lec02

lec02 - Lecture #2 1) Linear Transformations The next topic...

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Unformatted text preview: Lecture #2 1) Linear Transformations The next topic is functions from one vector space to another that preserve their linear structure. Definition: If A and B are non-empty sets, a function f from A to B assigns a single point in B to every point in A. Such a function is written as f: A ~ B. The set A is called the domain of the function and B is called the codomain. A function is also called a mapping. Definition: If V and W are vector spaces, T: V ~W is linear or a linear transformation if T(av1 + bvz) = aT(v1) + bT(v2), for all numbers a and b and for all vectors v andv in V. 1 2 Example: Let T: R“ ~ R“ be defined by T(x) = Ax, where A is an MxN matrix. Then T(ax + by) = A(ax + by) = an + bAy = aT(x) + bT(y), so that T is linear. Remark: if V is a finite dimensional vector space and W is a vector space, then to define a linear transformation from V to W, it is sufficient to define T(v ), for every member v ofa n n basis v , ,v for V. 1 N The Matrix Representation of Linear Transformations Matrices can be used to represent linear transformations from one finite dimensional vector space to another. Let T: V ~W be linear, let v1, ,vN be a basis for V, and let w , , w be a basis for W. If v e V, there are unique numbers x , , x such that 1 M 1 N v : x v + + x v . Since T(v) e W, there are unique numbers y1, , yM such that 1 1 N N T(v) = y w + + y w . Since T(v) e W, for each n, there are unique numbers a , ,a 1 1 M M n 1n Mn such that T(v ) = a w + + a w . By the linearity of T and what has been said, n 1n 1 Mn M M N N N M M N wa=T(v)=T():xv)= ZxT(v)=ZxZaw=Z[Zax)w, m=1 m m n=1 " " n=1 " ” n=1 nm=1 m“ m m=1 n=1 m" ” m N sothaty = [a x , for all m. Lety = (y1, , yM) and x = (x1, , xN) and letA: (a ) be m_ “:1 mn n mn the MxN matrix with (m, n)th entry a , for all m and n. Then y = Ax. The matrix A represents mn T in that there is one and only one linear transformation T corresponding to A and one and only one matrix A corresponding to the linear transformation T, given the bases v1, ,vN for V and w , , w for W. The advantage of the matrix representation is that it translates calculations 1 M involving T into simple arithmetic computations. Suppose that in addition to the linear transformation T: V aW we are given a linear transformation S: W ~Q where Q is a vector space with basis q , ,q. Let B: (b ) be the 1 J jm JXM matrix representing 8 with respect to the bases w , , w and q , , q, forW and Q, 1 M 1 J respectively, so that, for all m, J S(W) = 2b q. m i=1 1m] Define SOT: V 40 to be the linear transformation SoT(v) = S(T(v) ). The transformation SoT is called the composition of T and S. Then M where c. = 2 b a . The JxN matrix C: (c ) represents SoT. In "1:1 jm mn jn Definition: If A is an MxN matrix and B is a JxM matrix, the product of B and A is the M JxN matrix C = BA with typical entry c. = X b a in m=1 jm mn m: 23 110 23—1 001: , 010 001 100 Remark: If the MxN matrix A represents the linear transformation T and the JxM matrix B represents the linear transformation 8, then the JxN matrix C = BA represents the linear transformation SoT. Remark: The order in which matrices are multiplied does not affect the product in that if A is an MxN matrix, B is a JxM matrix, and C is a KxJ matrix, then (CB)A = C(BA), so that the