Linear-1 13

Linear-1 13

W=sp({1+x+x2,x+x2})U={p(x)∈R2[x]|p(1)=0p′(1)=0}Find basis and dimension of U+W,U∩WSolution:Let p(x)∈Up(x)=a+bx+cx2p(1)=a+b+c⟹a+b+c=0p′(x)=b+2cxp′(1)=b+2c⟹b+2c=0{a+b+c=0b+2c=0⟹{a=cb=−2c⟹U=sp({1−2x+x2})⟹U+W=sp({1−2x+x2,1+x+x2,x+x2})(1−21111011)→(1−21030011)→(1−21010001)⟹{1−2x+x2,1+x+x2,x+x2} is a linear independence⟹U+W=R2[x]dim(U+W)=dim(U)+dim(W)−dim(U∩W)⟹3=1+2−dim(U∩W)⟹dim(U∩W)=0⟹U∩W={0}
A,B∈Fn×nProve: dim(N(AB))≤dim(N(A))+dim(N(B))Proof:Let v∈N(B)Bv=0⟹ABv=0⟹N(B)⊆N(AB)⟹dim(N(B))≤dim(N(AB))Let B={v1,…,vk} be a basis of N(B)Let us add vectors {u1,…,ut} to B such thatB′={v1,…,vk,u1,…,ut} is a basis of N(AB)∀i∈[1,t]:ABui=0⟹{Bu1,Bu2,…,But}⊆N(A)Let α1,…,αt∈Fα1Bu1+⋯+αtBut=0⟹B(α1u1+⋯+αtut)=0⟹α1u1+⋯+αtut∈N(B)⟹∃{β1,…,βk}⊆F:α1u1+⋯+αtut=∑i=1kβkvk⟹α1u1+⋯+αtut−∑i=1kβkvk=0{v1,…,vk,u1,…,ut} is a linear independence⟹α1=⋯=αt=β1=⋯=βk=0⟹{Bu1,…,But} is a linear independence⟹dim(N(A))≥t⟹dim(N(A))+dim(N(B))≥k+t=dim(N(AB))
B={1,1+x,1+x2} is a basis of R2[x]C={(000−1),(1000),(0200),(0011)} is basis of R2×2Let T:R2[x]→R2×2 be a linear transformation[T]CB=(1020131151−1−1)Find basis and dimension of Im(T),ker(T)Solution:[T(1)]C=(1011)⟹T(1)=(0210)[T(1+x)]C=(011−1)⟹T(1+x)=(12−1−1)[T(1+x2)]C=(235−1)⟹T(1+x2)=(35−1−3)Another way (right way):[ker(T)]B=N([T]CB)(1020131151−1−1)→(1020130130−1−3)→(102013000000)⟹[ker(T)]B=sp({(−2−31)})[u]B=(−2−31)⟹u=−4−3x+1⟹ker(T)=sp({−4−3x+1})[Im(T)]C=C([T]CB)=sp({(1011),(011−1)})[v]C=(1011)⟹v=(0210)[w]C=(011−1)⟹w=(12−1−1)⟹Im(T)=sp({(0210),(12−1−1)})
Let V,W be vector spacesLet T,S:V→W be linear transformationsLet S be invertibleLet TS−1 be injectiveProve or disprove: T is invertibleProof:TS−1:W→W is a linear transformationTS−1 is injective and dim(W)=dim(W)⟹TS−1 is invertibleTS−1 is surjective⟹T is surjective(1)S is invertible⟹dim(V)=dim(W)(2)(1) and (2)⟹T is invertible
Let A,B∈Fn×nProve or disprove:1.rank(A)<n2 and rank(B)<n2⟹AB=02.AB=0⟹rank(A)≤n2 or rank(B)≤n2Disproof for 1:A=B=(100000000)AB=(100000000)≠0Proof for 2:AB=0⟹rank(AB)=0⟹dim(N(AB))=nn≤dim(N(A))+dim(N(B))∀v∈N(AB):ABv=0⟹Bv∈N(A)∀v∈Fn:Bv∈C(B)⟹C(B)⊆N(A)⟹dim(C(B))≤dim(N(A))⟹rank(B)≤n−rank(A)⟹rank(A)+rank(B)≤n⟹[rank(A)<n2rank(B)<n2
Let A∈R3×3 be anti-symmetricProve: A(111)=(abc)⟹a+b+c=0Proof:(0a2a3−a20a5−a3−a50)(111)=(a2+a3a5−a2−a3−a5){a2+a3=aa5−a2=b−a3−a5=c⟹a+b+c=a2+a3+a5−a2−a3−a5=0Note: this doesn’t work in Z2Because (100010001)∈Z23×3 is anti-symmetric
Let V,W be finitely generated vector space over FLet T:V→W be a linear transformationLet S:W→V be a linear transformation such that TST=TProve or disprove:1.ker(S)={0}2.ker(S)∩Im(T)={0}3.Prove that such S existsDisproof for 1:Let T=S=0⟹TST=T and ker(S)=W≠{0}Proof for 2:Let w∈ker(S)∩Im(T)S(w)=0,∃v∈V:T(v)=w⟹∃v∈V:(ST)(v)=0⟹(TST)(v)=0⟹T(v)=0⟹w=0⟹ker(S)∩Im(T)⊆{0}⟹ker(S)∩Im(T)={0}Proof for 3:Let {w1,…,wk} be a basis of Im(T)⟹∀i∈[1,k]:∃vi∈V:T(vi)=wiLet {w1,…,wk,u1,…,ut} be a basis of WBy the "definition theorem" exists linear transformation S:W→V such thatLet S:W→V,∀i∈[1,k+t]:S(wi)={vii≤k0i>kLet v∈VT(v)=w(TST)(v)=(TS)(T(v))=(TS)(w)=(TS)(∑i=1kαiwi)=T(∑i=1kαiS(wi))==T(∑i=1kαivi)=∑i=1kαiT(vi)=∑i=1kαiwi=w=T(v)