1m k m 0 c21k1 c22k 2 c2 m k m 0 cn1k1 cn 2 k 2 cnm k m 0 Thm 1.1: If the homogeneous system has fewer equations than variables, then it must have infinitely many solution. m > n k1u1+k2u2+…+kmum = 0 has nontrivial solution S1 is linearly dependent Elementary Linear Algebra: Section 4.5, p.225 Thm 4.11: (Number of vectors in a basis) If a vector space V has one basis with n vectors, then every basis for V has n vectors. (All bases for a finite-dimensional vector space has the same number of vectors.) Pf: S ={v1, v2, …, vn} S'={u1, u2, …, um} S is a basis S ' is L.I. S is L.I. S ' is a basis two bases for a vector space Thm.4.10 n m n m Thm . 4 . 10 n m Elementary Linear Algebra: Section 4.5, p.226 Finite dimensional: A vector space V is called finite dimensional, if it has a basis consisting of a finite number of elements. Infinite dimensional: If a vector space V is not finite dimensional, then it is called infinite dimensional. Dimension: The dimension of a finite dimensional vector space V is defined to be the number of vectors in a basis for V. V: a vector space dim(V) = #(S) S: a basis for V (the number of vectors in S) Elementary Linear Algebra: Section 4.5, p.227 dim(V) = n Notes: (1) dim({0}) = 0 = #(Ø) Generating Sets #(S) > n (2) dim(V) = n , SV Bases Linearly Independent Sets #(S) = n #(S) < n S:a generating set #(S) n S:a L.I. set #(S) n S:a basis #(S) = n (3) dim(V) = n , W is a subspace of V dim(W) n Elementary Linear Algebra: Section 4.5, Addition Ex: (1) Vector space Rn basis {e1 , e2 , , en} dim(Rn) = n (2) Vector space Mmn basis {Eij | 1im , 1jn} dim(Mmn)=mn (3) Vector space Pn(x) basis {1, x, x2, , xn} dim(Pn(x)) = n+1 (4) Vector space P(x) basis {1, x, x2, } dim(P(x)) = Elementary Linear Algebra: Section 4.5, Addition Ex 9: (Finding the dimension of a subspace) (a) W={(d, c–d, c): c and d are real numbers} (b) W={(2b, b, 0): b is a real number} Sol: (Note: Find a set of L.I. vectors that spans the subspace) (a) (d, c– d, c) = c(0, 1, 1) + d(1, – 1, 0) S = {(0, 1, 1) , (1, – 1, 0)} (S is L.I. and S spans W) S is a basis for W dim(W) = #(S) = 2 (b) 2b, b,0 b2,1,0 S = {(2, 1, 0)} spans W and S is L.I. S is a basis for W dim(W) = #(S) = 1 Elementary Linear Algebra: Section 4.5, p.228 Ex 11: (Finding the dimension of a subspace) Let W be the subspace of all symmetric matrices in M22. What is the dimension of W? Sol: a b W a , b, c R b c a b 1 0 0 1 0 0 a b c b c 0 0 1 0 0 1 1 0 0 1 0 0 S , , spans W and S is L.I. 0 0 1 0 0 1 S is a basis for W dim(W) = #(S) = 3 Elementary Linear Algebra: Section 4.5, p.229 Thm 4.12: (Basis tests in an n-dimensional space) Let V be a vector space of dimension n. (1) If S v1 , v 2 ,, v n is a linearly independent set of vectors in V, then S is a basis for V. (2) If S v1 , v 2 ,, v n spans V, then S is a basis for