程序代写代做代考 C kernel Sections 7.2 and 7.3

Sections 7.2 and 7.3
Week 4
Ali Mousavidehshikh
Department of Mathematics University of Toronto
Ali Mousavidehshikh
Week 4

Outline
1 Sections 7.2 and 7.3
Sections 7.2 and 7.3
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
Let T : V → W be a linear transformation. Define kerT = {u ∈ V : T(u) = 0} and
imT ={T(u):u∈V}=T(V).
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
Let T : V → W be a linear transformation. Define kerT = {u ∈ V : T(u) = 0} and
imT ={T(u):u∈V}=T(V).
Notice that ker T is a subspace of V and im T is a subspace of W.
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
Let T : V → W be a linear transformation. Define kerT = {u ∈ V : T(u) = 0} and
imT ={T(u):u∈V}=T(V).
Notice that ker T is a subspace of V and im T is a subspace of W.
The dim(ker T ) is called the nullity of T and is denoted by nullity(T). The dim(im T) is called the rank of T and is denoted by rank T.
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
Givenanm×nmatrixA,imTA =ColAand rank TA = rank A.
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
Givenanm×nmatrixA,imTA =ColAand
rank TA = rank A.
Proof. Write A = [c1 c2 · · · cn] in terms of its columns. Then imTA={Ax:x∈Rn}={􏰀ni=1xici :xi ∈Rn}=ColA, and the rest follows.
Define P : Mnn → Mnn via P(A) = A − AT . Show that P is a linear transformation and find its kernel and image (the image is the space of skew symmetric matrices).
A linear transformation T : V → W is said to be one-to-one (or injective) if it satisfies the following property: If
T(v) = T(u), then v = u.
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
LetS:R3→R2 viaS(a,b,c)=(a+b,a−b).ShowthatS is a linear transformation. Is T one-to-one? Is T onto?
Week 4
Ali Mousavidehshikh

Sections 7.2 and 7.3
LetS:R3→R2 viaS(a,b,c)=(a+b,a−b).ShowthatS is a linear transformation. Is T one-to-one? Is T onto?
Let A be an invertible matrix and define T : Mmn → Mmn via T(X)=UX forallX ∈Mmn. ShowthatT isalinear transformation. Is T one-to-one? Is T onto?
Let A be an invertible matrix and define T : Mmn → Mmn via T(X)=AX forallX ∈Mmn. ShowthatT isalinear transformation. Is T one-to-one? Is T onto?
LetAbeanm×nmatrix,andletTA :Rn →Rm bethe linear transformation induced by A, that is TA(x) = AX for all columns x ∈ Rn. Then
Ali Mousavidehshikh
Week 4

