Math 304–504Linear AlgebraLecture 21:Matrix of a linear transformation.Linear mapping = linear transformation = linear functionDefinition. Given vector spaces V1and V2, amapping L : V1→ V2is linear ifL(x + y) = L(x) + L(y),L(rx) = rL(x)for any x, y ∈ V1and r ∈ R.Basic pr operties of linear mappings:• L(r1v1+ · · · + rkvk) = r1L(v1) + · · · + rkL(vk)for all k ≥ 1, v1, . . . , vk∈ V1, and r1, . . . , rk∈ R.• L(01) = 02, where 01and 02are zero vectors inV1and V2, respectively.• L(−v) = −L(v) for any v ∈ V1.Matrix transformationsAny m×n matrix A gives rise to a transformationL : Rn→ Rmgiven by L(x) = Ax, where x ∈ Rnand L(x) ∈ Rmare regarded as colum n vectors.This transformation is linear.Example. Lxyz=1 0 23 4 70 5 8xyz.Let e1= (1, 0, 0), e2= (0, 1, 0), e3= (0, 0, 1) bethe standard basis for R3. Then vectorsL(e1), L(e2), L(e3) are columns of the matrix.Theorem 1 Suppose that {v1, v2, . . . , vn} is abasis for a vector space V . Then(i) any linear mapping L : V → W is uniquelydetermined by vectors L(v1), L(v2), . . . , L( vn);(ii) for any vectors w1, w2, . . . , wn∈ W there existsa linear mapping L : V → W such that L(vi) = wi,1 ≤ i ≤ n.Theorem 2 Suppose L : Rn→ Rmis a linear map.Then there exists an m×n matrix A such thatL(x) = Ax for all x ∈ Rn. The columns of A arevectors L(e1), L(e2), . . . , L( en), where e1, e2, . . . , enis the standard basis for Rn.Basis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique repr es entationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates o f v with respect to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)provides a one-to-one correspo ndence between Vand Rn. Besides, this mapping is linear.Matrix of a linear mappingLet V , W be vector spaces and f : V → W be a linear map.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let w1, w2, . . . , wmbe a basis for W and g2: W → Rmbethe coordinate mapping corresponding to this basis.Vf−→ Wg1yyg2Rn−→ RmThe composition g2◦f ◦g−11is a linear mapping of Rnto Rm.It is represented as v 7→ Av, where A is an m×n matrix.A is called the matrix of f with respect to bases v1, . . . , vnand w1, . . . , wm. Columns of A are coordinates of vectorsf (v1), . . . , f (vn) with respect to the basis w1, . . . , wm.Examples. • D : P3→ P2, (Dp)(x) = p′(x).Let ADbe the matrix of D with respect to the bases1, x, x2and 1, x. Columns of ADare coordinatesof pol ynomials D1, Dx, Dx2w.r.t. the basis 1, x.D1 = 0, Dx = 1, