# Chapter 2 Linear Transformations

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1 Chapter 2 Linear Transformations Linear Transformations Loosely speaking, a linear transformation is a function from one vector space to another that preserves the vector space operations. Let us be more precise. Definition Let and be vector spaces over a field. A function is a linear transformation if for all scalars and vectors,. A linear transformation is called a linear operator on. The set of all linear transformations from to is denoted by and the set of all linear operators on is denoted by. We should mention that some authors use the term linear operator for any linear transformation from to. Definition The following terms are also employed: 1) homomorphism for linear transformation 2) endomorphism for linear operator 3) monomorphism (or embedding) for injective linear transformation 4) epimorphism for surjective linear transformation 5) isomorphism for bijective linear transformation. 6) automorphism for bijective linear operator. Example 2.1 1) The derivative is a linear operator on the vector space of all infinitely differentiable functions on.

2 56 Advanced Linear Algebra 2) The integral operator defined by is a linear operator on. 3) Let be an matrix over. The function defined by, where all vectors are written as column vectors, is a linear transformation from to. This function is just multiplication by. 4) The coordinate map of an -dimensional vector space is a linear transformation from to. The set is a vector space in its own right and has the structure of an algebra, as defined in Chapter 0. Theorem 2.1 1) The set is a vector space under ordinary addition of functions and scalar multiplication of functions by elements of. 2) If and then the composition is in. 3) If is bijective then. 4) The vector space is an algebra, where multiplication is composition of functions. The identity map is the multiplicative identity and the zero map is the additive identity. Proof. We prove only part 3). Let be a bijective linear transformation. Then is a well-defined function and since any two vectors and in have the form and, we have which shows that is linear. One of the easiest ways to define a linear transformation is to give its values on a basis. The following theorem says that we may assign these values arbitrarily and obtain a unique linear transformation by linear extension to the entire domain. Theorem 2.2 Let and be vector spaces and let be a basis for. Then we can define a linear transformation by specifying the values of arbitrarily for all and extending the domain of to using linearity, that is,

3 Linear Transformations 57 This process uniquely defines a linear transformation, that is, if satisfy for all then. Proof. The crucial point is that the extension by linearity is well-defined, since each vector in has a unique representation as a linear combination of a finite number of vectors in. We leave the details to the reader. Note that if and if is a subspace of, then the restriction of to is a linear transformation from to. The Kernel and Image of a Linear Transformation There are two very important vector spaces associated with a linear transformation from to. Definition Let. The subspace ker is called the kernel of and the subspace im is called the image of. The dimension of ker is called the nullity of and is denoted by null. The dimension of im is called the rank of and is denoted by rk. It is routine to show that ker is a subspace of and im is a subspace of. Moreover, we have the following. Theorem 2.3 Let. Then 1) is surjective if and only if im 2) is injective if and only if ker Proof. The first statement is merely a restatement of the definition of surjectivity. To see the validity of the second statement, observe that ker Hence, if ker then, which shows that is injective. Conversely, if is injective and ker then and so. This shows that ker. Isomorphisms Definition A bijective linear transformation is called an isomorphism from to. When an isomorphism from to exists, we say that and are isomorphic and write. Example 2.2 Let dim. For any ordered basis of, the coordinate map that sends each vector to its coordinate matrix

4 58 Advanced Linear Algebra is an isomorphism. Hence, any -dimensional vector space over isomorphic to. is Isomorphic vector spaces share many properties, as the next theorem shows. If and we write Theorem 2.4 Let be an isomorphism. Let. Then 1) spans if and only if spans. 2) is linearly independent in if and only if is linearly independent in. 3) is a basis for if and only if is a basis for. An isomorphism can be characterized as a linear transformation maps a basis for to a basis for. that Theorem 2.5 A linear transformation is an isomorphism if and only if there is a basis of for which is a basis of. In this case, maps any basis of to a basis of. The following theorem says that, up to isomorphism, there is only one vector space of any given dimension. Theorem 2.6 Let and be vector spaces over. Then if and only if dim dim. In Example 2.2, we saw that any -dimensional vector space is isomorphic to. Now suppose that is a set of cardinality and let be the vector space of all functions from to with finite support. We leave it to the reader to show that the functions defined for all, by if if form a basis for, called the standard basis. Hence, dim. It follows that for any cardinal number, there is a vector space of dimension. Also, any vector space of dimension is isomorphic to. Theorem 2.7 If is a natural number then any -dimensional vector space over is isomorphic to. If is any cardinal number and if is a set of cardinality then any -dimensional vector space over is isomorphic to the vector space of all functions from to with finite support.

