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