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 ² " b #³ ~ ²"³ b ²#³ for all scalars Á - and vectors " , # = . A linear transformation ¢ = ¦ = is called a linear operator on == . The set of all linear transformations from to >²=Á>³ is denoted by B and the set of all linear operators on = is denoted by B²= ³.
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 s . 56 Advanced Linear Algebra
2) The integral operator ¢-´%µ¦-´%µ defined by % ²³ ~ ²!³! is a linear operator on -´%µ . 3) Let (d- be an matrix over . The function ( ¢-¦- defined by (²#³ ~ (#, 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 BB²= Á > ³ is a vector space in its own right and ²= ³ has the structure of an algebra, as defined in Chapter 0.
Theorem 2.1 1) The set B²= Á > ³ is a vector space under ordinary addition of functions and scalar multiplication of functions by elements of - . 2) If B²<Á=³ and B ²=Á>³ then the composition B is in ²<Á>³ . 3) If B²=Á>³ is bijective then c ²>Á=³ B . 4) The vector space B²= ³ is an algebra, where multiplication is composition of functions. The identity map B²=³ is the multiplicative identity and the zero map B ²= ³ is the additive identity. Proof. We prove only part 3). Let ¢= ¦> be a bijective linear transformation. Then c¢> ¦= is a well-defined function and since any two vectors $ and $ in > have the form $ ~ ²# ³ and $ ~ ²# ³ , we have c c ²$ b$³~ ² ²#³b ²#³³ c ~ ² ²# b # ³³ ~# b# c c ~ ²$³b ²$³ which shows that c 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 8 ~¸#0¹ be a basis for = . Then we can define a linear transformation B ²= Á > ³ by specifying the values of 8²# ³ > arbitrarily for all # and extending the domain of to = using linearity, that is,
²# bÄb # ³~ ²#³bÄb ²#³ Linear Transformations 57
This process uniquely defines a linear transformation, that is, if Á B ²=Á>³ satisfy ²#³~ ²#³ for all # 8 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 8 . We leave the details to the reader.
Note that if B²=Á>³ and if : is a subspace of = , then the restriction O: 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 B²=Á>³ . The subspace ker²³~¸#= ²#³~¹ is called the kernel of and the subspace im²³~¸²#³#=¹ is called the image of . The dimension of ker²³ is called the nullity of and is denoted by null²³ . The dimension of im ²³ is called the rank of and is denoted by rk²³ .
It is routine to show that ker²³ is a subspace of = and im ²³ is a subspace of > . Moreover, we have the following.
Theorem 2.3 Let B²=Á>³ . Then 1) is surjective if and only if im²³~> 2) is injective if and only if ker² ³ ~ ¸¹ Proof. The first statement is merely a restatement of the definition of surjectivity. To see the validity of the second statement, observe that ²"³ ~ ²#³ ¯ ²" c #³ ~ ¯ " c # ker ² ³ Hence, if ker² ³ ~ ¸¹ then ²"³ ~ ²#³ ¯ " ~ # , which shows that is injective. Conversely, if is injective and " ker ² ³ then ²"³ ~ ²³ and so " ~ . This shows that ker ² ³ ~ ¸¹ . 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 dim²= ³ ~ . For any ordered basis 8 of = , the coordinate map 8¢= ¦- that sends each vector #= to its coordinate matrix 58 Advanced Linear Algebra
´#µ8 - is an isomorphism. Hence, any -dimensional vector space over - is isomorphic to - .
Isomorphic vector spaces share many properties, as the next theorem shows. If B²=Á>³ and := we write ²:³ ~ ¸ ² ³ :¹
Theorem 2.4 Let B²=Á>³ be an isomorphism. Let := . Then 1) := spans if and only if ²:³> spans . 2) : is linearly independent in = if and only if ²:³ is linearly independent in > . 3) := is a basis for if and only if ²:³> is a basis for .
An isomorphism can be characterized as a linear transformation ¢= ¦> that maps a basis for => to a basis for .
Theorem 2.5 A linear transformation B²=Á>³ is an isomorphism if and only if there is a basis 88 of =²³> 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 dim²= ³ ~ dim ²> ³.