meaning of the product CBA is unambiguous. Remark: If A is an NXN matrix and I is the NxN identity matrix, then IA = A = A I. Remark: The NxN identity matrix I represents the identity function id :V ~ V from an V N-dimensional vector space V to itself, where id (v) = v, for all veV. V Invertible Matrices and Linear Transformations Definition: An NxN matrix A is invertible if there an NxN matrix A‘1 such that A“A= AA‘1 = I, where I is the NxN identity matrix. Lemma 2.1: If A and B are invertible NxN matrices, then AB is invertible and (AB)‘1 = B“A‘1. PrOOf? (B‘1A'1) (AB) = B~1("’F‘A) B = B"IB = 8‘18 2 1. (AB)( B“A“) = A(BB-‘)A-1= AIA-‘z AA—1= 1_ I Definition: A function f: V —) W is invertible, if there exists f": W ~ V such that M“1 = id and f'1 of = id , where id and id are the identity functions on W and V, respectively. W V W V That is, f(f"(w)) = w, for all w e W, and f“(f(v)) = v, for all v e V. Definition: A function f: V ~ W is onto, it for every w e W, there exists a v e V such that f(v) = w. Definition: A function f: V a W is one to one, if for every 1 e V and 7 e V such that _v_¢v,f(v)¢f(v). Remarks: 1) f: V ~ W is onto if and only if there exists a function 9: W « V such that f(g(w)) = w, for all w e W. 2) f: V a W is one to one if and only if there exists a function h: f(V) a V such that h(f(v)) = v, for all v e V, where f( V) = {f(v) | v e V} is the image or range of f. 3) f: W a V is invertible if and only if it is one to one and onto. Theorem 2.2: If T : V a W is an invertible linear transformation from the vector space V to the vector space W, then T‘1 is linear. Proof: Let w and w belong to W and let c 1 2 1 ). SinceTis linear, T(cv +cv) =cT(v 11 22 1 andc be numbers. Letv = T"(w) and 2 1 1 )=cw +cw. Hence 1 v =T“(w 2 1 2 2 ) + c2T(v 1 2 2 cT“(w) +cT“(w) =cv +cv =T“‘°T(cv +cv) =T“(cw +cw), 1 2 2 1 1 1122 122 1122 and so T‘1 is linear. I Theorem 2.3: Let T : V ~ V be a linear transformation and let v , , v be a basis for V. 1 N if A is the NxN matrix representing T with respect to the basis v , , v , then T is invertible 1 N if and only if A is invertible, and the matrix representation of T“ is A“. Proof: If T is invertible and B is the NxN matrix representing T“ with respect to the basis v , , v , then idV = T“ J, BA represents T“ 0T, and the NxN matrix I represents id . N V 1 Hence BA = I. Similarly AB = I, so that B : A“. If A is invertible, then A“ is the matrix representation with respect to the basis v , , vN of a linear transformation S: V ~ V. Since A“A= AA“ = I, if follows that 1 SOT = ToS = id . Hence 8 =T“, and so T is invertible. l V Proposition 2.4: Let T:V ~ W be an invertible linear transformation from the vector space V to the vector space W. The vectors v1, , v are a basis for V if and only if N T(v ), , T(vN) are a basis for W. 1 Proof: Please prove this theorem yourself as an exercise. Corollary 2.5: LetT:V a W be an invertible linear transformation from the finite dimensional vector space V to the vector space W. Then V and W have the same dimension. Theorem 2.6: If A is an NxN matrix, then the following are equivalent: 1) A is invertible, 2) there is an NxN matrix B such that BA = I, and 3) the system Ax = O has no non-zero solution. Proof: (1) implies (2): Let B = A“. (2) implies (3): If BA = I and Ax = 0, then 0 = BAx = Ix = x. (3) implies (1): By corollary 1.5, A is row equivalent to the NxN identity matrix 1. Each elementary row operation