Linear-1 7

1a

V1={(xyz)∈R3|z=2y∧y=x}Find the basis and dimension of V1Solution:V1={(xyz)∈R3|y=x∧z=2x}={(xx2x)|x∈R}=sp({(112)}){(112)} is a linear independence⟹Basis of V1 is {(112)} and dim(V1)=1

1b

V2={(xyz)∈R3|y=z−x}Find the basis and dimension of V2Solution:V2={(xx−zz)|x,z∈R}={x(110)+z(0−11)|x,z∈R}=sp({(110),(0−11)}){(110),(0−11)} is a linear independence⟹Basis of V2 is {(110),(0−11)} and dim(V2)=2

1c

V3={A∈R2×2|AT=A}Find the basis and dimension of V3Solution:V3={(abbc)|a,b,c∈R}={a(1000)+b(0110)+c(0001)|a,b,c∈R}==sp({(1000),(0110),(0001)}){(1000),(0110),(0001)} is a linear independence⟹Basis of V3 is {(1000),(0110),(0001)} and dim(V3)=3

2

V=R3[x]U={p(x)∈V|p(x)=x⋅p′(x)}

2a

Prove: U is a vector subspace of VProof:0V=p0p0(x)=0⟹p0′(x)=0⟹p0(x)=x⋅p0′(x)⟹0V∈ULet p1,p2∈V,α∈R(p1(x)+αp2(x))′=p1′(x)+αp2′(x)=xp1(x)+αxp2(x)=x(p1(x)+αp2(x))⟹∀p1,p2∈V,∀α∈R:p1+αp2∈U⟹U is a vector spaceU⊆V⟹U is a vector subspace of V

2b

Find the basis and dimension of USolution:U={a+bx+cx2+dx3|a,b,c,d,x∈Ra+bx+cx2+dx3=x(b+2cx+3dx2)}=={a+bx+cx2+dx3|a,b,c,d,x∈Ra=0c=2cd=3d}={bx|b,x∈R}∀p∈V:p(x)=(1xx2x3)⋅(abcd)Let us denote p(x) as (abcd)∈R4⟹U={b(0100)|b∈R}=sp({(0100)}){(0100)} is a linear independence⟹Basis of U is {(0100)} and dim(U)=1

2c

Find a vector subspace W of V such that U⊕W=VSolution:Let U+W=V⟹∀v∈V:v=u+wLet us use notation from the previous solution:p(x)=(abcd)∈R4⟹(abcd)=α(0100)⏟u+w⟹w=(abcd)−(0α00)=(ab−αcd)Let U∩W={0}⟹∀u∈U,w∈W:[u=w→u=w=0](0α00)=(ab−αcd)⟺{a=0α=b−αc=0d=0u=w=0⟹α=0⟹b−α=0For U+W to be a direct sum, b−α should be zero⟹w=(a0cd)⟹W={a(1000)+c(0010)+d(0001)|a,c,d∈R}W=sp({(1000),(0010),(0001)})

3

V=R2[x]V⊇W1=sp({1−x2,x−x2})V⊇W2={p(x)|p′(1)=0}Find the basis and dimension of W1∩W2,W1+W2W1=sp({(10−1),(01−1)})W2={a+bx+cx2|b+2c=0}={(abc)|a,b,c∈Rc=−b2}=sp({(100),(02−1)})Solution for W1+W2v∈W1+W2⟹v∈sp({(10−1),(01−1),(100),(02−1)})(02−1)=−1(10−1)+2(01−1)+1(100)⟹W1+W2=sp({(10−1),(01−1),(100),(02−1)})=sp({(10−1),(01−1),(100)}){(10−1),(01−1),(100)} is a linear independence⟹Basis of W1+W2 is {(10−1),(01−1),(100)} and dim(W1+W2)=3Solution for W1∩W2v∈W1∩W2⟹v=α(10−1)+β(01−1)=γ(100)+δ(02−1)α(10−1)+β(01−1)=γ(100)+δ(02−1)⟹α(10−1)+β(01−1)−γ(100)−δ(02−1)=0(10−100010−20−1−1010)→(10−100010−200−1−110)→(10−100010−2000−1−10)Let δ=−s{α−γ=0β−2δ=0−γ−δ=0⟹{α=sβ=−2sγ=sδ=−s⟹v=s(10−1)−2s(01−1)=(s−2ss)⟹W1∩W2=sp({(1−21)}){(1−21)} is a linear independence⟹Basis of W1∩W2 is {(1−21)} and dim(W1∩W2)=1