V. dim(V) = n Generating Sets Bases Linearly Independent Sets #(S) > n #(S) = n Elementary Linear Algebra: Section 4.5, p.229 #(S) < n Keywords in Section 4.5: basis:基底 dimension:維度 finite dimension:有限維度 infinite dimension:無限維度 Let A be an m×n matrix. Row space: The row space of A is the subspace of Rn spanned by the row vectors of A. RS ( A) {1 A(1) 2 A( 2) ... m A( m) | 1 , 2 ,..., m R} Column space: The column space of A is the subspace of Rm spanned by the column vectors of A. CS A {1 A(1) 2 A(2) n A( n) 1 , 2 ,n R} Null space: The null space of A is the set of all solutions of Ax=0 and it is a subspace of Rn. NS ( A) {x Rn | Ax 0} Elementary Linear Algebra: Section 4.6, p.233 4.6 Rank of a Matrix and Systems of Linear Equations row vectors: Row vectors of A a11 a12 a1n A1 A a a22 a2 n 2 21 A A am1 am 2 amn m [a11 , a12 ,, a1n ] A(1) [a21 , a22 ,, a2n ] A(2) [am1 , am2 ,, amn ] A( n ) column vectors: a11 a12 a1n a a a 22 2n A 21 A1 A2 An am1 am 2 amn Elementary Linear Algebra: Section 4.6, p.232 Column vectors of A a11 a12 a1n a a a 21 22 2 n am1 am 2 amn || || || (1) (2) (n) A A A Thm 4.13: (Row-equivalent matrices have the same row space) If an mn matrix A is row equivalent to an mn matrix B, then the row space of A is equal to the row space of B. Notes: (1) The row space of a matrix is not changed by elementary row operations. RS(r(A)) = RS(A) r: elementary row operations (2) Elementary row operations can change the column space. Elementary Linear Algebra: Section 4.6, p.233 Thm 4.14: (Basis for the row space of a matrix) If a matrix A is row equivalent to a matrix B in row-echelon form, then the nonzero row vectors of B form a basis for the row space of A. Elementary Linear Algebra: Section 4.6, p.234 Ex 2: ( Finding a basis for a row space) 1 0 Find a basis of row space of A = 3 3 2 Sol: 1 0 3 A= 3 2 a1 3 1 1 1 0 6 4 2 0 4 a2 a3 3 0 1 1 2 a4 Elementary Linear Algebra: Section 4.6, p.234 3 1 1 1 0 6 4 2 0 4 1 0 .E . G B = 0 0 0 b1 3 0 1 1 2 3 1 3 w 1 1 1 0 w 2 0 0 1 w 3 0 0 0 0 0 0 b2 b3 b4 a basis for RS(A) = {the nonzero row vectors of B} (Thm 4.14) = {w1, w2, w3} = {(1, 3, 1, 3), (0, 1, 1, 0), (0, 0, 0, 1)} Notes: (1) b3 2b1 b 2 a3 2a1 a 2 (2) {b1 , b 2 , b 4 } is L.I. {a1 , a 2 , a 4 } is L.I. Elementary Linear Algebra: Section 4.6, p.234 Ex 3: (Finding a basis for a subspace) Find a basis for the subspace of R3 spanned by v1 v3 v2 S {(1, 2, 5), (3, 0, 3), (5, 1, 8)} Sol: 1 2 5 v 1 A = 3 0 3 v 2 5 1 8 v 3 G.E. 1 2 5 w1 B 0 1 3 w 2 0 0 0 a basis for span({v1, v2, v3}) = a basis for RS(A) = {the nonzero row vectors of B} = {w1, w2} = {(1, –2, – 5) , (0, 1, 3)} Elementary Linear Algebra: Section 4.6, p.235 (Thm 4.14) Ex 4: (Finding a basis for the column space of a matrix) Find a basis for the column space of the matrix A given in Ex 2. 