Linear-1 12

1

B={v1=(101),v2=(210),v3=(031)} is a basis of R3C={u1=(02−1),u2=(202),u3=(1−21)} is a basis of R3E is a standard basis of R3

1a

Find [I]EB,[I]ECSolution:[I]EB=([v1]E||[v2]E||[v3]E||)=(v1||v2||v3||)=(120013101)In a similar way: [I]EC=(u1||u2||u3||)=(02120−2−121)⟹[I]EB=(120013101),[I]EC=(02120−2−121)

1b

Find [I]CESolution:[I]CE=([I]EC)−1=(02120−2−121)−1(02110020−2010−121001)→R2=R2+2R3R3=−R3R1↔R3(1−2−100−1040012021100)→R3=12(2R3−R2)R2=14R2(1−2−100−1010014120011−12−1)→R1=R1+R3R1=R1+2R2(10010−1010014120011−12−1)⟹[I]CE=(10−1014121−12−1)

1c

Find [I]CBSolution:[I]CB=[I]CE[I]EB=(10−1014121−12−1)(120013101)=(02−1121454032−52)⟹[I]CB=(02−1121454032−52)

2

Let V,W be vector spaces over RLet T:V→W be a linear transformationLet B={v1,v2,v3,v4} be a basis of VLet C={u1,u2,u3} be a basis of W[T]CB=(316910262143)Find bases of ker(T) and Im(T) via vectors of B and CSolution:[ker(T)]B=N([T]CB)(316910262143)→(1026010−9010−9)→(1026010−90000)Let x3=s,x4=t⟹{x1=−2s−6tx2=9tx3=sx4=t⟹N([T]CB)=sp({(−2010),(−6901)})⟹ker(T)=sp({−2v1+v3,−6v1+9v2+v4})[Im(T)]C=C([T]CB)=sp({(312),(101)})⟹ker(T)=sp({3u1+u2+2u3,u1+u3})

3

Let V,W be vector spaces over FLet T:V→W be a linear transformationLet B be a basis of VLet C be a basis of WLet A1=[T]CB∈Fm×nLet A2,A3∈Fm×n

3a

Prove: ∃C′ basis of W:[T]C′B=A2⟺A1 is row-equivalent to A2Proof:Let ∃C′ basis of W:[T]C′B=A2A1=[T]CB=[I]CC′⋅[T]C′B=[I]CC′⋅A2[I]CC′ is invertible⟹A1 is row-equivalent to A2Let A1 be row-equivalent to A2⟹A1=X⋅A2 where X is invertibleX is invertible,X∈Fm×m⟹∃C′ basis of W:[I]CC′=X[T]CB=[I]CC′⋅A2⟹[I]C′C⋅[T]CB=[I]C′C⋅[I]CC′⋅A2=A2⟹[T]C′B=A2

3b

Prove: ∃B′ basis of V:[T]CB′=A3⟺A1T is row-equivalent to A3TProof:Let ∃B′ basis of V:[T]CB′=A3[T]CB=[T]CB′⋅[I]B′B⟹([T]CB)T=([I]B′B)T([T]BB′)T⟹A1T=([I]B′B)TA3T[I]B′B is invertible⟹([I]B′B)T is invertible⟹A1T is row-equivalent to A3TLet A1T be row-equivalent to A3T⟹A1T=X⋅A3T⟹A1=A3⋅XT where X is invertibleX is invertible,X∈Fn×n⟹XT is invertible,XT∈Fn×n⟹∃B′ basis of V:[I]B′B=XT[T]CB=A3⋅[I]B′B⟹[T]CB⋅[I]BB′=A3⋅[I]B′B⋅[I]BB′=A3⟹[T]CB′=A3

4

B={(101),(011),(001)}C={(100),(011),(012)}Let T:R3→R3 be a linear transformation[T]CB=(1000a1a−21a) where a∈R is a parameter

4a

Find all values of a such that T is invertibleSolution:T is invertible ⟺[T]CB is invertible⟺rank([T]CB)=3(1000a1a−21a)→(1000a101a)→(10001a0a1)→(10001a001−a2)rank([T]CB)=3⟺1−a2≠0⟺(1−a)(1+a)≠0⟺{a≠1a≠−1

4b

For a=3 find T−1((xyz))Solution:a=3⟹[T]CB=(100031113)[T((101))]C=(101)⟹T((101))=(112)[T((011))]C=(031)⟹T((011))=(045)[T((001))]C=(013)⟹T((001))=(047)}⟹[T]EB=(100144257)[T−1(v)]E=[T−1]EE[v]E=([T]EE)−1[v]E=([T]EB[I]BE)−1[v]E[(100)]B=(10−1),[(010)]B=(01−1),[(001)]B=(001)⟹[I]BE=(100010−1−11)[T]EB[I]BE=(100144257)(100010−1−11)=(100−304−5−27)(100100−304010−5−27001)→(1001000−27501004310)→(1001000201474−100134140)→(1001000101878−1200134140)⟹([T]EB[I]BE)−1[v]E=(1001878−1234140)(xyz)=(x18x+78y−12z34x+14y)⟹T−1((xyz))=(x18x+78y−12z34x+14y)

4c

For a=1 find bases of ker(T) and Im(T)Solution:[T]CB=(100011−111)[ker(T)]B=N([T]CB)(100011−111)→4a(100011000)⟹{x1=0x2=−sx3=s⟹N([T]CB)=sp({(0−11)})⟹ker(T)=sp({(0−10)})[ker(T)]C=C([T]CB)=sp({(10−1),(011)})⟹ker(T)=sp({(1−1−2),(023)})

5

Let T:R2×2→R2×2,T(A)=A−ATB={(1000),(0100),(0010),(0001)}C={(1000),(0110),(0001),(0−110)}B,C are bases of R2×2

5a

Find [T]CB,[T]BCSolution:B is a standard basis,B=E⟹[T]BB=[T]EET(e1)=T(e4)=0T(e2)=−T(e3)=(01−10)⟹[T]BB=[T]EE=(000001−100−1100000)T((1000))=T((0110))=T((0001))=0T((0−110))=(0−220)⟹[T]BC=[T]EC=(0000000−200020000)

5b

Find basis and dimension of Im(T),ker(T)Solution:[Im(T)]E=C([T]EC)=sp({(0−220)})⟹Im(T)=sp({(0−220)}){(0−220)} is a linear independence⟹{(0−220)} is a basis of Im(T) and dim(Im(T))=1[ker(T)]E=N([T]EE)(000001−100−1100000)→(000001−1000000000)⟹{x1=ux2=tx3=tx4=s⟹N([T]EE)=sp({(1000),(0110),(0001)})⟹ker(T)=sp({(1000),(0110),(0001)}){(1000),(0110),(0001)} is a linear independence⟹{(1000),(0110),(0001)} is a basis of ker(T) and dim(ker(T))=3

5c

Find [T2]CCSolution:T2(A)=T(A)−(T(A))T=(A−AT)−(A−AT)T=A−AT−AT+A==2A−2AT=2T(A)Let C={c1,c2,c3,c4}T2(c1)=T2(c2)=T2(c3)=0T2(c4)=2T(c4)=(0−440)⟹[T2(c4)]C=(0004)⟹[T2]CC=(0000000000000004)