# Review: Vector space

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1 Math 2F Linear Algebra Lecture 13 1 Basis and dimensions Slide 1 Review: Subspace of a vector space. (Sec. 4.1) Linear combinations, l.d., l.i. vectors. (Sec. 4.3) Dimension and Base of a vector space. (Sec. 4.4) Review: Vector space Slide 2 A vector space is a set of elements of any kind, called vectors, on which certain operations, called addition and multiplication by numbers, can be performed. The main idea in the definition of vector space is to do not specify the nature of the elements nor do we tell how the operations are to be performed on them. Instead, we require that the operations have certain properties, which we take as axioms of a vector space. Examples include spaces of arrows, matrices, functions.

2 Math 2F Linear Algebra Lecture 13 2 Review: Subspace Slide 3 Definition 1 (Subspace) A subspace W of a vector space V is a subset of V that is closed under the addition and scalar multiplication operations on V. That is, W V, and for all u, v W and a IR holds that u + v W, au W. Slide 4 Examples The set W IR 3 given by W = {x IR 3 : x = (x 1, x 2, ), is a subspace of IR 3. The set Ŵ IR3 given by Ŵ = {x IR 3 : x = (x 1, x 2, 1), in contrast is not a subspace of IR 3. x 1, x 2 IR}, x 1, x 2 IR},

3 Math 2F Linear Algebra Lecture 13 3 Examples The set W = {(x 1, x 2 ) IR 2 : x 1 } is not a subspace of IR 2. Slide 5 The segment W {x IR : 1 x 1} is not a subspace of IR. The line W = {x IR 3 : x = (1, 2, 3)t} is a subspace or IR 3. The line W = {x IR 3 : x = (1, 2, ) + (1, 2, 3)t} is not a subspace or IR 3. Linear combinations Definition 2 (Linear combination) Let V be a vector space, v 1,, v k V be arbitrary vectors, and a 1,, a k IR be arbitrary scalars. We call a linear combination of v 1,, v k the vector Slide 6 k a i v i = a 1 v a k v k. Linear combinations give rise to the concept of span of a set of vectors. Definition 3 (Span) The span of v 1,, v k V is the set of all linear combinations of these vectors, that is, { k } Span{v 1,, v k } = a i v i : a i IR, i k.

4 Math 2F Linear Algebra Lecture 13 4 Notice that every linear combination of v 1,, v k also belongs to V, so Span{v 1,, v k } V. One can also check that it is a subspace. Claim 1 Let V be a vector space and v 1,, v k V. Then, Span{v 1,, v k } V is a subspace. Slide 7 Different sets may span the same subspace. Example: The space IR 2 is spanned by {e 1, e 2 }, and also by {e 1, e 2, e 1 + 2e 2 }, and also by {, e 1 + 2e 2, e 1, e 1 + e 2 }. The space P 2 is spanned by {1, t, t 2 }, and also by {1, t, t 2, (t 2 1), (t + 1) 2 }. It is then useful to remove redundant vectors from linear combinations. Definition 4 (l.d.) Let V be a vector space. The vectors v 1,, v k V are linearly dependent, l.d., if there exist scalars a 1,, a k R with at least one of them nonzero, such that Slide 8 k a i v i =. Definition 5 (l.i.) Let V be a vector space. The vectors v 1,, v k V are linearly independent, l.i., if they are not l.d., that is, the only choice of scalars a 1,, a k R for which the equation k a i v i = holds is a i = for all 1 i k. Linearly independent set of vectors contain no redundant vectors, in the sense of linear combinations.