In Example 2.2, we saw that any -dimensional vector space is isomorphic to ) -. Now suppose that ) is a set of cardinality and let ²- ³ be the vector space of all functions from )- to with finite support. We leave it to the reader ) to show that the functions ²- ³ defined for all ) , by %~if ²%³ ~ F %£if
)) form a basis for ²- ³ , called the standard basis . Hence, dim ²²- ³ ³ ~(( ) .
It follows that for any cardinal number , there is a vector space of dimension . ) Also, any vector space of dimension is isomorphic to ²- ³ .
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 ²- ³ of all functions from ) to - with finite support. Linear Transformations 59
The Rank Plus Nullity Theorem Let B²=Á>³ . Since any subspace of = has a complement, we can write =~ker ²³l ker ²³ where ker²³ is a complement of ker ²³ in = . It follows that dim²= ³ ~ dim ² ker ² ³³ b dim ² ker ² ³ ³ Now, the restriction of to ker²³ ¢²³¦>ker is injective, since ker² ³ ~ ker ² ³ q ker ² ³ ~ ¸¹ Also, im²³²³ im . For the reverse inclusion, if ²#³²³ im then since #~"b$"²³$²³ for ker and ker , we have ²#³ ~ ²"³ b ²$³ ~ ²$³ ~ ²$³ im ² ³ Thus im²³~²³ im . It follows that ker²³ im ²³ From this, we deduce the following theorem.
Theorem 2.8 Let B²=Á>³ . 1) Any complement of ker²³ is isomorphic to im ²³ 2) ()The rank plus nullity theorem dim² ker ² ³³ b dim ²im ² ³³ ~ dim ²= ³ or, in other notation, rk² ³ b null ² ³ ~dim ²= ³ Theorem 2.8 has an important corollary.
Corollary 2.9 Let B ²=Á>³ , where dim ²=³~ dim ²>³B . 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 d matrix ( over - the multiplication map
(²#³ ~ (# 60 Advanced Linear Algebra is a linear transformation. In fact, any linear transformation B²-Á-³ has this form, that is, is just multiplication by a matrix, for we have
²³ 2323² ³ Ä ² ³ ~ ² ³ Ä ² ³ ~ ² ³ and so ~ ( where
( ~23 ² ³ Ä ² ³ Theorem 2.10 1) If ( is an d matrix over - then B( ²- Á - ³ . 2) If B²-Á-³ then ~( where
(~² ²³Ä ²³³ The matrix ( is called the matrix of .
Example 2.3 Consider the linear transformation ¢- ¦- defined by ²%Á&Á'³~²%c&Á'Á%b&b'³ Then we have, in column form
vy% v %c& y v c% yvy &~~ ' & wz' w %b&b' z w zwz ' and so the standard matrix of is
vyc (~ wz
If (CÁ then since the image of ( is the column space of ( , we have dim²²³³b²(³~²-³ ker( rk dim This gives the following useful result.
Theorem 2.11 Let (d- be an matrix over . 1) (¢- ¦- is injective if and only if rk ²(³~ n. 2) (¢- ¦- is surjective if and only if rk ²(³~ m. Change of Basis Matrices
Suppose that 89~²ÁÃÁ ³ and ~²ÁÃÁ ³ are ordered bases for a vector space =´#µ´#µ . It is natural to ask how the coordinate matrices 89 and are c related. The map that takes ´#µ89899 to ´#µ is Á ~ 8 and is called the change of basis operator (or change of coordinates operator ). Since 89Á is an operator on - , it has the form ( where Linear Transformations 61
(~² 89Á ²³ÁÃÁ 89 Á ²³³ c c ~ ² 989888 ²´ µ ³Á à Á ²´ µ ³³ ~ ²´ µ99 Á à Á ´ µ ³³
We denote (4 by 89, and call it the change of basis matrix from 89 to .
Theorem 2.12 Let 89~²ÁÃÁ ³ and be ordered bases for a vector space c =~-. Then the change of basis operator 89Á 9 8 is an automorphism of , whose standard matrix is
489, ~²´ µ 9 ÁÃÁ´ µ 9 ³³ Hence
´#µ9898 ~ 4Á ´#µ
c and 4~498Á 89, .
Consider the equation
(~489Á or equivalently,
(~²´ µ99 ÁÃÁ´ µ ³³ Then given any two of (dÁ (an invertible matrix)8 (an ordered basis for --) and 9 (an order basis for ), the third component is uniquely determined by this equation. This is clear if 89 and are given or if (( and 9 are given. If c and 89 are given then there is a unique for which (~498Á and so there is a unique 9 for which (~489Á .
Theorem 2.13 If we are given any two of the following: 1) An invertible d matrix ( . 2) An ordered basis 8 for - . 3) An ordered basis 9 for - . then the third is uniquely determined by the equation
(~489Á
The Matrix of a Linear Transformation Let ¢ = ¦ > be a linear transformation, where dim ²= ³ ~ and dim²> ³ ~ and let 89 ~ ² Á Ã Á ³ be an ordered basis for = and an ordered basis for > . Then the map
¢ ´#µ89 ¦ ´ ²#³µ is a representation of as a linear transformation from -- to , in the sense 62 Advanced Linear Algebra that knowing 89 (along with and , of course) is equivalent to knowing . Of course, this representation depends on the choice of ordered bases 89 and .
Since is a linear transformation from -- to , it is just multiplication by an d matrix ( , that is
´ ²#³µ98 ~ (´#µ
Indeed, since ´ µ8 ~ , we get the columns of ( as follows:
²³ (~(~(´#µ~´²³µ89
Theorem 2.14 Let B²=Á>³ and let 8 ~²ÁÃÁ³ and 9 be ordered bases for => and , respectively. Then 8 can be represented with respect to and 9 as matrix multiplication, that is
´²#³µ~´µ9898, ´#µ where
´µ 89, ~²´²³µÄ´²³µ³99 is called the matrix of 89 with respect to the bases and . When =~> and 89~´µ´µ, we denote 88, by 8 and so
´²#³µ~´µ´#µ888
Example 2.4 Let +¢FF ¦ be the derivative operator, defined on the vector space of all polynomials of degree at most ~~²Á%Á%³ . Let 89 . Then
vy vy vy ´+²³µ~´µ~99, ´+²%³µ~´µ~ 99 Á´+²%³µ~´%µ~ 9 9 wz wz wz and so
vy ´+µ8 ~ wz
Hence, for example, if ²%³~ b%b% then
v yvyvy ´+ ²%³µ988 ~ ´+µ ´ ²%³µ ~ ~ w zwzwz and so + ²%³ ~ b % .
The following result shows that we may work equally well with linear transformations or with the matrices that represent them (with respect to fixed Linear Transformations 63 ordered bases 89 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 89~²ÁÃÁ ³ and ~²ÁÃÁ ³ , respectively. 1) The map B¢²=Á>³¦ CÁ ²-³ defined by
²³~´µ 89,
is an isomorphism and so BC²= Á > ³ Á ²- ³ . 2) If B²<Á=³ and B ²=Á>³ and if 89 , and : are ordered bases for <=, and > , respectively then
´µ8:,,, ~´µ´µ 9: 89 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
´ b ! µ89Á ´ µ 8 ~ ´² b ! ³² ³µ 9 ~ ´ ² ³ b ! ² ³µ9 ~ ´ ² ³µ99 b !´ ² ³µ ~ ´µ89Á ´µ 8 b!´µ 89 Á ´µ 8 ~² ´µ89ÁÁ b!´µ 89 ³´µ 8 and since ´ µ8 ~ is a standard basis vector, we conclude that
´ b ! µ89ÁÁÁ ~ ´ µ 89 b !´ µ 89
²³ and so C is linear. If (Á , we define by the condition ´ ²³µ 9 ~( , whence ²³~( and is surjective. Since dim ² B ²=Á>³³~ dim ² CÁ ²-³³ , the map is an isomorphism. To prove part 2), we have
´ µ8:ÁÁÁ ´#µ 8 ~ ´ ² ²#³³µ : ~ ´ µ 9:, ´ ²#³µ 9 ~ ´ µ 9: ´ µ 89 ´#µ 8
Change of Bases for Linear Transformations
Since the matrix ´µ89, that represents depends on the ordered bases 89 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 ´µ89ÁÁ and ´µ 8ZZ 9 of with respect to two different pairs ZZ ²Á³89 and ² 8 Á 9 ³ of ordered bases. Multiplication by ´µ 89ZZÁ sends ´#µ 8 Z to 64 Advanced Linear Algebra
Z ´²#³µ889Z . This can be reproduced by first switching from to , then applying Z ´µ9989Á and finally switching from to , that is,
c ´µ8ZZ,, 9 ~4 99ÁÁÁ Z ´µ 89 4 8 Z 8 ~4 99 Z ´µ 89 , 488Á Z Theorem 2.16 Let B²=>³ , and let ²Á³ 89 and ²Á³ 89ZZ be pairs of ordered bases of => and , respectively. Then
´µ89ZZÁÁÁÁ ~4 99 Z ´µ 89 4 88 Z (2.1) When B²=³ is a linear operator on = , it is generally more convenient to represent by matrices of the form ´µ8 , where the ordered bases used to represent vectors in the domain and image are the same. When 89~ , Theorem 2.16 takes the following important form.