5 Linear Transformations 59 The Rank Plus Nullity Theorem Let. Since any subspace of has a complement, we can write kerker where ker is a complement of ker in. It follows that Now, the restriction of to ker dim dimker dimker ker is injective, since ker ker ker Also, im im. For the reverse inclusion, if im then since for ker and ker, we have im Thus im im. It follows that ker im From this, we deduce the following theorem. Theorem 2.8 Let. 1) Any complement of ker is isomorphic to im 2) ( The rank plus nullity theorem) dimker dimim dim or, in other notation, rk null dim Theorem 2.8 has an important corollary. Corollary 2.9 Let, where dimdim. Then is injective if and only if it is surjective. Note that this result fails if the vector spaces are not finite-dimensional. Linear Transformations from to Recall that for any matrix over the multiplication map

6 60 Advanced Linear Algebra is a linear transformation. In fact, any linear transformation has this form, that is, is just multiplication by a matrix, for we have and so where Theorem ) If is an matrix over then. 2) If then where The matrix is called the matrix of. Example 2.3 Consider the linear transformation Then we have, in column form defined by and so the standard matrix of is If then since the image of is the column space of, we have dimker rk dim This gives the following useful result. Theorem 2.11 Let be an matrix over. 1) is injective if and only if rk n. 2) is surjective if and only if rk m. Change of Basis Matrices Suppose that and are ordered bases for a vector space. It is natural to ask how the coordinate matrices and are related. The map that takes to is and is called the change of basis operator (or change of coordinates operator). Since is an operator on, it has the form where

7 Linear Transformations 61 We denote by, and call it the change of basis matrix from to. Theorem 2.12 Let and be ordered bases for a vector space. Then the change of basis operator is an automorphism of, whose standard matrix is, Hence and., Consider the equation or equivalently, Then given any two of (an invertible matrix) (an ordered basis for ) and (an order basis for ), the third component is uniquely determined by this equation. This is clear if and are given or if and are given. If and are given then there is a unique for which and so there is a unique for which. Theorem 2.13 If we are given any two of the following: 1) An invertible matrix. 2) An ordered basis for. 3) An ordered basis for. then the third is uniquely determined by the equation The Matrix of a Linear Transformation Let be a linear transformation, where dim and dim and let be an ordered basis for and an ordered basis for. Then the map is a representation of as a linear transformation from to, in the sense

8 62 Advanced Linear Algebra that knowing (along with and, of course) is equivalent to knowing. Of course, this representation depends on the choice of ordered bases and. Since is a linear transformation from to, it is just multiplication by an matrix, that is Indeed, since, we get the columns of as follows: Theorem 2.14 Let and let and be ordered bases for and, respectively. Then can be represented with respect to and as matrix multiplication, that is where,, is called the matrix of with respect to the bases and. When and, we denote by and so, Example 2.4 Let be the derivative operator, defined on the vector space of all polynomials of degree at most. Let. Then, and so Hence, for example, if then and so. The following result shows that we may work equally well with linear transformations or with the matrices that represent them (with respect to fixed

9 Linear Transformations 63 ordered bases and ). This applies not only to addition and scalar multiplication, but also to matrix multiplication. Theorem 2.15 Let and be vector spaces over, with ordered bases and, respectively. 1) The map defined by, is an isomorphism and so. 2) If and and if, and are ordered bases for, and, respectively then,,, Thus, the matrix of the product (composition) is the product of the matrices of and. In fact, this is the primary motivation for the definition of matrix multiplication. Proof. To see that is linear, observe that for all and since is a standard basis vector, we conclude that and so is linear. If, we define by the condition, whence and is surjective. Since dim dim, the map is an isomorphism. To prove part 2), we have, Change of Bases for Linear Transformations Since the matrix, that represents depends on the ordered bases and, it is natural to wonder how to choose these bases in order to make this matrix as simple as possible. For instance, can we always choose the bases so that is represented by a diagonal matrix? As we will see in Chapter 7, the answer to this question is no. In that chapter, we will take up the general question of how best to represent a linear operator by a matrix. For now, let us take the first step and describe the relationship between the matrices and of with respect to two different pairs and of ordered bases. Multiplication by sends to