5 Math 2F Linear Algebra Lecture 13 5 Examples: If one element in S V is a scalar multiple of another, S is l.d.. If S, then S is l.d.. Slide 9 Let V be the space of continuous functions on [, 2π]. The set of vectors v 1 = cos 2 (t), v 2 = sin 2 (t), v 3 = 1 is l.d.. Indeed, Pythagoras theorem says that v 1 + v 2 v 3 =. Consider the space P 2. The set {1, t, t 2 } is l.i.. Indeed, the equation a + a 1 t + a 2 t 2 =, implies that all coefficient vanishes. Evaluate the equation at t =, then a =. So we have a 1 t + a 2 t 2 =. If the function vanishes, its derivative also vanishes. So a 1 + a 2 t =. Evaluate this at t =, then a 1 =. Repeat the procedure one more time, then a 2 =. Dimensions and base Definition 6 (Finite dimension and base) A vector space V has finite dimension is there exists a maximal set of l.i. vectors {v 1,, v n }. Slide 1 The number n is called the dimension of V, and we denote by n = dim V. The l.i. vectors {v 1,, v n } are called a base of V. If there is no maximal set of l.i. vectors, then V is called infinite dimensional. That is, the set {v 1,, v n } is maximal if it is l.i. but the set {v 1,, v n, v n+1 } is l.d. for all v n+1 V.

6 Math 2F Linear Algebra Lecture 13 6 Examples: The set {e 1, e 2 } IR 2 is a basis. Another basis is {e 1 + e 2, e 1 e 2 }. The set {e 1, e 2, e 1 + e 2 } is not a basis. The set {e 1, 2e 1 } is not a basis. Notice that a basis of a vector space is not unique. Slide 11 The space IR n is finite dimensional, of dimension n, because the vectors {e 1,, e n } are l.i., and any set of n + 1 vectors in IR n is l.d.. The space P, polynomials on [, 1] is infinite dimensional. The infinite set {1, t, t 2, t 3, } is l.i.. Theorem 1 A set of vectors E V is a base if the vectors in E are l.i. and span V. Components or coordinates Slide 12 Theorem 2 Let V be a finite dimensional vector space with n = dim V. If {u 1,, u n } is a basis of V, then each vector v V has a unique decomposition v = n a i u i. The n scalars a i are called components or coordinates of v with respect to this basis.

7 Math 2F Linear Algebra Lecture 14 7 Slide 13 Exercises Let u 1 = (1, 1) and u 2 = (1, 1), be a basis of IR 2. Find the components of x = (1, 2) in that basis. The vectors u 1, u 2 form a basis so there exists constants c 1, c 2 such that x = c 1 u 1 + c 2 u 2. That is, so c 1 = 3/2 and c 2 = 1/2, and then x = 3/2u 1 1/2u 2. c 1 c 2 = 1. 2 Then one has to solve the augmented matrix , Proof: The set {u 1,, u n } is a basis so there exists scalars a i, for 1 i n such that the following decomposition holds, v = n a i u i. This decomposition is unique. Because, if there is another decomposition v = n b i u i. then the difference has the form n (a i b i )u i =. Because the vectors {u 1,, u n } are l.i. this implies that a i = b i for all i n.

8 Math 2F Linear Algebra Lecture 14 8 Null and Column spaces Null space and range. (Sec. 4.2) Slide 14 Examples. Linear transformations. Null space and range for L.T.. Fundamental theorem of algebra. (Sec. 4.6) Null and range spaces: Matrix version Slide 15 Definition 7 Let A be an m n. The null space of A, denoted as N(A) V, is the set of all elements of V solution of Av =, that is, N(A) = {v V : Av = }. Definition 8 Let A be an m n. The column space of A, denoted as Col(A), is the span of the columns of A. Its dimension is called the rank of A, that is rank(a) = dim Col(A).

9 Math 2F Linear Algebra Lecture 14 9 Exercises Find the null space of A = Slide 16 The null space are all x IR 3 solutions of Ax =. One can check that these solutions have the form 3 e = 1 x 3. 1 Therefore, N(A) = Span{( 3, 1, 1)}. Exercises (Problem 2, Sec. 4.2) Find k 1 and k 2 such that N(A) is a subspace of IR k1, and Col(A) is a subspace of IR k2, with A = [1, 3, 9,, 5]. The solutions x of the equation Ax = form a hyperplane in IR 5, given by the equation Slide 17 x 1 3x 2 + 9x 3 5x 5 =. So, the solutions can be written as x = x x 3 + x x 5. 1 N(A) = Span{(3, 1,,, ), ( 9,, 1,, ), (,,, 1, ), (5,,,, 1)} IR 4. Col(A) IR.

