Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 8/20/2025
Problem 1. (Full Walkthrough)
Let B={(1,0,1),(1,1,0),(0,1,1)},C={(1,1,0),(0,1,1),(1,0,0)},[F]C←B=10221−10−11,v=(3,2,−1).Solution.1) Change v from standard to B-coordinates.PB=101110011,PB−1=2121−21−21212121−2121.[v]B=PB−1[v]ε=2121−21−21212121−212132−1=03−1.2) Apply the given matrix to get [F(v)]C.[F(v)]C=[F]C←B[v]B=10221−10−1103−1=64−4.3) Convert [F(v)]C to standard coordinates.PC=110011100,F(v)=PC[F(v)]C.F(v)=11001110064−4=2104.F(v)=(2,10,4)
Problem 2. (Composition)
Let B={(1,0),(1,1)},C={(2,1),(1,0)},D={(1,2),(0,1)}be bases of R2.Suppose[F]C←B=[1032],[G]D←C=[2111].(a) Compute [G∘F]D←B.(b) For v=(4,−1) in standard coordinates, find (G∘F)(v) in standard coordinates.Answer: [G∘F]D←B=[2185],(G∘F)(4,−1)=(2,4).
Problem 3. Apply Transformation
Let B={(1,2,0),(0,1,1),(1,0,1)},C={(1,0,0),(1,1,0),(1,1,1)},[F]C←B=2−10012101,v=(1,3,2).Compute F(v) in standard coordinates.Answer: F(v)=(319,314,311).
Problem 4. Recover Standard Matrix
Let B={(1,1),(0,1)},C={(−1,1),(1,0)}.Suppose [F]C←B=[0121].Compute the standard matrix of F (with respect to the standard bases).Answer: [F]ε←ε=[2−2−12].
Problem 5. Change of Basis
Let B={(1,2),(1,0)},C={(1,1),(0,1)}.Suppose the standard matrix of F is [F]ε←ε=[3012].(a) Compute [F]C←B.(b) For v=(2,−1) in standard coordinates, find [F(v)]C.Answer: [F]C←B=[5−13−3],[F(v)]C=[5−7].
Problem 6. Matrix Representation Identity
Let F:V→U be linear, S={v1,…,vm} a basis of V, and S′={u1,…,un} a basis of U.Prove that for every v∈V,[F(v)]S′=[F]S,S′[v]S.Proof.Write [v]S=c1⋮cm so that v=j=1∑mcjvj.For each j,expand F(vj) in the basis S′:F(vj)=i=1∑naijui, so [F(vj)]S′=a1j⋮anj.Define the n×m matrix A=[[F(v1)]S′⋯[F(vm)]S′]=(aij).By definition A=[F]S,S′.By linearity of F,F(v)=j=1∑mcjF(vj)⇒[F(v)]S′=j=1∑mcj[F(vj)]S′.In matrix form,[F(v)]S′=Ac1⋮cm=[F]S,S′[v]S.This is exactly the desired identity.□
Problem 7. Change of Basis Problem with Domain and Codomain
Let B={(1,0),(1,1)},B′={(2,1),(1,2)},C={(1,1),(0,1)},C′={(1,0),(1,1)}.Suppose [F]C←B=[2013].(a) Compute [F]C′←B′.(b) For v=(3,1) in standard basis , find [F(v)]C′.Answer: [F]C′←B′=[−36−66],[F(v)]C′=[−38].
Problem 8. Proof Change-of-Basis Matrix decomposition.
Prove that PB←B′=MB−1MB′.Proof.Let B={b1,…,bn},B′={b1′,…,bn′} be two bases of Rn.Define MB=[b1b2⋯bn],MB′=[b1′b2′⋯bn′].By definition, [v]B=PB←B′[v]B′for all v.In standard coordinates: v=MB[v]B=MB′[v]B′.Substitute [v]B=PB←B′[v]B′ into v=MB[v]B:MBPB←B′[v]B′=MB′[v]B′.Since this holds for all [v]B′,MBPB←B′=MB′.Multiply by MB−1:PB←B′=MB−1MB′.PB←B′=MB−1MB′.