10 64 Advanced Linear Algebra. This can be reproduced by first switching from to, then applying and finally switching from to, that is,,,, Theorem 2.16 Let, and let and be pairs of ordered bases of and, respectively. Then (2.1) When is a linear operator on, it is generally more convenient to represent by matrices of the form, where the ordered bases used to represent vectors in the domain and image are the same. When, Theorem 2.16 takes the following important form. Corollary 2.17 Let and let and be ordered bases for. Then the matrix of with respect to can be expressed in terms of the matrix of with respect to as follows Equivalence of Matrices (2.2) Since the change of basis matrices are precisely the invertible matrices, (2.1) has the form where and are invertible matrices. This motivates the following definition. Definition Two matrices and are equivalent if there exist invertible matrices and for which We remarked in Chapter 0 that is equivalent to if and only if can be obtained from by a series of elementary row and column operations. Performing the row operations is equivalent to multiplying the matrix on the left by and performing the column operations is equivalent to multiplying on the right by. In terms of (2.1), we see that performing row operations (premultiplying by ) is equivalent to changing the basis used to represent vectors in the image and performing column operations (postmultiplying by ) is equivalent to changing the basis used to represent vectors in the domain.

11 Linear Transformations 65 According to Theorem 2.16, if and are matrices that represent with respect to possibly different ordered bases then and are equivalent. The converse of this also holds. Theorem 2.18 Let and be vector spaces with dim and dim. Then two matrices and are equivalent if and only if they represent the same linear transformation, but possibly with respect to different ordered bases. In this case, and represent exactly the same set of linear transformations in. Proof. If and represent, that is, if and,, for ordered bases and then Theorem 2.16 shows that and are equivalent. Now suppose that and are equivalent, say where and are invertible. Suppose also that represents a linear transformation for some ordered bases and, that is, Theorem 2.13 implies that there is a unique ordered basis for for which and a unique ordered basis for for which. Hence Hence, also represents. By symmetry, we see that and represent the same set of linear transformations. This completes the proof. We remarked in Example 0.3 that every matrix is equivalent to exactly one matrix of the block form block Hence, the set of these matrices is a set of canonical forms for equivalence. Moreover, the rank is a complete invariant for equivalence. In other words, two matrices are equivalent if and only if they have the same rank. Similarity of Matrices When a linear operator is represented by a matrix of the form, equation (2.2) has the form where is an invertible matrix. This motivates the following definition.

12 66 Advanced Linear Algebra Definition Two matrices and are similar if there exists an invertible matrix for which The equivalence classes associated with similarity are called similarity classes. The analog of Theorem 2.18 for square matrices is the following. Theorem 2.19 Let be a vector space of dimension. Then two matrices and are similar if and only if they represent the same linear operator, but possibly with respect to different ordered bases. In this case, and represent exactly the same set of linear operators in. Proof. If and represent, that is, if and for ordered bases and then Corollary 2.17 shows that and are similar. Now suppose that and are similar, say Suppose also that represents a linear operator for some ordered basis, that is, Theorem 2.13 implies that there is a unique ordered basis for for which. Hence Hence, also represents. By symmetry, we see that and represent the same set of linear operators. This completes the proof. We will devote much effort in Chapter 7 to finding a canonical form for similarity. Similarity of Operators We can also define similarity of operators. Definition Two linear operators are similar if there exists an automorphism for which The equivalence classes associated with similarity are called similarity classes.

13 Linear Transformations 67 The analog of Theorem 2.19 in this case is the following. Theorem 2.20 Let be a vector space of dimension. Then two linear operators and on are similar if and only if there is a matrix that represents both operators (but with respect to possibly different ordered bases). In this case, and are represented by exactly the same set of matrices in. Proof. If and are represented by, that is, if for ordered bases and then Let be the automorphism of defined by, where and. Then and so from which it follows that and are similar. Conversely, suppose that and are similar, say Suppose also that is represented by the matrix, that is, for some ordered basis. Then If we set then is an ordered basis for and Hence It follows that and so also represents. By symmetry, we see that and are represented by the same set of matrices. This completes the proof.