1 0 A 3 3 2 3 1 1 1 0 6 4 2 0 4 3 0 1 1 2 Sol. 1: 1 3 AT 1 3 0 3 3 2 1 0 1 0 4 0 .E. G B 0 1 6 2 4 0 1 1 2 0 Elementary Linear Algebra: Section 4.6, p.236 0 3 3 2 w1 1 9 5 6 w 2 0 1 1 1 w3 0 0 0 0 CS(A)=RS(AT) a basis for CS(A) = a basis for RS(AT) = {the nonzero vectors of B} = {w1, w2, w3} 1 0 3, 3 2 0 0 1 0 9 , 1 (a basis for the column space of A) 5 1 6 1 Note: This basis is not a subset of {c1, c2, c3, c4}. Elementary Linear Algebra: Section 4.6, p.236 Sol. 2: 1 0 A 3 3 2 c1 Leading 1 3 1 0 4 0 c2 3 1 0 1 0 G.E . B 0 6 1 2 1 0 0 4 2 c3 c4 v1 1 3 1 3 1 1 0 0 0 1 0 0 0 0 0 0 v2 v3 v4 => {v1, v2, v4} is a basis for CS(B) {c1, c2, c4} is a basis for CS(A) Notes: (1) This basis is a subset of {c1, c2, c3, c4}. (2) v3 = –2v1+ v2, thus c3 = – 2c1+ c2 . Elementary Linear Algebra: Section 4.6, p.236 Thm 4.16: (Solutions of a homogeneous system) If A is an mn matrix, then the set of all solutions of the homogeneous system of linear equations Ax = 0 is a subspace of Rn called the nullspace of A. Pf: NS ( A) {x R n | Ax 0} NS ( A) R n NS ( A) ( A0 0) Let x1, x 2 NS ( A) (i.e. Ax1 0, Ax 2 0) Then (1) A(x1 x 2) Ax1 Ax 2 0 0 0 Addition (2) A(cx1) c( Ax1) c(0) 0 Scalar multiplica tion Thus NS ( A) is a subspace of R n Notes: The nullspace of A is also called the solution space of the homogeneous system Ax = 0. Elementary Linear Algebra: Section 4.6, p.239 Ex 6: (Finding the solution space of a homogeneous system) 1 2 2 1 Find the nullspace of the matrix A. A 3 6 5 4 1 2 0 3 Sol: The nullspace of A is the solution space of Ax = 0. 1 2 2 1 1 2 0 3 . J .E A 3 6 5 4 G 0 0 1 1 1 2 0 3 0 0 0 0 x1 = –2s – 3t, x2 = s, x3 = –t, x4 = t x1 2s 3t 2 3 x s 1 0 s t sv1 tv 2 x 2 x3 t 0 1 x t 0 1 4 NS ( A) {sv1 tv 2 | s, t R} Elementary Linear Algebra: Section 4.6, p.240 Thm 4.15: (Row and column space have equal dimensions) If A is an mn matrix, then the row space and the column space of A have the same dimension. dim(RS(A)) = dim(CS(A)) Rank: The dimension of the row (or column) space of a matrix A is called the rank of A and is denoted by rank(A). rank(A) = dim(RS(A)) = dim(CS(A)) Elementary Linear Algebra: Section 4.6, pp.237-238 Nullity: The dimension of the nullspace of A is called the nullity of A. nullity(A) = dim(NS(A)) Note: rank(AT) = rank(A) Pf: rank(AT) = dim(RS(AT)) = dim(CS(A)) = rank(A) Elementary Linear Algebra: Section 4.6, p.238 Thm 4.17: (Dimension of the solution space) If A is an mn matrix of rank r, then the dimension of the solution space of Ax = 0 is n – r. That is n = rank(A) + nullity(A) Notes: (1) rank(A): The number of leading variables in the solution of Ax=0. (The number of nonzero rows in the row-echelon form of A) (2) nullity (A): The number of free variables in the solution of Ax = 0. Elementary Linear Algebra: Section 4.6, p.241 Notes: If A is an mn matrix and rank(A) = r, then Fundamental Space Dimension RS(A)=CS(AT) r CS(A)=RS(AT) r NS(A) n–r NS(AT) m–r Elementary Linear Algebra: Section 4.6, Addition Ex 7: (Rank and nullity of a matrix) Let the column vectors of the matrix A be denoted by a1, a2, a3, a4, and a5. 