Problem 9. Prove the Proposition
Proposition.For any basis S={s1,…,sn} of Rn,and any vector v∈Rn, we have [v]std=Pstd←S[v]S.Proof.By definition, write v=c1s1+⋯+cnsn,[v]S=c1⋮cn.Form the matrix MS=[s1s2⋯sn], whose columns are the basis vectors in S.Then in standard coordinates we have v=MS[v]S.Hence the change-of-basis matrix from S to the standard basis is Pstd←S=MS.Therefore [v]std=MS[v]S=Pstd←S[v]S.[v]std=Pstd←S[v]S.
Problem 10. Change of the Basis with Codomain & Domain - 2
Let B={(1,1),(1,0)},B′={(1,2),(2,1)},C={(0,1),(1,1)},C′={(1,0),(0,1)}(the standard basis).Suppose [F]C←B=[1−123].(a) Compute [F]C′←B′.(b) For v=(2,−1) in the standard basis, find [F(v)]C′.Answer: [F]C′←B′=[−5−525],[F(v)]C′=[1015].
Problem 11. Canonical Form of a 4→3 Linear Map
Let F:R4→R3 be linear with (standard) matrix [F]std=101213011112.(a) Find bases B of R4 and C of R3 such that [F]C←B=[Ir000]. Give one valid pair (B,C).(b) For v=(3,0,−1,2) (in the standard basis), compute [v]B and [F(v)]C.Answer: r=2,B={(1,0,0,0),(0,1,0,0),(2,−1,1,0),(1,−1,0,1)},C={(1,0,1),(2,1,3),(0,1,0)}.[F]C←B=100010000000,[v]B=31−12,[F(v)]C=310.
Problem 12. Canonical Form of a 5→4 Linear Map
Let F:R3→R2 be linear with (standard) matrix [F]std=[100111].(a) Produce bases B of R3 and C of R2 so that [F]C←B=[Ir0]. Give one valid pair (B,C).(b) For v=(2,−1,3) (in the standard basis), compute [v]B and [F(v)]C.Answer: r=2,B={(1,0,0),(0,1,0),(−1,−1,1)},C={(1,0),(0,1)}.[F]C←B=[100100],[v]B=523,[F(v)]C=[52].
Problem 13
Let V,W be finite-dimensional vector spaces over a field F, and let F:V→W be linear. Let B,B′ be bases of V and C,C′ be bases of W.Denote by [F]CB and [F]C′B′ the matrices of F in these bases, and by [v]B,[v]B′ the coordinate columns of v∈V.Prove that kerF is basis-independent, i.e. v∈kerF⟺[F]CB[v]B=0⟺[F]C′B′[v]B′=0,and conclude that kerF⊆V and dim(kerF) do not depend on the chosen bases.Proof outline.(1) Let S be the change-of-basis matrix on V from B′ to B,so [v]B=S[v]B′for all v∈V,and S is invertible.(2) Let P be the change-of-basis matrix on W from C to C′,so [w]C′=P[w]Cfor all w∈W,and P is invertible.(3) Show the representation-change formula [F]C′B′=P[F]CBS.(Apply [⋅]C′=P[⋅]C to [F(v)] and use [v]B=S[v]B′.)(4) For any v∈V,[F]C′B′[v]B′=P[F]CBS[v]B′=P[F]CB[v]B.Since P is invertible, [F]C′B′[v]B′=0⟺[F]CB[v]B=0.(5) But [F]CB[v]B=[F(v)]C,so [F]CB[v]B=0⟺F(v)=0.Thus{v∈V:[F]CB[v]B=0}={v∈V:F(v)=0}={v∈V:[F]C′B′[v]B′=0}=kerF.(6) Hence kerF is the same subset/subspace of V regardless of bases, and its dimension (the nullity) is invariant under any change of basis.