Sections 7.2 and 7.3
Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ and dimV = dim(kerT)+dim(im T). Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ and dimV =dim(kerT)+dim(imT). Thatis, dim V = nullity T + rank T . Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ and dimV =dim(kerT)+dim(imT). Thatis, dim V = nullity T + rank T . IfS:R3→R2 isgivenbyS(a,b,c)=(a+b,a−b),verify that 3 = dim(ker S) + dim(im S). Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ and dimV =dim(kerT)+dim(imT). Thatis, dim V = nullity T + rank T . IfS:R3→R2 isgivenbyS(a,b,c)=(a+b,a−b),verify that 3 = dim(ker S) + dim(im S). Theorem. Let T : V → W be a linear transformation, and let {e1,...,er,er+1,...,en} be a basis of V such that {er+1,...,en} is a basis for kerT. Then {T(e1),...,T(er)} is a basis for im(T), and hence rank T = r. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem (Dimension Theorem): Let T : V → W be a linear transformation and assume dim(ker T ) < ∞ and dim(im T ) < ∞. Then dim V < ∞ and dimV =dim(kerT)+dim(imT). Thatis, dim V = nullity T + rank T . IfS:R3→R2 isgivenbyS(a,b,c)=(a+b,a−b),verify that 3 = dim(ker S) + dim(im S). Theorem. Let T : V → W be a linear transformation, and let {e1,...,er,er+1,...,en} be a basis of V such that {er+1,...,en} is a basis for kerT. Then {T(e1),...,T(er)} is a basis for im(T), and hence rank T = r. Example. If T : V → W is a linear transformation where V is finite dimensional, then dim(ker T ) ≤ dim V and rank T ≤ dim V . Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D) = 1. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D ) = 1. The dimension theorem gives rank D = dim Pn − dim(ker D) = n + 1 − 1 = n = dim(Pn−1). Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D ) = 1. The dimension theorem gives rank D = dim Pn − dim(ker D) = n + 1 − 1 = n = dim(Pn−1). Since im(D) is a subspace of Pn−1 and has the same dimension, Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D ) = 1. The dimension theorem gives rank D = dim Pn − dim(ker D) = n + 1 − 1 = n = dim(Pn−1). Since im(D) is a subspace of Pn−1 and has the same dimension, im(D) = Pn−1. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D ) = 1. The dimension theorem gives rank D = dim Pn − dim(ker D) = n + 1 − 1 = n = dim(Pn−1). Since im(D) is a subspace of Pn−1 and has the same dimension, im(D) = Pn−1. A linear transformation T : V → W that is also a bijection is called an isomorphism. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Example. Let D : Pn → Pn−1 be the differentiation map defined via D(p(x)) = p′(x). Compute ker D and conclude that D is onto. Proof . Because p′(x) = 0 means p(x) is constant, dim(ker D ) = 1. The dimension theorem gives rank D = dim Pn − dim(ker D) = n + 1 − 1 = n = dim(Pn−1). Since im(D) is a subspace of Pn−1 and has the same dimension, im(D) = Pn−1. A linear transformation T : V → W that is also a bijection is called an isomorphism. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Examples. (1) 1V : V → V is an isomorphism. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Examples. (1) 1V : V → V is an isomorphism. (2) T : Mmn → Mnm given by T(A) = AT is an isomorphism. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Examples. (1) 1V : V → V is an isomorphism. (2) T : Mmn → Mnm given by T(A) = AT is an isomorphism. (3)T :M22 →P3 givenby 􏰑􏰔a b􏰕􏰒 2 3 T c d =a+bx+cx +dx isanisomorphism. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Examples. (1) 1V : V → V is an isomorphism. (2) T : Mmn → Mnm given by T(A) = AT is an isomorphism. (3)T :M22 →P3 givenby 􏰑􏰔a b􏰕􏰒 2 3 T c d =a+bx+cx +dx isanisomorphism. Notice that V ∼= W says there exists an isomorphism from V to W (or vice versa). It does not say that every linear transformation must be an isomorphism. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Two vector space V and W are said to be isomorphic (written V ∼= W ) if and only if there exists an isomorphism T:V→W. Examples. (1) 1V : V → V is an isomorphism. (2) T : Mmn → Mnm given by T(A) = AT is an isomorphism. (3)T :M22 →P3 givenby 􏰑􏰔a b􏰕􏰒 2 3 T c d =a+bx+cx +dx isanisomorphism. Notice that V ∼= W says there exists an isomorphism from V to W (or vice versa). It does not say that every linear transformation must be an isomorphism. (Example)f :R→Rgivenbyf(x)=x2 isnotan isomorphism, but R ∼= R via the identity map. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. (3) There exists a basis {e1,...,en} of V such that {T(e1),...,T(en)} is a basis of W. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. (3) There exists a basis {e1,...,en} of V such that {T(e1),...,T(en)} is a basis of W. Theorem: If V and W are finite dimensional, then V ∼=W ⇔dimV =dimW. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. (3) There exists a basis {e1,...,en} of V such that {T(e1),...,T(en)} is a basis of W. Theorem: If V and W are finite dimensional, then V ∼=W ⇔dimV =dimW. Corollary: (1) V ∼= V , Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. (3) There exists a basis {e1,...,en} of V such that {T(e1),...,T(en)} is a basis of W. Theorem: If V and W are finite dimensional, then V ∼=W ⇔dimV =dimW. Corollary: (1) V ∼= V , (2) V ∼= W ⇔ W ∼= V , Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 Theorem: If V and W are finite dimensional spaces, the following conditions are equivalent for a linear transformation T:V→W. (1) T is an isomorphism. (2) If {e1,...,en} is any basis of V, then {T(e1),...,T(en)} is a basis of W. (3) There exists a basis {e1,...,en} of V such that {T(e1),...,T(en)} is a basis of W. Theorem: If V and W are finite dimensional, then V ∼=W ⇔dimV =dimW. Corollary: (1) V ∼= V , (2) V ∼= W ⇔ W ∼= V , (3)U∼=V andV ∼=W ⇒U∼=W. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 It follows that if dimV = n, then V ∼= Rn. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 It follows that if dimV = n, then V ∼= Rn. Let V denote the space of all 2 × 2 symmetric matrices. Find anisomorphismT :P2 →V suchthatT(1)=I,whereI is the 2 × 2 identity matrix. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 It follows that if dimV = n, then V ∼= Rn. Let V denote the space of all 2 × 2 symmetric matrices. Find anisomorphismT :P2 →V suchthatT(1)=I,whereI is the 2 × 2 identity matrix. Solution. Show this by using the basis {1, x, x2}. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 It follows that if dimV = n, then V ∼= Rn. Let V denote the space of all 2 × 2 symmetric matrices. Find anisomorphismT :P2 →V suchthatT(1)=I,whereI is the 2 × 2 identity matrix. Solution. Show this by using the basis {1, x, x2}. Theorem: If V and W have the same dimension n, a linear transformation T : V → W is an isomorphism if and only if it is either one-to-one or onto. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 It follows that if dimV = n, then V ∼= Rn. Let V denote the space of all 2 × 2 symmetric matrices. Find anisomorphismT :P2 →V suchthatT(1)=I,whereI is the 2 × 2 identity matrix. Solution. Show this by using the basis {1, x, x2}. Theorem: If V and W have the same dimension n, a linear transformation T : V → W is an isomorphism if and only if it is either one-to-one or onto. Example. Prove that T : R3 → P2 given by T(a,b,c)=a+bx+cx2 isanisomorphism. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Theorem: Let V and W be finite dimensional. The following are equivalent for a linear transformation T : V → W . Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Theorem: Let V and W be finite dimensional. The following are equivalent for a linear transformation T : V → W . (1) T is an isomorphism. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Theorem: Let V and W be finite dimensional. The following are equivalent for a linear transformation T : V → W . (1) T is an isomorphism. (2) There exists a linear transformation S : W → V such that ST=1V andTS=1W. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Theorem: Let V and W be finite dimensional. The following are equivalent for a linear transformation T : V → W . (1) T is an isomorphism. (2) There exists a linear transformation S : W → V such that ST=1V andTS=1W.Moreover,inthiscaseSisalsoan isomorphism and is uniquely determined by T: Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 ThecompositionofV→T W→S UisthemapST:V→U given by ST(v) = S(T(v)) for all v ∈ V. Theorem: Let V →T W →S U →R Z be linear transformations. Then (1) ST is a linear transformation. (2)T1V =Tand1WT=T. (3) (RS)T = R(ST). Theorem: Let V and W be finite dimensional. The following are equivalent for a linear transformation T : V → W . (1) T is an isomorphism. (2) There exists a linear transformation S : W → V such that ST=1V andTS=1W.Moreover,inthiscaseSisalsoan isomorphism and is uniquely determined by T: that is, T(v) = w if and only if v = S(w). Notation: S = T−1. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1+x,−1}. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Moreover, T −1 sends the basis {1 + x , −1} to the basis {1,x}. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Moreover, T −1 sends the basis {1 + x , −1} to the basis {1,x}. Thatis,T−1(1+x)=1andT−1(−1)=x. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Moreover, T −1 sends the basis {1 + x , −1} to the basis {1,x}. Thatis,T−1(1+x)=1andT−1(−1)=x. Because a + bx = b(1 + x) + (b − a)(−1), we obtain T−1(a + bx) = T−1(b(1 + x) + (b − a)(−1)) = b + (b − a)x. Week 4 Ali Mousavidehshikh Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Moreover, T −1 sends the basis {1 + x , −1} to the basis {1,x}. Thatis,T−1(1+x)=1andT−1(−1)=x. Because a + bx = b(1 + x) + (b − a)(−1), we obtain T−1(a + bx) = T−1(b(1 + x) + (b − a)(−1)) = b + (b − a)x. Define T : R3 → R3 by T(x,y,z) = (z,x,y). Find T−1. Ali Mousavidehshikh Week 4 Sections 7.2 and 7.3 DefineT :P1 →P1 byT(a+bx)=(a−b)+ax. Showthat T has an inverse, and find T−1. Proof . Notice that T is linear. Moreover, T (1) = 1 + x and T(x) = −1, so T carries the basis {1,x} to the basis {1 + x , −1}. Hence, T is an isomorphism, and T −1 exists. Moreover, T −1 sends the basis {1 + x , −1} to the basis {1,x}. Thatis,T−1(1+x)=1andT−1(−1)=x. Because a + bx = b(1 + x) + (b − a)(−1), we obtain T−1(a + bx) = T−1(b(1 + x) + (b − a)(−1)) = b + (b − a)x. Define T : R3 → R3 by T(x,y,z) = (z,x,y). Find T−1. Hint. Show that T3 = 1R3. Ali Mousavidehshikh Week 4

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