14 68 Advanced Linear Algebra Invariant Subspaces and Reducing Pairs The restriction of a linear operator to a subspace of is not necessarily a linear operator on. This prompts the following definition. Definition Let. A subspace of is said to be invariant under or - invariant if, that is, if for all. Put another way, is invariant under if the restriction is a linear operator on. If then the fact that is -invariant does not imply that the complement is also -invariant. (The reader may wish to supply a simple example with.) Definition Let. If and if both and are -invariant, we say that the pair reduces. A reducing pair can be used to decompose a linear operator into a direct sum as follows. Definition Let. If reduces we write and call the direct sum of and. Thus, the expression means that there exist subspaces and of for which reduces and and The concept of the direct sum of linear operators will play a key role in the study of the structure of a linear operator. Topological Vector Spaces This section is for readers with some familiarity with point-set topology. The standard topology on is the topology for which the set of open rectangles 's are open intervals in is a basis (in the sense of topology), that is, a subset of is open if and only if it is a union of sets in. The standard topology is the topology induced by the Euclidean metric on.

15 Linear Transformations 69 The standard topology on has the property that the addition function and the scalar multiplication function are continuous. As such, is a topological vector space. Also, any linear functional is a continuous map. More generally, any real vector space endowed with a topology is called a topological vector space if the operations of addition and scalar multiplication are continuous under. Let be a real vector space of dimension and fix an ordered basis for. Consider the coordinate map and its inverse We claim that there is precisely one topology on for which becomes a topological vector space and for which all linear functionals are continuous. This is called the natural topology on. In fact, the natural topology is the topology for which (and therefore also ) is a homeomorphism, for any basis. (Recall that a homeomorphism is a bijective map that is continuous and has a continuous inverse.) Once this has been established, it will follow that the open sets in are precisely the images of the open sets in under the map. A basis for the natural topology is given by 's are open intervals in 's are open intervals in First, we show that if is a topological vector space under a topology then is continuous. Since where is defined by it is sufficient to show that these maps are continuous. (The sum of continuous maps is continuous.) Let be an open set in. Then is open in. We need to show that the set

16 70 Advanced Linear Algebra is open in, so let. Thus,. It follows that, which is open, and so there is an open interval and an open set of for which Then the open set, where the factor is in the th position, has the property that. Thus and so is open. Hence,, and therefore also, is continuous. Next we show that if every linear functional on is continuous under a topology on then the coordinate map is continuous. If denote by the th coordinate of. The map defined by is a linear functional and so is continuous by assumption. Hence, for any open interval the set is open. Now, if are open intervals in then is open. Thus, is continuous. Thus, if a topology has the property that is a topological vector space and every linear functional is continuous, then and are homeomorphisms. This means that, if it exists, must be unique. It remains to prove that the topology on that makes a homeomorphism has the property that is a topological vector space under and that any linear functional on continuous. As to addition, the maps and are homeomorphisms and the map is continuous and so the map, being equal to, is also continuous. As to scalar multiplication, the maps and are homeomorphisms and the map is continuous and so the map, being equal to, is also continuous. Now let be a linear functional. Since is continuous if and only if is continuous, we can confine attention to. In this case, if is the

17 Linear Transformations 71 standard basis for have and, then for any we Now, if then and so, which implies that is continuous. According to the Riesz representation theorem and the Cauchy Schwarz inequality, we have Hence, implies and so by linearity, implies and so is continuous. Theorem 2.21 Let be a real vector space of dimension. There is a unique topology on, called the natural topology for which is a topological vector space and for which all linear functionals on are continuous. This topology is determined by the fact that the coordinate map is a homeomorphism. Linear Operators on A linear operator on a real vector space can be extended to a linear operator on the complexification by defining Here are the basic properties of this complexification of. Theorem 2.22 If then 1), 2) 3) 4). Let us recall that for any ordered basis for and any vector we have cpx Now, if is a basis for, then the th column of is cpx cpx which is the th column of the coordinate matrix of cpx. Thus we have the following theorem. with respect to the basis