Corollary 2.17 Let B²=³ and let 8 and 9 be ordered bases for = . Then the matrix of 9 with respect to can be expressed in terms of the matrix of with respect to 8 as follows
c ´µ9898 ~4Á ´µ489Á (2.2)
Equivalence of Matrices Since the change of basis matrices are precisely the invertible matrices, (2.1) has the form
c ´µ89ZZÁÁ ~7´µ 89 8 where 78 and are invertible matrices. This motivates the following definition.
Definition Two matrices () and are equivalent if there exist invertible matrices 78 and for which )~7(8c 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 7( and performing the column operations is equivalent to multiplying on the right by 8c .
In terms of (2.1), we see that performing row operations (premultiplying by 7 ) is equivalent to changing the basis used to represent vectors in the image and performing column operations (postmultiplying by 8c ) is equivalent to changing the basis used to represent vectors in the domain. 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 ²= ³ ~ and dim²> ³ ~ . Then two d matrices ( and ) are equivalent if and only if they represent the same linear transformation B²=Á>³ , but possibly with respect to different ordered bases. In this case, () and represent exactly the same set of linear transformations in B²= Á > ³ . Proof. If () and represent , that is, if
(~´µ89,, and )~´µ 8ZZ 9 for ordered bases 898ÁÁZZ and 9 then Theorem 2.16 shows that ( and ) are equivalent. Now suppose that () and are equivalent, say )~7(8c where 78 and are invertible. Suppose also that ( represents a linear transformation B²=Á>³ for some ordered bases 8 and 9 , that is,
(~´ µ89Á Theorem 2.13 implies that there is a unique ordered basis 8Z for = for which Z 8~488Á Z and a unique ordered basis 9 for > for which 7 ~499Á Z . Hence
)~499ÁÁÁZZZZ ´ µ 89 4 88 ~´ µ 89 Á 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
0Ác 1~ >? cÁ cÁc 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 B²=³ is represented by a matrix of the form ´µ 8 , equation (2.2) has the form
c ´µ88Z ~7´µ7 where 7 is an invertible matrix. This motivates the following definition. 66 Advanced Linear Algebra
Definition Two matrices () and are similar if there exists an invertible matrix 7 for which )~7(7c 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 =d be a vector space of dimension . Then two matrices () and are similar if and only if they represent the same linear operator B²=³ , but possibly with respect to different ordered bases. In this case, ( and ) represent exactly the same set of linear operators in B ²= ³ . Proof. If () and represent B ²=³ , that is, if
(~´µ89 and )~´µ for ordered bases 89 and then Corollary 2.17 shows that () and are similar. Now suppose that () and are similar, say )~7(7c Suppose also that ( represents a linear operator B ²= ³ for some ordered basis 8 , that is,
(~´ µ8 Theorem 2.13 implies that there is a unique ordered basis 9 for = for which 7~489Á . Hence
c )~489Á ´ µ 8 489Á ~´ µ 9 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 Á B ²= ³ are similar if there exists an automorphism B²=³ for which ~ c The equivalence classes associated with similarity are called similarity classes. 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 C 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 C . Proof. If and are represented by ( C , that is, if
´ µ89 ~(~´ µ for ordered bases 89 and then
´µ~´µ9898989 ~4ÁÁ ´µ4
Let B²=³ be the automorphism of = defined by ²³~ , where 89~¸ÁÃÁ ¹ and ~¸ÁÃÁ ¹ . Then
´µ~²´²³µÄ´²³µ³~²´µÄ´µ³~4 99Á 99989 and so
c c ´µ~´µ 99999 ´µ´µ~´ µ from which it follows that and are similar. Conversely, suppose that and are similar, say ~ c
Suppose also that C is represented by the matrix ( , that is,
(~´ µ8 for some ordered basis 8 . Then
c c ´µ 8888 ~´ µ ~´µ´µ´µ8
If we set ~ 9 ² ³ then ~ ² Á Ã Á ³ is an ordered basis for = and
´µ 88 ~²´²³µÄ´²³µ³~²´µÄ´µ³~4Á 88898 Hence
c ´µ8988 ~4Á ´µ498Á It follows that
c (~´µ8898 ~4Á ´µ489Á ~´µ 9 and so ( also represents . By symmetry, we see that and are represented by the same set of matrices. This completes the proof. 68 Advanced Linear Algebra
Invariant Subspaces and Reducing Pairs The restriction of a linear operator B²=³ to a subspace : of = is not necessarily a linear operator on : . This prompts the following definition.