4a

Let V be a vector spaceV={A∈R3×3|A=AT}Are there four vector subspaces V1,V2,V3,V4 of V such that{0}⊂V1⊂V2⊂V3⊂V4⊂V?Solution:Yes, there are such vector subspaces of V:Let V1=sp({(100000000)}),V2=sp({(100000000),(010100000)})V3=sp({(100000000),(010100000),(001010100)})V4=sp({(100000000),(010100000),(001010100),(000001010)}){0}⊂V1⊂V2⊂V3⊂V4⊂V

4b

Let V be a vector spaceV={A∈R2×2|A=AT}Are there four vector subspaces V1,V2,V3,V4 of V such that{0}⊂V1⊂V2⊂V3⊂V4⊂V?Solution:V={(abbc)|a,b,c∈R}=sp({(1000),(0110),(0001)})⟹dim(V)=3Let U⊂V,BU is a basis of ULet dim(U)=dim(V)=3|BU|=dim(V) and BU is a linear independence⟹sp(BU)=V⟹U=V−Contradiction!⟹[U⊂V⟹dim(U)<dim(V)]V4⊂V⟹dim(V4)≤2V3⊂V4⟹dim(V3)≤1V2⊂V3⟹dim(V2)≤0⟹V2={0}⟹∄V1,V2,V3,V4:{0}⊂V1⊂V2⊂V3⊂V4⊂V

5

W=sp({(021),(120)}),V=sp({(54−5),(010)})

5a

Find W+VSolution:v∈W+V⟹v∈sp({(021),(120),(54−5),(010)})(54−5)=−5(021)+5(120)+4(010)⟹W+V=sp({(021),(120),(54−5),(010)})=sp({(021),(120),(010)})(010221100)→(010001100)→I3⟹{(021),(120),(010)} is a linear independence⟹W+V=sp({(021),(120),(010)})

5b

Find W∩VSolution:v∈W∩V⟹v=α(021)+β(120)=γ(54−5)+δ(010)⟹α(021)+β(120)−γ(54−5)−δ(010)=0(01−5022−4−11050)→(105001−5002−14−1)→(105001−500041)Let γ=s⟹{α=−5sβ=5sγ=sδ=−4sv=−5s(021)+5s(120)=(5s0−5s)⟹W∩V=sp({(10−1)})

5c

Is W+V a direct sum?Solution:W∩V≠{0}⟹W+V is not a direct sum

6

W=sp({(abcd)∈R4|a=d,b=2c}),V=sp({(abcd)∈R4|b−2c+d=0})W=sp({(a2cca)|a,c∈R})=sp(sp({(1001),(0210)}))=sp({(1001),(0210)})V=sp({(a2c−dcd)|a,c,d∈R})=sp(sp({(1000),(0210),(0−101)}))==sp({(1000),(0210),(0−101)})

6a

Find W+VSolution:v∈W+V⟹v∈sp({(1001),(0210),(1000),(0210),(0−101)})sp({(1001),(0210),(1000),(0210),(0−101)})=sp({(1001),(0210),(1000),(0−101)})(1010020−101001001)→(1010010000−11020−1)→(10100100001−10001)→I4⟹{(1001),(0210),(1000),(0−101)} is a linear independence⟹W+V=sp({(1001),(0210),(1000),(0−101)})

6b

Find W∩VSolution:v∈W∩V⟹v=α(1001)+β(0210)=γ(1000)+δ(0210)+ω(0−101)⟹α(1001)+β(0210)−γ(1000)−δ(0210)−ω(0−101)=0(10−1000020−210010−1001000−10)→(10−1000010−1000010−10020−210)→(10−1000010−1000010−10000010)Let δ=s⟹{α=0β=sγ=0δ=sω=0⟹v=s(0210)⟹W∩V=sp({(0210)})

6c

Is W+V a direct sum?Solution:W∩V≠{0}⟹W+V is not a direct sum

7a

Prove or disprove: V=U⊕W⟹dim(V)=dim(U)+dim(W)Proof:V=U⊕W⟹U+W=V,U∩W={0}dim(U+W)=dim(U)+dim(W)−dim(U∩W)U∩W={0}⟹dim(U∩W)=0⟹dim(U+W)=dim(V)=dim(U)+dim(W)

7b

Prove or disprove: dim(V)=dim(U)+dim(W)⟹V=U⊕WDisproof:V=R3,U=sp({(100)}),W=sp({(100),(010)})dim(U)=1,dim(W)=2dim(V)=3=dim(U)+dim(W)U+W=sp({(100),(100),(010)})=sp({(100),(010)})=W≠V