18 72 Advanced Linear Algebra Theorem 2.23 Let where is a real vector space. The matrix of with respect to the basis cpx is equal to the matrix of with respect to the basis cpx Hence, if a real matrix represents a linear operator on then also represents the complexification of on. Exercises 1. Let have rank. Prove that there are matrices and, both of rank, for which. Prove that has rank if and only if it has the form where and are row matrices. 2. Prove Corollary 2.9 and find an example to show that the corollary does not hold without the finiteness condition. 3. Let. Prove that is an isomorphism if and only if it carries a basis for to a basis for. 4. If and we define the external direct sum by Show that is a linear transformation. 5. Let. Prove that. Thus, internal and external direct sums are equivalent up to isomorphism. 6. Let and consider the external direct sum. Define a map by. Show that is linear. What is the kernel of? When is an isomorphism? 7. Let be a subset of. A subspace of is -invariant if is - invariant for every. Also, is -irreducible if the only -invariant subspaces of are and. Prove the following form of Schur's lemma. Suppose that and and is -irreducible and is -irreducible. Let satisfy, that is, for any there is a such that. Prove that or is an isomorphism. 8. Let where dim. If rk rk show that im ker. 9. Let, and. Show that rk minrkrk 10. Let and. Show that null nullnull 11. Let where is invertible. Show that rk rk rk

19 Linear Transformations Let. Show that rk rkrk 13. Let be a subspace of. Show that there is a for which ker. Show also that there exists a for which im. 14. Suppose that. a) Show that for some if and only if im im. b) Show that for some if and only if ker ker. 15. Let. Define linear operators on by for. These are referred to as projection operators. Show that 1) 2), where is the identity map on. 3) for where is the zero map. 4) imim 16. Let dim and suppose that satisfies. Show that rk dim. 17. Let be an matrix over. What is the relationship between the linear transformation and the system of equations? Use your knowledge of linear transformations to state and prove various results concerning the system, especially when. 18. Let have basis. Suppose that for each we define by if if Prove that the are invertible and form a basis for. 19. Let. If is a -invariant subspace of must there be a subspace of for which reduces? 20. Find an example of a vector space and a proper subspace of for which. 21. Let dim. If, prove that implies that and are invertible and that for some polynomial. 22. Let where dim. If for all show that, for some, where is the identity map. 23. Let. Let be a field containing. Show that if and are similar over, that is, if where then and are also similar over, that is, there exists for which. Hint: consider the equation as a homogeneous system of linear equations with coefficients in. Does it have a solution? Where? 24. Let be a continuous function with the property that Prove that is a linear functional on.

20 74 Advanced Linear Algebra 25. Prove that any linear functional is a continuous map. 26. Prove that any subspace of is a closed set or, equivalently, that is open, that is, for any there is an open ball centered at with radius for which. 27. Prove that any linear transformation is continuous under the natural topologies of and. 28. Prove that any surjective linear transformation from to (both finitedimensional topological vector spaces under the natural topology) is an open map, that is, maps open sets to open sets. 29. Prove that any subspace of a finite-dimensional vector space is a closed set or, equivalently, that is open, that is, for any there is an open ball centered at with radius for which. 30. Let be a subspace of with dim. a) Show that the subspace topology on inherited from is the natural topology. b) Show that the natural topology on is the topology for which the natural projection map continuous and open. 31. If is a real vector space then is a complex vector space. Thinking of as a vector space over, show that is isomorphic to the external direct product. 34. (When is a complex linear map a complexification?) Let be a real vector space with complexification and let. Prove that is a complexification, that is, has the form for some if and only if commutes with the conjugate map defined by. 35. Let be a complex vector space. a) Consider replacing the scalar multiplication on by the operation where and. Show that the resulting set with the addition defined for the vector space and with this scalar multiplication is a complex vector space, which we denote by. b) Show, without using dimension arguments, that. 36. a) Let be a linear operator on the real vector space with the property that. Define a scalar multiplication on by complex numbers as follows for and. Prove that under the addition of and this scalar multiplication is a complex vector space, which we denote by. b) What is the relationship between and? Hint: consider and.

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