on A corresponds to left multiplication by an invertible matrix P. i now check this statement for each such operation. a) Multiplication of the kth row of A by the non-zero number 0 corresponds to replacing A by PA, where P = (p ) is the NxN matrix defined by mn 1,ifm=n¢k, p : c,ifm=n=k, rm 0, ifmatn. P—t (qmn) is the matrix defined by 1,ifm= n¢k, q mn c‘1,ifm=n=k, O, ifm¢n. b) Interchange of rows k and r of A corresponds to replacing A by PA, where P = (p > is the NxN matrix defined by 1, ifm=n,m¢k, m¢r, p Z 1, ifm=kandn= r, m" 1,ifm=randn=k, O,otherwise. P“=P. c) Replacement of the kth row of A by row k plus c times row r corresponds to replacing A by PA, where P = (p ) is the NxN matrix defined by mn 1, if m = n, pmn_ c, ifm=k,n= r, O, othenNise. P'1 = (qmn) is the matrix defined by 1,ifm=n, q mm —c, ifm: k,n= r, 0, otherwise. It follows that I: P P ..... ..PQA, where P is invertible, for all q. Let P = P P ..... ..P , so thatI q 1 2 1 2 Q = PA. Then P is invertible, since P‘1 = P: ..... ..P;‘. Then P" = P‘1PA = (P‘1P) A = IA= A, so that P = A‘1 and A is invertible. I Corollary 2.7: if A is an NxN matrix and BA = I, for some NxN matrix B, then B = A“. Proof: By the theorem, A is invertible. Therefore the equation BA = 1 implies that (BA)A“= A"‘, so that B = BI = B(AA“) = (BA)A“= A“. l The Range, Rank, Kernel, and Nullity of Linear Transformations For the rest of this lecture, let V and W by finite dimensional vector spaces. Definition: If f: A ~ B is a function, the image or range of f is {f( x) | x e A}. Definition: If T: V ~ W is a linear transformation, the null space or kernel of T is {v e V| T(v) = 0}. Theorem 2.8: If T: V ~ W is a linear transformation, then the range of T is a subspace of W and the kernel of T is a subspace of V. Proof: If T(v1) and T( v2) belong to the range ofT and c1 and 02 are numbers, then c T(v ) + c T(v ) belongs to the range of T, because, by the linearity of T, c T(v ) + c T(v ) 1 1 2 2 1 2 2 = T( c v + 0 v2) . Hence the range of T is a subspace of W. 1 1 2 1 If v and v belong to the kernel of T and c1 and c are numbers, then c v + c v belongs 1 2 2 1 1 2 2 to the kernel ofT, because T(c1v1+ cevz) = c1T(v ) + czT(v ) = C10 + 020 = O. I 1 2 Definition: lf T: V ~ W is a linear transformation, the rank of T is the dimension of the range of T and nullity of T is the dimension of the null space of T. Theorem 2.9: Let T: V « W be a linear transformation, then rank T + nullity T = dim V. Proof: Let N = dim V, let K be the nullity of T. I must show that rank T = N — K. Let v , , v be a basis for the null space of T. Letv , , v , v , , v be an extension of 1 K 1 K K+1 N v , , v to a basis for V. | show that the T( vK 1), , T(vN) is a basis for the range of T. The 1 k + vectors T(v1), , T(VN) span the range of T. Since T(v1) = T(v2) = = T(vK) = 0, it follows that T(v ), , T(v) span the range of T. lshow that T(v ), , T(v) are N K+1 N K+1 linearly independent, so that T(vK 1), , T(vN) is a basis for the range of T and hence rank T = N N — K. Suppose that Z c T(v ) = 0. Then the linearity of T implies that n=K+1 ” ” N N N T[ 2 cv )2 Z cT(v) = 0, so that 2 cv belongstothe kernelofT. Sincev , , v is n=K+1 n " n=K+1 n n n:K+1 n n 1 K N K a basis for the kernel of T, Z c v = 2 b v , for some numbers b1, , b . Therefore n=K+1 " " n=1 ” ” K K N 2 b v — Z c v = 0. Since v , , v form a basis for V, they are linearly independent and ":1 n n n=K+1 n n 1 N henceb=.....