Definition Let B²=³ . A subspace : of = is said to be invariant under or -invariant if ²:³ : , that is, if ² ³ : for all : . Put another way, : is invariant under if the restriction O:: is a linear operator on .
If =~:l; then the fact that :; is -invariant does not imply that the complement is also s-invariant. (The reader may wish to supply a simple example with =~ .)
Definition Let B²=³=~:l; . If and if both : and ; are -invariant, we say that the pair ²:Á ; ³ reduces .
A reducing pair can be used to decompose a linear operator into a direct sum as follows.
Definition Let B ²= ³ . If ²:Á ; ³ reduces we write
~OlO:; and call the direct sum of OO:; and . Thus, the expression ~l means that there exist subspaces :;= and of for which ²:Á;³ reduces and
~O:; and ~O 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 s is the topology for which the set of open rectangles
)~¸0 dÄd0 0's are open intervals in s ¹ is a basis (in the sense of topology), that is, a subset of s 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 s . Linear Transformations 69
The standard topology on s has the property that the addition function 7s¢ d s ¦ s ¢ ²#Á $³ ¦ # b $ and the scalar multiplication function Cs¢d s ¦ s ¢² Á#³¦ # are continuous. As such, s is a topological vector space . Also, any linear functional ¢ss ¦ is a continuous map.
More generally, any real vector space = endowed with a topology J is called a topological vector space if the operations of addition 7¢= d= ¦= and scalar multiplication Cs¢d=¦= are continuous under J .
Let = be a real vector space of dimension and fix an ordered basis 8 ~²#ÁÃÁ# ³ for = . Consider the coordinate map ~¢=¦¢#¦´#µ88 s and its inverse
c s8 ~¢¦=¢²ÁÃÁ³¦#8
We claim that there is precisely one topology JJ~== 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 88 (and therefore also ) is a homeomorphism, for any basis 8 . (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 J are precisely the images of the open sets in s under the map 8 . A basis for the natural topology is given by
¸²0dÄd0³0 s8 's are open intervals in ¹
~0DE # 's are open intervals in s
0 First, we show that if = is a topological vector space under a topology J then is continuous. Since ~¢¦= where s is defined by
² ÁÃÁ ³~# it is sufficient to show that these maps are continuous. (The sum of continuous maps is continuous.) Let 6 be an open set in J . Then Csc²6³ ~ ¸² Á %³ d = % 6¹ is open in s d= . We need to show that the set 70 Advanced Linear Algebra
c s ²6³ ~ ¸² Á à Á ³ # 6¹ c is open in s , so let ² ÁÃÁ ³ ²6³ . Thus, # 6 . It follows that c ² Á # ³ Cs ²6³, which is open, and so there is an open interval 0 and an open set )J of = for which
c ² Á # ³ 0 d ) C ²6³ Then the open set < ~ssss dÄd d0d dÄd , where the factor 0 is in the ²<³6 th position, has the property that . Thus c ² ÁÃÁ ³< ²6³ c and so ²6³ is open. Hence, , and therefore also , is continuous.
