Almaz Khusnutdinov•Math enthusiast, love coding and mathematics
Published 8/22/2025
Problem 1. Change of the Basis - Apply transformation to the basis-vector
Let E={(1,0),(0,1)},B={(1,2),(3,1)}be bases of R2.SupposePB←B′=[21−11].(a) Compute the new basis B′={b1′,b2′} via B′=BPB←B′ and find PB′←B.(b) If [x]B′=[3−2],find x in standard coordinates.(c) With respect to which basis, vectors in B’ expressed?Answer: B′={(5,5),(2,−1)},PB←B′=[21−11],PB′←B=31[1−112],x=(11,17).Since is expressed in standard coordinates, the resulting vectors of B′ are also expressed in standard coordinates
Problem 2. Change of the Basis - Find new matrix [T]
Let E={(1,0),(0,1)},B={(1,2),(2,1)},C={(1,0),(1,1)}be bases of R2.Suppose [T]E←E=[3012].(a) Compute [T]C←B.(b) Given [v]B=[2−1],find T(v) in standard coordinates.(c) Are matrices [T]Eand[T]C←Esimilar? Are they equivalent? And why?Hint: [T]C←B=PC←E[T]E←EPE←B.Answer: [T]C←B=[1452],T(v)=(3,6).They are not similar. However they are equivalent.
Problem 3. Change of the Basis - Apply transformation to the basis-vector #2
Let E={(1,0,0),(0,1,0),(0,0,1)},B={(1,0,1),(0,1,2),(1,1,0)}be bases of R3.SupposePB←B′=101110011.(a) Compute the new basis B′={b1′,b2′,b3′} via B′=BPB←B′ and find PB′←B.(b) If [x]B′=1−23,find x in standard coordinates.Answer: B′={(2,1,1),(1,1,3),(1,2,2)},PB←B′=101110011,PB′←B=2111−1−1111−11,x=(3,5,1).
Problem 4. Playing with Change of Basis
Let B={(1,0),(2,1)},C={(2,1),(1,1)}be bases of R2.(a) Compute the change-of-basis matrices PC←B and PB←C.(b) Verify PC←BPB←C=I2.(c) For w=(3,−1) in standard coordinates, compute [w]B and [w]C.Answer: PC←B=[1−110],PB←C=[01−11],PC←BPB←C=I2,[w]B=[5−1],[w]C=[4−5].
Problem 5. Playing with Change of Basis #2
Let B={(1,0,1),(0,1,1),(1,1,0)},C={(1,0,0),(1,1,0),(1,1,1)},D={(0,1,0),(0,0,1),(1,0,0)}be bases of R3.Suppose [F]C←B=101210011,[G]D←C=210011101.(a) Compute [G∘F]D←B.(b) For u=(2,−1,3) in standard coordinates, find (G∘F)(u) in standard coordinates.Answer: [G∘F]D←B=311431112,(G∘F)(2,−1,3)=(1,8,2).
Problem 6. Matrix Equivalence - finding PAQ.
Let A=100110000be a 3×3 matrix.(a) Find invertible P∈GL3(R),Q∈GL3(R)such that PAQ=100010000.(b) Using your P,Q,confirm rank(A)=2,dimkerA=1,dimker(A⊤)=1.Answer: P=I3,Q=100−110001,PAQ=100010000;rank(A)=2,dimkerA=3−2=1,dimker(A⊤)=3−2=1.
Problem 7. Similar Matrices - Finding Similar
Let A=[2003],S=[1110](invertible).(a) Compute B=S−1AS.(b) Verify that A and B have the same trace, determinant, and eigenvalues.Answer: S−1=[011−1],B=[3−102].trA=trB=5,detA=detB=6,σ(A)=σ(B)={2,3}.
Problem 8. Matrix Equivalence - finding PAQ #2
Let A=[122436]be the (standard-basis) 2×3 matrix of T:R3→R2.Let B=⎩⎨⎧100,−210,−301⎭⎬⎫ be a basis of R3,C={[12],[01]} a basis of R2.(Thus the (C,B)-matrix of T is PAQ, where Q=[I]B and P=[I]C−1.)(a) Find invertible P∈GL2(R),Q∈GL3(R)such that PAQ=[100000].(b) Compute rank(A),dimkerA,dimker(A⊤).Answer: Q=100−210−301,AQ=[120000];P=[1−201],PAQ=[100000].rank(A)=1,dimkerA=3−1=2,dimker(A⊤)=2−1=1.
Problem 9. Similar Matrices - Finding Similar #2
Let A=[1000],B=[0010].(a) Are A and B similar? Justify using a similarity invariant.(b) Find invertible P,Qsuch that B=PAQ(matrix equivalence).Answer: (a)Not similar since trA=1=0=trB(trace is similarity-invariant).(b)P=I2,Q=[0110]⇒AQ=[0010]=B.
Problem 10. Construct Bases based on Conditions & Verify Smith Normal Form PAQ
Let A=121240361480be the (standard-basis) 3×4 matrix of T:R4→R3.(a) Construct a basis B of R4 whose last two vectors form a basis of kerA,and a basis C of R3 whose first two vectorsform a basis of imA.Let Q=[I]B∈GL4(R),P=[I]C−1∈GL3(R).Show that PAQ=100010000000.(b) Compute rank(A),dimkerA,dimker(A⊤).Answer: kerA=span{(−1,−1,1,0)⊤,(0,−2,0,1)⊤}.Choose B={e1,e3,b3(−1,−1,1,0)⊤,b4(0,−2,0,1)⊤},C={Ae1,Ae3,c3010}.⇒Q=[I]B=10000010−1−1100−201,AQ=121361000000.[I]C=[Ae1Ae3c3]=121361010,P=[I]C−1=−2121−200123−210.PAQ=100010000000.rank(A)=2,dimkerA=4−2=2,dimker(A⊤)=3−2=1.