=b=0=c =....=C. I 1 K K+1 N Theorem 2.10: Let T: V ~ W be a linear transformation and suppose that A is the MxN matrix representing T with respect to the bases v , , v for V and w , , w for W. Then 1 N 1 M rank T equals the column rank of A and nullity T equals N minus the column rank of A, which equals N minus the row rank of A. Proof: Let SV: RN 4 V be the linear transformation such that S (9”) : v , for n = 1, , V n n N, where eN is the nth standard basis vector for B”. Let S : RM ~ W be the linear transformation W n such that SW(eM) = w , for m = 1, , M, where e'V' is the mth standard basis vector for R“. m m m The transformations S and SW are invertible. Let Q: S‘1 oToS . Then Q is a linear v W V transformation from R“ to. RM, and the matrix representation of Q with respect to the standard bases of RN and R” is A. The range of Q is the column space of A, so that rank 0 equals the column rank of A. The range ofT equals the range ofToS equals 31 (range Q), so that by corollary 2.5, rank Q = rank V W ToSv, which equals rank T. Putting all this together, we see that the column rank of A equals rank 0 equals rank T, as was to be proved. By theorem 2.9, nullity Q = N — rank Q. It was just proved that rank Q equals the column rank of A. By theorem 1.18, the column rank of A equals the row rank of A. Therefore nullity Q equals N minus the row rank of A. Since Q(x) = 0 if and only if ToSV( x) = O, the null space ofTequals SV( null space of Q), so that by corollary 2.5 nullity T = nullity O, which equals N minus the row rank of A. I If T: V ~ W is a linear transformation, then for any w e W, T"(w) = v + T"(0), where v is any vector in V such that T(v) = w and where v + T'1(O) = {v + 2| 2 e T"( O) }. In order to see that this is so, let 2 e T"(w). Then T(z — v) = T(z) — T(v) = w -— w = 0. Hence 2 = v + (z — v) e v + T"(O). Similarly any point in v + T“(O) belongs to T“( w). it is possible to visualize the meaning of the assertion T“(w) = v + T“( 0), by graphing a linear transformation T: R2 ~ R in the manner shown below. The function portrayed in the diagram may be thought of as a projection of R2 onto the vertical axis of R2 followed by a linear function from the vertical axis onto R. This way of visualizing a linear function may help you understand the implicit function theorem, when we get to it. T (0) = kernel of T Singular and Non-Singular Linear Transformations Definition: if T: V ~ W is a linear transformation, T is non—singular if the kernel of T equals {0}. mark: The linear transformation T is non-singular if and only if T is one to one, since T(v) = T(v2) if and only if T(v1 — v2) = o, 1 Lemma 2.11: if T: V «W is a linear transformation, then T is non-singular if and only if T(v ), , T(v ) are linearly independent whenever v1, , vN are linearly independent. 1 N Proof: Suppose that T is non-singular. Since T is linear, the equation 0 T(v ) + + c T(v ) = 0 implies that T(c1v1 + + chN) = O, which implies that 1 1 N N c v + + c v = 0, since T is non-singular; Since v , ,v are linearly independent, it 1 1 N 1 N follows that c = c = 0 and hence that T(v ), , T(v ) are linearly independent. 