Next we show that if every linear functional on = is continuous under a topology J on =#= then the coordinate map is continuous. If denote by ´#µ88Á the th coordinate of ´#µ . The map s ¢ = ¦ defined by ²#³ ~ ´#µ 8Á is a linear functional and so is continuous by assumption. Hence, for any open interval 0 s the set
(Á ~¸#= ´#µ8 0¹ is open. Now, if 0 are open intervals in s then
c ²0 dÄd0 ³~¸#= ´#µ8 0 dÄd0 ¹~ ( is open. Thus, is continuous.
Thus, if a topology J has the property that = is a topological vector space and every linear functional is continuous, then and ~ c are homeomorphisms. This means that J , if it exists, must be unique.
It remains to prove that the topology J on = that makes a homeomorphism has the property that = is a topological vector space under J and that any linear functional = on continuous.
As to addition, the maps s¢= ¦ and ² d ³¢= d= ¦ ss d are homeomorphisms and the map 7sZ¢d¦ s s is continuous and so the map 7 7 ¢=d=¦=, being equal to c k Z k²d³ , is also continuous.
As to scalar multiplication, the maps s¢= ¦ and ²d s ³¢d=¦ ss d are homeomorphisms and the map CsZ¢d s ¦ s is continuous and so the map C ¢=d=¦= , being equal to C ckk²d³ Z , is also continuous.
Now let k be a linear functional. Since is continuous if and only if c is continuous, we can confine attention to =~s . In this case, if ÁÃÁ is the Linear Transformations 71
standard basis for ss and ((²³ 4 , then for any %~² ÁÃÁ ³ we have
((²%³ ~cc ²³ ((( ²³ ( 4 ((
Now, if ((%²%³°4 then (( °4 and so ( ( , which implies that is continuous.
According to the Riesz representation theorem and the Cauchy–Schwarz inequality, we have
))))))²%³ H %
Hence, %¦ implies ²%³¦ and so by linearity, %¦% implies ²% ³ ¦ % 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 s¢= ¦ is a homeomorphism. Linear Operators on = d A linear operator on a real vector space = can be extended to a linear operator dd on the complexification = by defining d²" b #³ ~ ²"³ b ²#³
Here are the basic properties of this complexification of .
Theorem 2.22 If Á B ²= ³ then 1) ²s ³dd ~ , 2) ²b³~ddd b 3) ²³~ddd 4) ´ ²#³µddd ~ ²# ³.
Let us recall that for any ordered basis 8 for =#= and any vector we have
´# b µcpx²³8 ~ ´#µ8
Now, if 8 is a basis for = , then the th column of ´µ8 is d ´²³µ~´²³bµ8 cpx²³88 ~´ ²b³µ cpx ²³ which is the th column of the coordinate matrix of d with respect to the basis cpx²³8 . Thus we have the following theorem. 72 Advanced Linear Algebra
Theorem 2.23 Let B²=³ where = is a real vector space. The matrix of d with respect to the basis cpx²³8 is equal to the matrix of with respect to the basis 8
d ´µcpx²³8 ~´µ8 Hence, if a real matrix (=( represents a linear operator on then also represents the complexification dd of on = . Exercises
1. Let (CCÁ have rank . Prove that there are matrices ? Á and @CÁ, 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 B²=Á>³ . Prove that is an isomorphism if and only if it carries a basis for => to a basis for . 4. If B²=Á>³ and B ²=Á>³ we define the external direct sum ^²==Á>>³ B ^ ^ by
²^ ³²²# Á # ³³ ~ ² ²# ³Á ²# ³³ Show that ^ is a linear transformation. 5. Let =~:l; . Prove that :l;:^ ; . Thus, internal and external direct sums are equivalent up to isomorphism. 6. Let =~(b) and consider the external direct sum ,~(^ ) . Define a map ^¢( ) ¦= by ²#Á$³~#b$ . Show that is linear. What is the kernel of ? When is an isomorphism? 7. Let JB be a subset of ²= ³ . A subspace : of = is J-invariant if : is - invariant for every J= . Also, is J-irreducible if the only J -invariant subspaces of = are ¸¹ and = . Prove the following form of Schur's lemma. Suppose that JB=>²=³ and J ²>³ B and = is J = -irreducible and > is JBJJ>=> -irreducible. Let ²=Á>³ satisfy ~ , that is, for any J=> there is a J such that ~ . Prove that ~ or is an isomorphism. 8. Let B ²=³ where dim ²=³B . If rk ² ³~ rk ² ³ show that im² ³ qker ² ³ ~ ¸¹ . 9. Let B ²< , = ³ and B ²= Á > ³ . Show that rk²³¸²³Á²³¹min rk rk 10. Let B²<Á=³ and B ²=Á>³ . Show that null²³²³b²³ null null 11. Let Á B ²= ³ where is invertible. Show that rk²³~²³~²³ rk rk Linear Transformations 73
