Math 221: LINEAR ALGEBRA Chapter 7. Linear Transformations §7-3. Isomorphisms and Composition Le Chen 1 Emory University, 2020 Fall (last updated on 10/21/2020) Creative Commons License (CC BY-NC-SA) 1 Slides are adapted from those by Karen Seyffarth from University of Calgary.
What is an isomorphism? Example P 1 = { ax + b | a , b ∈ R } , has addition and scalar multiplication defined as follows: ( a 1 x + b 1 ) + ( a 2 x + b 2 ) = ( a 1 + a 2 ) x + ( b 1 + b 2 ) , k ( a 1 x + b 1 ) = ( ka 1 ) x + ( kb 1 ) , for all ( a 1 x + b 1 ) , ( a 2 x + b 2 ) ∈ P 1 and k ∈ R . The role of the variable x is to distinguish a 1 from b 1 , a 2 from b 2 , ( a 1 + a 2 ) from ( b 1 + b 2 ) , and ( ka 1 ) from ( kb 1 ) .
Example (continued) This can be accomplished equally well by using vectors in R 2 . �� a � � � R 2 = � � a , b ∈ R � b where addition and scalar multiplication are defined as follows: � a 1 � a 2 � a 1 + a 2 � a 1 � ka 1 � � � � � + = , k = b 1 b 2 b 1 + b 2 b 1 kb 1 � a 1 � a 2 � � ∈ R 2 and k ∈ R . for all , b 1 b 2
Definition Let V and W be vector spaces, and T : V → W a linear transformation. T is an isomorphism if and only if T is both one-to-one and onto (i.e., ker ( T ) = { 0 } and im ( T ) = W). If T : V → W is an isomorphism, then the vector spaces V and W are said to be isomorphic, and we write V ∼ = W.
Example The identity operator on any vector space is an isomorphism.
Example The identity operator on any vector space is an isomorphism. Example T : P n → R n +1 defined by a 0 a 1 T ( a 0 + a 1 x + a 2 x 2 + · · · + a n x n ) = a 2 . . . a n for all a 0 + a 1 x + a 2 x 2 + · · · + a n x n ∈ P n is an isomorphism. To verify this, prove that T is a linear transformation that is one-to-one and onto.
M M Proving isomorphism of vector spaces Problem Prove that M 22 and R 4 are isomorphic.
Proving isomorphism of vector spaces Problem Prove that M 22 and R 4 are isomorphic. Proof. Let T : M 22 → R 4 be defined by a � a � a � � b b b T = for all ∈ M 22 . c d c c d d
Proving isomorphism of vector spaces Problem Prove that M 22 and R 4 are isomorphic. Proof. Let T : M 22 → R 4 be defined by a � a � a � � b b b T = for all ∈ M 22 . c d c c d d It remains to prove that 1. T is a linear transformation; 2. T is one-to-one; 3. T is onto.
Solution (continued – 1. linear transformation) � a 1 � b 1 a 2 � b 2 � Let A = , B = ∈ M 22 and let k ∈ R . Then a 3 a 4 b 3 b 4 a 1 b 1 a 2 b 2 T ( A ) = and T ( B ) = . a 3 b 3 a 4 b 4
Solution (continued – 1. linear transformation) � a 1 � b 1 a 2 � b 2 � Let A = , B = ∈ M 22 and let k ∈ R . Then a 3 a 4 b 3 b 4 a 1 b 1 a 2 b 2 T ( A ) = and T ( B ) = . a 3 b 3 a 4 b 4 ⇓ a 1 + b 1 a 1 b 1 � a 1 + b 1 a 2 + b 2 � a 2 + b 2 a 2 b 2 T ( A + B ) = T = = + = T ( A )+ T ( B ) a 3 + b 3 a 4 + b 4 a 3 + b 3 a 3 b 3 a 4 + b 4 a 4 b 4
Solution (continued – 1. linear transformation) � a 1 � b 1 a 2 � b 2 � Let A = , B = ∈ M 22 and let k ∈ R . Then a 3 a 4 b 3 b 4 a 1 b 1 a 2 b 2 T ( A ) = and T ( B ) = . a 3 b 3 a 4 b 4 ⇓ a 1 + b 1 a 1 b 1 � a 1 + b 1 a 2 + b 2 � a 2 + b 2 a 2 b 2 T ( A + B ) = T = = + = T ( A )+ T ( B ) a 3 + b 3 a 4 + b 4 a 3 + b 3 a 3 b 3 a 4 + b 4 a 4 b 4 ⇓ T preserves addition.
Solution (continued – 1. linear transformation) Also ka 1 a 1 � ka 1 � ka 2 ka 2 a 2 T ( kA ) = T = = k = kT ( A ) ka 3 ka 4 ka 3 a 3 ka 4 a 4
Solution (continued – 1. linear transformation) Also ka 1 a 1 � ka 1 � ka 2 ka 2 a 2 T ( kA ) = T = = k = kT ( A ) ka 3 ka 4 ka 3 a 3 ka 4 a 4 ⇓ T preserves scalar multiplication. Since T preserves addition and scalar multiplication, T is a linear transformation.
