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)=∫_0x▒f(s^2+3)ds=[s2/3+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