1 0 1 0 2 0 1 3 1 3 A 2 1 1 1 3 9 0 12 0 3 a1 a2 a3 a4 a5 (a) Find the rank and nullity of A. (b) Find a subset of the column vectors of A that forms a basis for the column space of A . (c) If possible, write the third column of A as a linear combination of the first two columns. Elementary Linear Algebra: Section 4.6, p.242 Sol: Let B be the reduced row-echelon form of A. 0 1 0 2 1 0 1 3 1 3 A 2 1 1 1 3 0 3 9 0 12 a1 a2 a3 a4 a5 (a) rank(A) = 3 1 0 B 0 0 b1 0 2 0 1 1 3 0 4 0 0 1 1 0 0 0 0 b2 b3 b4 b5 (the number of nonzero rows in B) nuillity ( A) n rank ( A) 5 3 2 Elementary Linear Algebra: Section 4.6, p.242 (b) Leading 1 {b1 , b 2 , b 4 } is a basis for CS ( B) {a1 , a 2 , a 4 } is a basis for CS ( A) 1 0 1 0 1 1 a1 , a 2 , and a 4 , 2 1 1 0 3 0 (c) b3 2b1 3b 2 a3 2a1 3a 2 Elementary Linear Algebra: Section 4.6, p.243 Thm 4.18: (Solutions of a nonhomogeneous linear system) If xp is a particular solution of the nonhomogeneous system Ax = b, then every solution of this system can be written in the form x = xp + xh , wher xh is a solution of the corresponding homogeneous system Ax = 0. Pf: Let x be any solution of Ax = b. A(x x p ) Ax Ax p b b 0. (x x p ) is a solution of Ax = 0 Let x h x x p x x p xh Elementary Linear Algebra: Section 4.6, p.243 Ex 8: (Finding the solution set of a nonhomogeneous system) Find the set of all solution vectors of the system of linear equations. x1 Sol: 3x1 x2 x1 2 x2 2 x3 x4 5 x3 5 x4 5 8 9 1 5 1 5 1 0 2 1 0 2 3 1 5 G 0 1 . J .E 0 8 1 3 7 1 2 0 0 0 5 9 0 0 0 s t Elementary Linear Algebra: Section 4.6, p.243 x1 2s x s x 2 x3 s x4 0 s t 5 2 1 5 3t 7 1 3 7 s t 0t 0 1 0 0 t 0 0 1 0 su1 tu 2 x p 5 7 i.e. x p is a particular solution vector of Ax=b. 0 0 xh = su1 + tu2 is a solution of Ax = 0 Elementary Linear Algebra: Section 4.6, p.244 Thm 4.19: (Solution of a system of linear equations) The system of linear equations Ax = b is consistent if and only if b is in the column space of A. Pf: Let a11 a A 21 am1 a12 a1n a22 a2 n , am 2 amn x1 x x 2, xn and b1 b b 2 bm be the coefficient matrix, the column matrix of unknowns, and the right-hand side, respectively, of the system Ax = b. Elementary Linear Algebra: Section 4.6, pp.244-245 Then a11 a12 a1n x1 a11 x1 a x a x a a 22 2n 2 Ax 21 21 1 a a a m2 mn xn m1 am1 x1 a11 a12 a1n a a a x1 21 x2 22 xn 2 n . a a m1 m2 amn a12 x2 a22 x2 a m 2 x2 a1n xn a2 n xn amn xn Hence, Ax = b is consistent if and only if b is a linear combination of the columns of A. That is, the system is consistent if and only if b is in the subspace of Rm spanned by the columns