12. Let Á B ²= Á > ³ . Show that rk²b³²³b²³ rk rk 13. Let := be a subspace of . Show that there is a B ²=³ for which ker²B ³~:. Show also that there exists a ²=³ for which im ² ³~: . 14. Suppose that Á B ²= ³ . a) Show that ~²³²³ for some ²=³B if and only if im im . b) Show that ~²³²³ for some ²=³B if and only if ker ker . 15. Let = ~ : l : . Define linear operators on = by ² b ³ ~ for ~Á. These are referred to as projection operators . Show that 1) ~ 2) b~0 , where 0 is the identity map on = . 3) ~ for £ where is the zero map. 4) =~im ² ³l im ² ³ 16. Let dim²= ³ B and suppose that B ²= ³ satisfies ~ . Show that rk ² ³ dim ²= ³ . 17. Let (d- 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 8 ~¸# ÁÃÁ# ¹ . Suppose that for each Á we define BÁ ²=³ by
#£ if Á²# ³ ~ F #b#if ~
Prove that the BÁ are invertible and form a basis for ²= ³ . 19. Let B²=³: . If is a -invariant subspace of = must there be a subspace ; of = for which ²:Á ; ³ reduces ? 20. Find an example of a vector space =:= and a proper subspace of for which =:. 21. Let dim²= ³ B . If , B ²= ³ prove that ~ implies that and are invertible and that ~ ²³ for some polynomial ²%³-´%µ . 22. Let B²=³ where dim ²=³B . If ~ for all B ²=³ show that ~, for some - , where is the identity map. 23. Let (Á ) C ²- ³ . Let 2 be a field containing - . Show that if ( and ) c are similar over 2)~7(77²2³( , that is, if where C then and ) are also similar over - , that is, there exists 8 C ²- ³ for which )~8(8c. Hint : consider the equation ?(c)?~ as a homogeneous system of linear equations with coefficients in - . Does it have a solution? Where? 24. Let ¢ss ¦ be a continuous function with the property that ²%b&³~ ²%³b²&³ Prove that is a linear functional on s . 74 Advanced Linear Algebra
25. Prove that any linear functional ¢ss ¦ is a continuous map. 26. Prove that any subspace : of s is a closed set or, equivalently, that :~s ±: is open, that is, for any %: there is an open ball )² Á³ centered at %)²%Á³: 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 finite- dimensional 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 )² Á ³ centered at % with radius for which )²%Á ³ :. 30. Let : be a subspace of = with dim ²= ³ B . 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 d is a complex vector space. Thinking of dd d = as a vector space ²= ³ss over s , show that ²= ³ is isomorphic to the external direct product ==^ . 34. (When is a complex linear map a complexification?) Let = be a real vector space with complexification =dd and let B ²= ³ . Prove that is a complexification, that is, B has the form d for some ²=³ if and only if commutes with the conjugate map ¢=dd ¦= defined by ²"b#³~"c#. 35. Let > be a complex vector space. a) Consider replacing the scalar multiplication on > by the operation ²'Á $³ ¦ '$ where 'd 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 > . d b) Show, without using dimension arguments, that ²>s ³ >^ > . 36. a) Let be a linear operator on the real vector space < with the property that ~c . Define a scalar multiplication on < by complex numbers as follows ² b ³ h # ~ # b ²#³ for Á s and # < . Prove that under the addition of < and this scalar multiplication < is a complex vector space, which we denote by < . d b) What is the relationship between <= and ? Hint: consider < ~ =^ = and ²"Á #³ ~ ²c#Á "³ . http://www.springer.com/978-0-387-27677-9