10 Math 2F Linear Algebra Lecture 14 1 linear transformations Definition 9 Let V, W be vector spaces. The function T : V W is called a linear transformation if it has the following properties: T (v 1 + v 2 ) = T (v 1 ) + T (v 2 ) for all v 1, v 2 V ; T (av) = at (v), for all v V and a IR. Slide 18 These properties means that T preserves addition and multiplication by numbers. The two properties combine in one formula T (av + bu) = at (v) + bt (u), or equivalently, ( k ) k T a i v i = a i T (v i ). Examples: The identity transformation, that is T : V V, given by T (v) = v. Slide 19 A stretching by a IR, that is, T : V V, given by T (v) = av. A projection, that is T : IR 3 IR 3 given by T (x 1 e 1 + x 2 e 2 + x 3 e 3 ) = x 1 e 1 + x 2 e 2. The derivative. Let V be the vector space of differentiable functions on (, 1), and let W be the vector space of continuous function on (, 1). Then, T : V W given by T (f) = f is a linear transformation.

11 Math 2F Linear Algebra Lecture Null space and range of T Slide 2 Definition 1 (Null space of T ) Let T : V W be a linear transformation. The null space of T, denoted as N(T ) V, is the set of all elements of V that T maps onto the W. That is, N(T ) = {v V : T (v) = }. Definition 11 (Range of T ) Let T : V W be a linear transformation. The range of T, denoted as T (V ) W, is the set of all elements of W of the form w = T (v) for some v V. Theorem 3 (N(T ) subspace of V ) Let T : V W be a linear transformation. The null space of T is a subspace of V. Slide 21 Theorem 4 (T (V ) subspace of W ) Let T : V W be a linear transformation. The range of T, denoted by T (V ) W, is a subspace of W. The nullity is the dimension of N(T ). The rank is the dimension of T (V ). Theorem 5 Let T be the linear transformation associated to A. Then N(T ) = N(A) and T (V ) = Col(A).

12 Math 2F Linear Algebra Lecture Proof of Theorem 3: Assume that u and v N(T ). Then, au + bv N(T ) because T (au + bv) = at (u) + bt (v) = a + b =. Proof of Theorem 4: Given two vectors T (u) and T (v) in the range of T, then at (u) + bt (v) also belongs to the range of T because T is linear, that is, at (u) + bt (v) = T (au + bv). Examples: The identity transformation. The null space is N(I) = {}. The range space is T (V ) = V. Slide 22 A stretching by a IR. In the case a the null space is N(T ) = {} and T (V ) = V. In the case a = one has N(T ) = V and T (V ) = {}. A projection, T : IR 3 IR 3 given by T (x 1 e 1 + x 2 e 2 + x 3 e 3 ) = x 1 e 1 + x 2 e 2. Then N(T ) = {x 3 e 3 }. The range space is the plane spanned by {e 1, e 2 }. The derivative, given by T (f) = f. Then N(T ) are the constant functions. The range of T are the continuous functions on (, 1).