of A. Elementary Linear Algebra: Section 4.6, p.245 Note: If rank([A|b])=rank(A) Then the system Ax=b is consistent. Ex 9: (Consistency of a system of linear equations) x1 x2 x1 3x1 Sol: 2 x2 x3 1 x3 3 x3 1 1 1 1 1 1 0 . J .E. A 1 0 1 G 0 1 2 3 2 1 0 0 0 Elementary Linear Algebra: Section 4.6, p.245 1 3 1 1 1 1 1 0 . J .E. [ A b] 1 0 1 3 G 0 1 2 4 3 2 1 1 0 0 0 0 c1 c2 c3 b w1 w2 w3 v v 3w1 4w 2 b 3c1 4c 2 0c3 (b is in the column space of A) The system of linear equations is consistent. Check: rank ( A) rank ([ A b]) 2 Elementary Linear Algebra: Section 4.6, p.245 Summary of equivalent conditions for square matrices: If A is an n×n matrix, then the following conditions are equivalent. (1) A is invertible (2) Ax = b has a unique solution for any n×1 matrix b. (3) Ax = 0 has only the trivial solution (4) A is row-equivalent to In (5) | A | 0 (6) rank(A) = n (7) The n row vectors of A are linearly independent. (8) The n column vectors of A are linearly independent. Elementary Linear Algebra: Section 4.6, p.246 Keywords in Section 4.6: row space : 列空間 column space : 行空間 null space: 零空間 solution space : 解空間 rank: 秩 nullity : 核次數 4.7 Coordinates and Change of Basis Coordinate representation relative to a basis Let B = {v1, v2, …, vn} be an ordered basis for a vector space V and let x be a vector in V such that x c1v1 c2 v 2 cn v n . The scalars c1, c2, …, cn are called the coordinates of x relative to the basis B. The coordinate matrix (or coordinate vector) of x relative to B is the column matrix in Rn whose components are the coordinates of x. xB c1 c 2 cn Elementary Linear Algebra: Section 4.7, p.249 n Ex 1: (Coordinates and components in R ) Find the coordinate matrix of x = (–2, 1, 3) in R3 relative to the standard basis S = {(1, 0, 0), ( 0, 1, 0), (0, 0, 1)} Sol: x (2, 1, 3) 2(1, 0, 0) 1(0, 1, 0) 3(0, 0, 1), 2 [x]S 1. 3 Elementary Linear Algebra: Section 4.7, p.250 Ex 3: (Finding a coordinate matrix relative to a nonstandard basis) Find the coordinate matrix of x=(1, 2, –1) in R3 relative to the (nonstandard) basis B ' = {u1, u2, u3}={(1, 0, 1), (0, – 1, 2), (2, 3, – 5)} Sol: x c u c u c u (1, 2, 1) c (1, 0, 1) c (0, 1, 2) c (2, 3, 5) 1 1 c1 c1 1 0 0 1 1 2 2 2 3 3 2c3 3c3 5c3 1 2 3 1 2 1 2 c1 1 1 0 c2 i.e. 0 1 3 c2 2 1 2 5 c3 1 2c2 2 1 5 1 0 0 5 G.J.E. 3 2 0 1 0 8 [ x ] 8 2 0 0 1 2 5 1 Elementary Linear Algebra: Section 4.7, p.251 B Change of basis problem: You were given the coordinates of a vector relative to one basis B and were asked to find the coordinates relative to another basis B'. Ex: (Change of basis) Consider two bases for a vector space V B {u1 , u2 }, B {u1 , u2 } a c If [u1 ]B , [u2 ]B b d i.e., u1 au1 bu 2 , u2 cu1 du 2 Elementary Linear Algebra: Section 4.7, p.253 k1 Let v V , [ v]B k 2 v k1u1 k 2u2 k1 (au1 bu 2 ) k 2 (cu1 du 2 ) (k1a k 2 