1 N 1 N Ilz Suppose that T carries independent vectors to independent vectors. Let v at 0, where v e V. Since the vector v by itself is linearly independent, TM is independent and hence T(v) at O. (The vector 0 is linearly dependent.) Therefore the kernel of T is {O}. I Theorem 2.12: If T: V ~ W is a linear transformation and dim V = dim W, then the following are equivalent. a) T is invertible. b) T is non-singular. c) T is onto. d) if v1, , vN is a basis for V, then T( v), , T(vN) is a basis for W. 9) There is a basis v1, , vN for V such that T( v), , T(vN) is a basis for W. Proof: (a) implies (b). This assertion is obvious. (b) implies (0). Suppose that T is non-singular. Let v1, , vN be a basis for V. By lemma 2.11, T(v1), , T(vN) are linearly independent. Since dim W = N, corollary 1.11 implies that T(v ), , T(vN) is a basis for W. lf w e W, then w = c1T(v1) + + c T(v ), 1 N N for some numbers 0, , 0. Since T is linear, c T(v) + + c T(v) = T(c v + + c v ), 1 N 1 1 N N 1 1 N N so that w = T( c v + + chN) and hence T is onto. 1 1 (0) implies (d). Let v1, , vN be a basis forV. Since these vectors span V and T is onto, the vectors T(v ) , , T(vN) span W. Since dim W = N, corollary 1.8 implies that 1 T(v ), , T(vN) form a basis for W. 1 (d) implies (e). This assertion is obvious. (e) implies (a). Suppose there is a basis v , , v for V such that T(v ), , T(v ) is 1 N 1 N a basis for W. Then rank T = dim W = dim V. Therefore by theorem 2.9, nullity T = 0, so that T is one to one. Since rank T = dim W, T is onto. Therefore T is invertible. I The Inner Product: Definition: The standard inner product or dot product on Ft” is the function x.y from N RNxRN to R defined by x.y = Z x y . ":1 n n In this definition, the symbol x stands for the Cartesian product, defined as follows. Dem: if A and B are sets, the Cartesian product of A and B is AxB = {(a, b) | a eAandb e B}. Definition: if x e R”, the length of x or the norm of x is ||x|l = 4x.x = ,lx: + + x2. m: 1) if a and b are numbers and x, y, and z belong to R“, then x.y = y.x and x.(ay + bz) = ax.y + bx.z. 2) If x e R“ and y 6 RN and if e is the angle between x and y, then cos 6 = ___L. llel llyll 3) lx.y| S ||x|| ||y||. This is called the Cauchy-Schwarz inequality and follows from (2). 4) x is perpendicular or orthogonal to y if and only if cos 9 = O, which is true if and only if x.y = 0. N 5) x.x = Z x2 2 O and x.x = 0 implies that x = 0. n=1 n 6) If y e R“, the function f(x) = y.x is a linear transformation from R” to R. Orthonormal Bases and Orthogonal Complements Definition: A set of vectors v , , v in R” is said to be orthogonal if v .v = 0, 1 M n m whenever n at m. Theorem 2.13: Orthogonal non-zero vectors are linearly independent. M Proof: Let v , , vM be orthogonal and suppose that Z c v : O, for some numbers 1 :1 m m M M n c, . Fork=1, M,O=v.0=v.[2cv J: Ecv.v =cv.v. Sincev.v >0, 1 M k km=1mm m=1ml<m kkk kk c = 0. Hence 0 = O, for all k, and so v , , v are linearly independent. I k k 1 M Definition; A basis v , , v for a subspace V of R” is said to be orthonormal if it is 1 M orthogonal and if v .v = 1, for all m. m m Lemma 2.14: If v , , v is an orthonormal basis for V, then for any v e V, 1 M M M M Proof:lfv=Ecv,then,foranyk,v.v= ch .v=2cv.v=cv.v=c. ["31 m m k "1:1 m k m=1 m m k k k k k 10 Theorem 2.15: Every vector space V that is a subspace of Ft” has an orthonormal basis. Proof; Let y , , y be a basis for V. I define an orthonormal basis v , , v by 1 M 1 M induction on the index k of the successive basis vectors. Let v = Y1 . Then v .v = 1 . 