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Math 375 – 10/17

$Examples of Linear Transformations • Example 1 ○ V={all polynomials} ○ Consider D:V→V defined by § Given f∈V § Df=g if g(x)=f′(x) § e.g. D(1+x−3x^2 )=1−6x ○ Null Space § Null(D)={f∈V│Df=0} § ={f∈V│f^′ (x)=0} § ={f∈V│f is constant function} § ={f(x)=c│c∈R} § dim⁡Null(D)=1 § Basis for Null(D)={1} • Example 2 ○ V={all polynomials} ○ K:V→V ○ Kf=g⟺g(x)=∫_0^x▒f(s)ds ○ e.g. K(x^2+3)=∫_0^x▒f(s^2+3)ds=[s^2/3+3s]_0^x=1/3 x^3+3x Addition and Scalar Multiplication of Linear Transformations • Addition ○ V,W: vector spaces ○ T,S: V→W: linear transformations ○ T+S is the map V→W with (T+S)(x)=Tx+Sx • Example ○ V=W=R2 ○ T= rotation by 45° counter-clockwise ○ S= reflection in the y-axis ○ T+S=? • Theroem ○ Statement § If T,S:V→W is linear, so are (T+S) ○ Proof: closed under addition § (T+S)(x+y) § =T(x+y)+S(x+y) § =Tx+Ty+Sx+Sy § =(Tx+Sx)+(Ty+Sy) § =(T+S)(x)+(T+S)(y) ○ Proof: closed under scalar multiplication § (T+S)(cx) § =T(cx)+S(cx) § =c⋅T(x)+c⋅S(x) § =c[T(x)+S(x)] § =c(T+S)(x) • Scalar Multiplication ○ V,W: vector spaces ○ T,S: V→W: linear transformations ○ cT:V→W (c∈R is defined by ○ (cT)(x)=c(Tx), ∀x∈V • Theorem ○ Let V,W be two vector spaces ○ L(V,W)={all linear transformation from V to W} ○ Then L(V,W) is a vector space ○ e.g. T,S∈L(V,W)⇒c_1 T+c_2 S∈L(V,W), ∀c_1,c_2∈R Multiplication/Composition of Linear Transformations • Definition ○ U,V,W: vector spaces ○ T:U→V, S:V→W ○ Then ST:V→W is given by (ST)(x)=S(Tx) • Theorem ○ If S,T_1,T_2 is linear, then S(T_1+T_2 )=ST_1+ST_2 • Example ○ Given § V={all polynomials} § D,K:V→V § Df=f^′, (Kf)(x)=∫_0^x▒f(s)ds ○ DKf=? § Let g=Kf=∫_0^x▒f(s)ds § D(g(x))=d/dx g(x)=d/dx ∫_0^x▒f(s)ds=f(x) § Therefore DKf=f ○ KDf=? § KDf=∫_0^x▒(Df)(s)ds=∫_0^x▒〖f^′ (s)ds〗=f(x)−f(0) § Therefore KDf≠f Injective and Inverse • Injective ○ T is injective if and only if N(T)={0} ○ If T:V→W is injective then ○ Tx=y has exactly one solution for every y∈Range(T) ○ (Range(T)={Tx│x∈V}, "exactly one" because T is injective) • Inverse ○ T^(−1):Range(T)→V is given by ○ T^(−1) (y)=x, if y=Tx • Example ○ Given § V=R2, W=R2 § T:V→W § Tx=(x,x) ○ Whether T is inversable? § Tx=0⇒x=0⇒N(T)={0} § Range(T)={(x,x)│x∈R={(x,y)∈R2│x=y} § T^(−1):Range(T)→R § T^(−1) (x,x)=x • Theorem ○ Statement § T^(−1):Range(T)→V is linear § ⟺{█(T^(−1) (u+v)=T^(−1) (u)+T^(−1) (v)@T^(−1) (c⋅u)=c⋅T^(−1) (u)@∀u,v∈Range(T), c∈R┤ ○ Proof § If u∈Range(T) then there is an x∈V with u=Tx § By definition of T^(−1), x=T^(−1) (u) § Similarly, there is y∈V with v=Ty, and y=T^(−1) (u) § T(x+y)=Tx+Ty=u+v § ⇒u+v∈Range(T) § ⇒x+y=T^(−1) (u+v) • Theorem ○ Statement § Suppose V is a finite-dimensional linear space § T:V→V is injective, then § Range(T)=V ○ Proof § Rank–Nullity Theorem says that § dim⁡Null(T)+rank(T)=dim⁡V § T is injective ⇒Null(T)={0}⇒dim⁡Null(T)=0 § Therefore dim⁡Range(T)=dim⁡V § Also, Range(T) is a sunspace of V § ⇒Range(T)=V • Theorem ○ Suppose V is a finite-dimensional linear space ○ T:V→V is injective, then ○ Tx=y has a unique solution for every y∈V • Example ○ Given § V={all polynomials} § D,K:V→V § Df=f^′, (Kf)(x)=∫_0^x▒f(s)ds ○ Is K injective? § We have proven DKf=f § Suppose Kf=0, then D(Kf)=0 § But f=DKf, so f=0 § ⇒K is injective ○ Is K surjective? § Suppose K is surjective then § Given g∈V, we can solve Kf=g with f∈W § i.e. given g∈V, there is one f with § ∫_0^x▒f(s)ds=1, ∀x∈R § At x=0, we have § ∫_0^0▒f(s)ds=1 § Which makes a contradiction, therefore K is not surjective$