c)u1 (k1b k 2 d )u 2 k1a k2c a c k1 [ v ]B k k b k d b d 2 2 1 u1 B u2 B v B Elementary Linear Algebra: Section 4.7, p.253 Transition matrix from B' to B: Let B {u1 , u 2 ,..., u n } and B {u1 , u2 ..., un } be two bases for a vector space V If [v]B is the coordinate matrix of v relative to B [v]B‘ is the coordinate matrix of v relative to B' then [ v]B P[ v]B u1 B , u2 B ,..., un B vB where P u1 B , u2 B , ..., un B is called the transition matrix from B' to B Elementary Linear Algebra: Section 4.7, p.255 Thm 4.20: (The inverse of a transition matrix) If P is the transition matrix from a basis B' to a basis B in Rn, then (1) P is invertible –1 (2) The transition matrix from B to B' is P Notes: B {u1 , u 2 , ..., u n }, B' {u1 , u2 , ..., un } vB [u1 ]B , [u2 ]B , ..., [un ]B vB P vB vB [u1 ]B , [u 2 ]B , ..., [u n ]B vB P 1 vB Elementary Linear Algebra: Section 4.7, p.253 Thm 4.21: (Transition matrix from B to B') Let B={v1, v2, … , vn} and B' ={u1, u2, … , un} be two bases n for R . Then the transition matrix P–1 from B to B' can be found by using Gauss-Jordan elimination on the n×2n matrix B B as follows. B B I n P 1 Elementary Linear Algebra: Section 4.7, p.255 Ex 5: (Finding a transition matrix) B={(–3, 2), (4,–2)} and B' ={(–1, 2), (2,–2)} are two bases for R2 (a) Find the transition matrix from B' to B. 1 (b) Let [ v]B ' , find [ v]B 2 (c) Find the transition matrix from B to B' . Elementary Linear Algebra: Section 4.7, p.257 Sol: (a) 3 2 (b) Check : 4 2 B 1 2 2 2 B' 3 2 P 2 1 G.J.E. 1 0 3 2 0 1 2 1 I P (the transition matrix from B' to B) 1 3 2 1 1 [v ]B [v ]B P [v ]B 2 2 1 2 0 1 [v ]B v (1)( 1,2) (2)( 2,2) (3,2) 2 1 [v ]B v (1)(3,2) (0)( 4,2) (3,2) 0 Elementary Linear Algebra: Section 4.7, p.258 (c) 2 3 4 1 2 2 2 2 B' B 1 2 P 2 3 1 G.J.E. 1 0 1 2 0 1 2 3 -1 I P (the transition matrix from B to B') Check: 3 2 1 2 1 0 PP I2 2 1 2 3 0 1 1 Elementary Linear Algebra: Section 4.7, p.259 Ex 6: (Coordinate representation in P3(x)) (a) Find the coordinate matrix of p = 3x3-2x2+4 relative to the standard basis S = {1, x, x2, x3} in P3(x). (b) Find the coordinate matrix of p = 3x3-2x2+4 relative to the basis S = {1, 1+x, 1+ x2, 1+ x3} in P3(x). Sol: 4 0 (a) p (4)(1) (0)( x) (-2)( x 2 ) (3)( x 3 ) [ p]B 2 3 3 0 (b) p (3)(1) (0)(1 x) (-2)(1 x 2 ) (3)(1 x 3 ) [ p]B 2 3 Elementary Linear Algebra: Section 4.7, p.259 Ex: (Coordinate representation in M2x2) Find the coordinate matrix of x = 5 6 relative to 7 8 the standard basis in M2x2. B = 1 0, 0 1, 0 0, 0 0 0 0 1 0 0 1 0 0 Sol: 5 6 1 0 0 1 0 0 0 0 x 5 6 7 8 7 8 0 0 0 0 1 0 0 1 5 6 x B 7 8 Elementary Linear Algebra: Section 5.7, Addition Keywords in Section 4.7: coordinates of x relative to B:x相對於B的座標 coordinate matrix:座標矩陣 coordinate vector:座標向量 change of basis problem:基底變換問題 transition matrix from B' to B:從 B' 到 B的轉移矩陣