1 1 1 JV -y 1 1 Suppose that for k between 1 and M we are given v , , v such that v .v = 1, if n s k, 1 k n n v , , v are orthogonal, and v is a linear combination of y , , y, for n = 1, , k. Let 1 k n 1 n w = y — (y .v )v — —— (y .v )v. Thenw is a linear combination of y , , y . k+1 k+1 k+1 1 1 k+1 k k k+1 1 k+1 Also w i 0, for otherwise y1, , y would be linearly dependent, which would contradict k+1 k+1 the linear independence of y , , y . If n S k, then w .v = y .v — (y .v )v .v 1 M k+1 n k+1 n k+1 n n n W I =y .v —y .v =0. Letv = _____k_+__1____. Thenv .v =0, Ifnskandv .v :1. k+1 n k+1 n k+1 k+1 n k+1 k+1 w .w k+1 k+1 This completes the induction and hence the definition of v , , v . Since v , , v are 1 M 1 M independent and dim V = M, corollary 1.11 implies that these vectors form a basis for V.l The construction used in the previous proof is called the Gram-Schmidt orthogonalization process. Definition: If S is a subset of V, which is a subspace of Ft“, the orthogonal complement of SinVisSi={y EV | y.x=0, forallx ES}. Remark: Si is a subspace of V. Theorem 2.16: If W is a subspace of V and V is a subspace of R”, then dim W + dim W~L = dim V. Proof: Let dim V = M. By theorem 2.15, W has an orthonormal basis v , , v . If 1 K K = M = dim V, then dim W = dim V and so W = V. Hence Wi = {0} and dim Wi = O and so dim W + dim Wi = dim V = M. Suppose that K < M. Use the Gram-Schmidt orthogonalization process to extend v , , v to an orthonormal basis v , , vK, v , , vM of V. Then 1 K 1 K+1 v , , v belong to Wi. Since these vectors are non—zero and orthogonal, theorem 2.13 K+1 M implies that they are independent. I show that they span WL and so form a basis for Wi. Let M v eWi. By lemma 2.14,v= 2 (v.v)v. Sincev eWi, v.v =0, ifn=1, and so n=1 n n n M v = Z (v.v )v . This proves that v , ,vM form a basis for Wi and hence thatdim Wi = n = K+1 n n K+1 11 M—K=dimV—dimW. I Orthogonal Proiections Let W be a subspace of V, which is a subspace of RN. Definition: An orthogonal proiection n: V ~ W is a linear transformation from V to W such that v — TE(V) e Wi, for all v e V. That is, for all v e V, [v — n(v)].w = O, for all w e W. Example: Let V = R2. Let W = {(—t, t) | t is a number}. Then Wi = {(t, t) | t e R} and n(4, 14) = (—5, 5), since (4, 14) — (—5, 5) = (9, 9), which belongs to Wi. The example is illustrated in the diagram below. Theorem 2.17: If W is a subspace of V, which is a subspace of B“, there exists a unique orthogonal projection from V onto W. Proof: By theorem 2.15, W has an orthonormal basis v , , v . Use the Gram-Schmidt 1 K orthogonalization process to extend v1, , vK to an orthonormal basis v , , v , v , , v 1 K K+1 M for V. By an argument in the proof of theorem 2.16, v , , vM is a basis for Wi. If v e V, K+1 M K thenv: Z (v,v )v , by lemma 2.14. Letn(v) : [(v,v )v . Then m m m m=1 m=1 12 M v—n(v) = Z (v, v )v , which belongs to Wi. Since 7: is clearly linear, it is a linear m=K+1 m m projection and so a linear projection exists. In order to show that n is unique, let v e V and w e W be such that v — w e Wi. Then M M M M v—w= Z (v—w,v)v = Z (v,v)v — Z (w,v)v : Z (v,v)v,wherethe m=K+1 m m m=K+1 m m m=K+1 m m m=K+1 m m last equation follows because w belongs to W and v , , v belong to Wi. Therefore K+1 M M M K w=v—(v—-w)=Z(v,v)v— E(v,v)v=2(v,v)v =Tt(V). I m=1 m m m=K+1 m m m=1 m m 13 ...
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This note was uploaded on 02/22/2012 for the course EE 441 taught by Professor Neely during the Spring '08 term at USC.

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