Problem 11. Obtaining Linear Transformation based on the given domain-codomain Bases.
Let A=[101101]be the (standard-basis) 2×3 matrix of T:R3→R2.Let the domain basis B=⎩⎨⎧110,011,001⎭⎬⎫,and the codomain basis C={[10],[11]}.(a) Build Q=[I]B∈GL3(R)and P=[I]C−1∈GL2(R),then compute D=PAQ.(b) Explain why P and Q are written via [I] and interpret D.Answer: Q=[I]B=110011001,[I]C=[1011],P=[I]C−1=[10−11].AQ=[211201],D=PAQ=[11−12−11].(b) Explanation: For x∈R3,[x]std=[I]B[x]B=Q[x]B.For y=T(x)∈R2,[y]C=[I]C−1[y]std=P[y]std.Thus [y]C=PAQ[x]B,i.e. D=PAQ is the (C,B)-matrix of T.It represents the same linear map as A but in the C-coords for outputs and B-coords for inputs.
Problem 12. Obtaining Linear Transformation based on the given domain-codomain Bases #2
Let A=101011be the (standard-basis) 3×2 matrix of T:R2→R3.Let the domain basis B={[11],[−11]},and the codomain basis C=⎩⎨⎧110,011,101⎭⎬⎫.(a) Build Q=[I]B∈GL2(R)and P=[I]C−1∈GL3(R),then compute D=PAQ.(b) Explain why P and Q appear as [I]C−1 and [I]B, and interpret D in relation to A.Answer: Q=[I]B=[11−11],[I]C=110011101,P=[I]C−1=21−21212121−21−212121.AQ=112−110,D=PAQ=01101−1.(b) Explanation: Coordinate change is mediated by the identity map. For x∈R2,[x]std=[I]B[x]B=Q[x]B.For y=T(x)∈R3,[y]C=[I]C−1[y]std=P[y]std.Hence [y]C=PAQ[x]B,so D=PAQ=[T]C,Bis the matrix of the same linear map as A in the new bases (C,B).
Problem 13. Proof # 1 - Matrix of a linear map under given bases
Let T:Rn→Rm be linear. Let B be a basis of Rn,C a basis of Rm,and let e denote the standard bases.Define the change-of-coordinates matrices by [x]e=Pe←B[x]B,[y]C=PC←e[y]e.Prove the formula [T]C←B=PC←e[T]ePe←B.Outline of proof:(1)By definition of [T]C←B,[T(x)]C=[T]C←B[x]B for all x.(2)Change output coordinates: [T(x)]C=PC←e[T(x)]e.(3)Use the standard matrix of T:[T(x)]e=[T]e[x]e.(4)Change input coordinates: [x]e=Pe←B[x]B.(5)Combine (2)–(4): [T(x)]C=PC←e[T]ePe←B[x]B.(6)Compare with (1) and use arbitrariness of [x]B to conclude [T]C←B=PC←e[T]ePe←B.
Problem 14. Proof # 2 - Relating two basis-matrices via a change matrix
Let B={b1,…,bn},B′={b1′,…,bn′}be two bases of Rn.Let B∈Rn×n be the matrix with columns bi(so [x]e=B[x]B), and B′ analogously for B′.Let PB←B′ be the change-of-coordinates matrix so that [x]B=PB←B′[x]B′.Prove the formula B′=BPB←B′.Outline of proof:(1)For any x,[x]e=B[x]B=B′[x]B′.(2)Insert the change [x]B=PB←B′[x]B′ into the first expression: [x]e=BPB←B′[x]B′.(3)Compare with [x]e=B′[x]B′and use arbitrariness of [x]B′⇒B′=BPB←B′.(4)Interpretation: B and B′ collect basis vectors in standard coordinates; PB←B′=[I]B←B′ converts B′-coords to B-coords.
Problem 15
Let V,W be finite-dimensional real vector spaces, S={s1,…,sm} a basis of W.Let F,G:V→W be linear, and for w∈W write [w]S∈Rm for the coordinate column of w in the basis S.Prove that for all v∈V,[(F+G)(v)]S=[F(v)]S+[G(v)]S.Outline of proof:(1)Define the coordinate (identity) map φS:W→Rm,φS(w)=[w]S.It is a linear isomorphism since S is a basis.(1)(Linearity: if w=i∑aisi,z=i∑bisi,α,β∈R,(1)[αw+βz]S=(αai+βbi)i=α(ai)i+β(bi)i=α[w]S+β[z]S.)(2)Use linearity of F,G and of φS:[(F+G)(v)]S=φS((F+G)(v))=φS(F(v)+G(v))(2)=φS(F(v))+φS(G(v))=[F(v)]S+[G(v)]S.(3)Thus the coordinate representation respects addition; equivalently, [⋅]S (i.e., φS) is linear, so sums are preserved.
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Matrix Representations & Linear Mappings Part 1 - nerjik.com