13 Math 2F Linear Algebra Lecture Fundamental theorem of algebra Slide 23 Theorem 6 (Nullity plus rank) Let T : V W be a linear transformation, and V be finite dimensional. Then N(T ) and T (V ) are finite dimensional and the following relation holds, dim N(T ) + dim T (V ) = dim V. Theorem 7 (Nullity plus rank: Matrix form) Let A be an m n matrix with m, n finite Then, the following relation holds, dim N(A) + rank(a) = n. Proof of Theorem 6: Let n = dim V and let S = {e 1,, e k } be a basis for N(T ), so we say that the nullity is some number k. Because N(T ) is contained in V one knows that k n. Let us add l.i. vectors to S to complete a basis of V, say, {e 1,, e k, e k+1,, e k+r } for some number r such that k + r = n. We shall prove that {T (e k+1 ),, T (e k+r )} is a basis for T (V ), and then r = dim T (V ). This relation proves Theorem 6. The elements {T (e k+1 ),, T (e k+r )} are a basis of T (V ) because they span T (V ) and they are l.i.. They span T (V ) because for every w T (V ) we know that there exists v V such that w = T (v). If we write v = n v ie i, then we have ( n ) n k+r w = T e i = a i T (e i ) = a i T (e i ), i= i= i=k+1 then the {T (e k+1 ),, T (e k+r )} span T (V ). These vectors are also l.i., by the following argument. Suppose there are scalars c k+1,, c k+r such that Then, this implies T k+r i=k+1 ( k+r i=k+1 c i T (e i ) =. c i e i ) =, so the vector u = k+r i=k+1 c it (e i ) belongs to N(T ). But if u belongs to N(T ), then it must be written also as a linear combination of the elements of the base of N(T ), namely, the

14 Math 2F Linear Algebra Lecture vectors e 1,, e k, so there exists constants c 1,, c k such that u = k c i e i. Then, we can construct the linear combination = u u = k c i e i k+r i=k+1 Because the set {e 1, e k+r } is a basis o V we have that all the c i with 1 i k + r must vanish. Then, the vectors {T (e k+1 ),, T (e k+r )} are l.i.. Therefore they are a basis of T (V ), and then the dimension of dim T (V ) = r. This proves the Theorem. c i e i. The matrix of a linear transformation Slide 24 Theorem 8 Let T : V W be a linear transformation, where V and W are finite dimensional vector spaces with dim V = n and dim W = m. Let {v 1,, v n } be a basis of V, and {w 1,, w m } be a basis of W. Then, T has associated a matrix A = [a i ], 1 i n, given by m T (v i ) = a i = a ij w j. j=1 Idea of the proof: every vector in W is a linear combinations of the basis vectors w j for j m. In particular all the n vectors T (e i ), i n. therefore, there must exist the m n set of numbers a ij given above. These coefficients define the matrix A associated to T.

15 Math 2F Linear Algebra Lecture Example: In the case T : IR 2 IR 3, one chooses the basis vectors v 1 = (1, ), v 2 = (, 1), for IR 2, and the basis vectors w 1 = (1,, ), w 2 = (, 1, ), w 3 = (,, 1) for IR 3. Then, a linear transformation Slide 25 T (x 1, x 2 ) = (x 1 + 3x 2, x 1 + x 2, x 2 ), has associated the matrix A = [a 1, a 2 ] given by 1 3 a 1 = 1, a 2 = 1. 1 Invertible transformations Slide 26 One-to-one and onto. Isomorphisms (one-to-one and onto). Invertible transformations.

16 Math 2F Linear Algebra Lecture One-to-one and onto Definition 12 (one-to-one onto) Let T : V W be a linear transformation. Slide 27 T is one-to-one if v 1 v 2 implies T (v 1 ) T (v 2 ); T is onto if for all w W there exists v V such that T (v) = w. Theorem 9 Let T : V W be a linear transformation. T is one-to-one N(T ) = {}; T is onto T (V ) = W. Isomorphism Definition 13 (Isomorphism) The linear transformation T : V W is an isomorphism if T is one-to-one and onto. Isomorphic spaces are essentially the same. There is a one to one correspondence between elements from one space and of the other. Slide 28 Example: P 2 is isomorphic to R 3. A basis of P 2 is {1, t, t 2 } A general element v P has the form v = a + a 1 t + a 2 t 2. The linear transformation T : P 2 R 3 given by T (a + a 1 t + a 2 t 2 ) = a e + a 1 e 2 + a 2 e 3 IR 3 is an isomorphism. Theorem 1 Let T : V W be an isomorphism, with V and W of finite dimension. Then dim V = dim W.

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