0 ker ker Solution (continued – 2. One-to-one) By definition, ker ( T ) = { A ∈ M 22 | T ( A ) = 0 } � a 0 � a � � � b b 0 � = a , b , c , d ∈ R and = . � c d c 0 � � d 0 �
Solution (continued – 2. One-to-one) By definition, ker ( T ) = { A ∈ M 22 | T ( A ) = 0 } � a 0 � a � � � b b 0 � = a , b , c , d ∈ R and = . � c d c 0 � � d 0 � � a � b If A = ∈ ker T, then a = b = c = d = 0 , and thus ker ( T ) = { 0 22 } . c d
Solution (continued – 2. One-to-one) By definition, ker ( T ) = { A ∈ M 22 | T ( A ) = 0 } � a 0 � a � � � b b 0 � = a , b , c , d ∈ R and = . � c d c 0 � � d 0 � � a � b If A = ∈ ker T, then a = b = c = d = 0 , and thus ker ( T ) = { 0 22 } . c d ⇓ T is one-to-one.
M Solution (continued – 3. Onto) Let x 1 x 2 ∈ R 4 , X = x 3 x 4 and define matrix A ∈ M 22 as follows: � x 1 x 2 � A = . x 3 x 4
M Solution (continued – 3. Onto) Let x 1 x 2 ∈ R 4 , X = x 3 x 4 and define matrix A ∈ M 22 as follows: � x 1 x 2 � A = . x 3 x 4 Then T ( A ) = X, and therefore T is onto. Finally, since T is a linear transformation that is one-to-one and onto, T is an isomorphism.
Solution (continued – 3. Onto) Let x 1 x 2 ∈ R 4 , X = x 3 x 4 and define matrix A ∈ M 22 as follows: � x 1 x 2 � A = . x 3 x 4 Then T ( A ) = X, and therefore T is onto. Finally, since T is a linear transformation that is one-to-one and onto, T is an isomorphism. Therefore, M 22 and R 4 are isomorphic vector spaces. �
Example ( Other isomorphic vector spaces ) 1. For all integers n ≥ 0 , P n ∼ = R n +1 . 2. For all integers m and n, m , n ≥ 1 , M mn ∼ = R m × n . 3. For all integers m and n, m , n ≥ 1 , M mn ∼ = P mn − 1 . You should be able to define appropriate linear transformations and prove each of these statements.
Characterizing isomorphisms Theorem Let V and W be finite dimensional vector spaces and T : V → W a linear transformation. The following are equivalent. 1. T is an isomorphism. 2. If { � b 1 ,� b 2 , . . . ,� b n } is any basis of V, then { T ( � b 1 ) , T ( � b 2 ) , . . . , T ( � b n ) } is a basis of W. 3. There exists a basis { � b 1 ,� b 2 , . . . ,� b n } of V such that { T ( � b 1 ) , T ( � b 2 ) , . . . , T ( � b n ) } is a basis of W.
Characterizing isomorphisms Theorem Let V and W be finite dimensional vector spaces and T : V → W a linear transformation. The following are equivalent. 1. T is an isomorphism. 2. If { � b 1 ,� b 2 , . . . ,� b n } is any basis of V, then { T ( � b 1 ) , T ( � b 2 ) , . . . , T ( � b n ) } is a basis of W. 3. There exists a basis { � b 1 ,� b 2 , . . . ,� b n } of V such that { T ( � b 1 ) , T ( � b 2 ) , . . . , T ( � b n ) } is a basis of W. The proof relies on the following results of this chapter). ◮ One-to-one linear transformations preserve independent sets. ◮ Onto linear transformations preserve spanning sets.
dim dim Suppose V and W are finite dimensional vector spaces with dim ( V ) = dim ( W ) , and let { � b 1 ,� b 2 , . . . ,� { � f 1 ,� f 2 , . . . ,� b n } and f n } be bases of V and W respectively.
dim dim Suppose V and W are finite dimensional vector spaces with dim ( V ) = dim ( W ) , and let { � b 1 ,� b 2 , . . . ,� { � f 1 ,� f 2 , . . . ,� b n } and f n } be bases of V and W respectively. Then T : V → W defined by T ( � b i ) = � f i for 1 ≤ k ≤ n is a linear transformation that maps a basis of V to a basis of W. By the previous Theorem, T is an isomorphism.
Suppose V and W are finite dimensional vector spaces with dim ( V ) = dim ( W ) , and let { � b 1 ,� b 2 , . . . ,� { � f 1 ,� f 2 , . . . ,� b n } and f n } be bases of V and W respectively. Then T : V → W defined by T ( � b i ) = � f i for 1 ≤ k ≤ n is a linear transformation that maps a basis of V to a basis of W. By the previous Theorem, T is an isomorphism. Conversely, if V and W are isomorphic and T : V → W is an isomorphism, then (by the previous Theorem) for any basis { � b 1 ,� b 2 , . . . ,� b n } of V, { T ( � b 1 ) , T ( � b 2 ) , . . . , T ( � b n ) } is a basis of W, implying that dim ( V ) = dim ( W ) . This proves the next theorem.
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