Math 375 – 10/16

Question 1 • Given ○ Let V be a set ○ Let S,T:V→V be invertible functions • Prove ○ ST is also invertible and (ST)^(−1)=T^(−1) S^(−1) • Proof ○ (ST)(T^(−1) S^(−1) )=S(TT^(−1) ) S^(−1)=SIS^(−1)=SS^(−1)=I ○ (T^(−1) S^(−1) )(ST)=T^(−1) (S^(−1) S)T=T^(−1) IT=T^(−1) T=I Question 2 • Given ○ Let V and W be finite-dimensional vector spaces. • Proof ○ There exists a surjective linear map f:V→W if and only if dim⁡W≤dim⁡V

Math 375 – Homework 6

Math 375 – 10/12

$Injective • Definition ○ If V,W are vector space and T:V→W is linear ○ Then T is injective if for all x,y∈V ○ Tx=Ty⇒x=y • Theorem ○ T:V→W is injective if and only if for all x∈V ○ Tx=0⇒x=0 ○ i.e. if and only if N(T)={0} • Proof ○ Suppose Tx=0⇒x=0 for all x∈V ○ Let x,y∈V be giben, and assume § Tx=Ty ○ Since T is linear, we have § T(x−y)=Tx−Ty=0 ○ Therefore § x−y=0 § ⇒x=y Null Space • Definition ○ If T:V→W is linear then ○ Null(T)=N(T)=kern(T)≝{x∈V│Tx=0} • Theorem ○ N(T) is a linear subspace of V • Proof: ○ To prove N(T)∈V is a linear subspace ○ We need to check closure properties i.e. § x,y∈N(T)⇒x+y∈N(T) § x∈N(T), c∈R⇒cx∈N(T) ○ Check closure under addition § Let x,y∈N(T), then Tx=0, Ty=0 § We have T(x+y)=Tx+Ty=0+0=0 § Therefore x+y∈N(T) ○ Check closure under scalar multiplication § Let x∈N(T), then Tx=0 § Let c∈R, then T(cx)=c⋅Tx=c⋅0=0 § Therefore cx∈N(T) ○ In conclusion, N(T)∈V is a linear subspace Range • Definition ○ If T:V→W is linear then ○ Range(T)=R(T)={Tx│x∈V} • Theorem ○ R(T) is a linear subspace of W Examples • Example 1 ○ Let V=W=R2, T(x,y)=(x,y) ○ Injective? § Given (x,y)∈R2, and (x ̅,y ̅ )∈R2 § with T(x,y)=T(x ̅,y ̅ ) § By definition of T § (x,y)=(x ̅,y ̅ ) § So T is injective ○ Null Space? § Because T is injective § N(T)={0,0} ○ Range? § R(T)≝{T(x,y)│(x,y)∈R2 }=R2 • Example 2 ○ Let V=W=R2, T(x,y)=(x,0) ○ Injective? § No § T(1,0)=T(1,1)=(1,0) ○ Null Space? § N(T)={u│Tu=0}={(0,t)│t∈R ○ Range? § R(T)={T(x,y)│(x,y)∈R2 } § ={(t,0)│t∈R2 } § =x\-axis • Example 3 ○ Let V=R3, W=R2, T(x,y,z)=(x,y) ○ Injective? § No § T(1,1,0)=T(1,1,1)=(1,1) ○ Null Space? § N(T)={(0,0,t)│t∈R ○ Range? § R(T)={T(x,y,z)│(x,y,z)∈R3 } § ={(x,y)│(x,y)∈R2 }=R2 • Example 4 ○ Let V=R2, W=R3, T(x,y)=(x,y,z) ○ T is injective ○ N(T)={0,0} ○ R(T)={(x,y,0)│(x,y)∈R2 }=xy\-plane • Summary T V W N(T) dim⁡N(T) R(T) dim⁡R(T) T(x,y)=(x,y) R2 R2 {0} 0 R2 2 T(x,y)=(x,0) R2 R2 y\-axis 1 x\-axis 1 T(x,y,z)=(x,y) R3 R2 z\-axis 1 R2 2 T(x,y)=(x,y,z) R2 R3 {0} 0 xy\-plane 2 Rank–Nullity Theorem • Statement ○ If T:V→W is linear and if V is finite dimensional ○ Then dim⁡N(T)+dim⁡R(T)=dim⁡V • Proof ○ Let § dim⁡N(T)=k § dim⁡V=n § {e_1,…,e_k } be a basis for N(T) ○ Claim § {e_1,…,e_k }⊆V is independent § ⇒There is a basis {e_1,…,e_k,e_(k+1),…,e_n } of V so dim⁡V=n § {Te_(k+1),Te_(k+2),…,Te_n } is a basis for R(T) ○ Prove {Te_(k+1),Te_(k+2),…,Te_n } is independent § Suppose □ c_(k+1) Te_(k+1)+…+c_n Te_n=0 § Then □ T(c_(k+1) e_(k+1)+…+c_n e_n )=0 □ ⇒c_(k+1) e_(k+1)+…+c_n e_n∈N(T) § Since {e_1,…,e_k } is a basis for N(T) □ c_(k+1) e_(k+1)+…+c_n e_n=c_1 e_1+…+c_k e_k □ −c_1 e_1−…−c_k e_k+c_(k+1) e_(k+1)+…+c_n e_n=0 § Since {e_1,…,e_n } is independent □ c_1=c_2=…=c_n=0 § In particular □ c_(k+1) Te_(k+1)+…+c_n Te_n=0 □ implies c_(k+1)=c_(k+2)=…=c_n=0 § Therefore □ {Te_(k+1),Te_(k+2),…,Te_n } is independent ○ Prove {Te_(k+1),Te_(k+2),…,Te_n } spans R(T) § Every y∈R(T) is of the form □ y=Tx □ For some x∈V § {e_1,…,e_n } is a basis for V, so □ x=x_1 e_1+x_2 e_2+…+x_n e_n □ For some x_1,x_2,…,x_n∈R § Therefore □ y=Tx □ =T(x_1 e_1+x_2 e_2+…+x_n e_n ) □ =x_1 Te_1+…+x_k Te_k+x_(k+1) Te_(k+1)+…+x_n Te_n □ =x_(k+1) Te_(k+1)+…+x_n Te_n∈span{Te_(k+1),Te_(k+2),…,Te_n } ○ Conclusion § dim⁡〖R(T)〗=n−k=dim⁡V−dim⁡N(T) § ⇒dim⁡N(T)+dim⁡R(T)=dim⁡V dad try (X,y)=(X,0) ZX a cyst) (x.y) • _ _ - _ _ TX J PS TT ox, y) → by a, oaky) his$

Math 375 – 10/11

$Question • Given ○ V=C([−1,1]) ○ ⟨v,w⟩=∫_(−1)^1▒v(x)w(x)dx • Find the linear polynomial closest to f(x)=e^x • Answer ○ Let S=span{1,x} ○ Projection of f onto S is ○ ⟨1,e^x ⟩/⟨1,1⟩ ⋅1+⟨x,e^x ⟩/⟨x,x⟩ ⋅x ○ Therefore the linear polynomial closest to f(x)=e^x is ○ g(x)=3/e x+(e−e^(−1))/2$

Shawn Zhong

Shawn Zhong

Math 375 – 10/17

Math 375 – 10/16

Math 375 – Homework 6

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Math 375 – 10/11

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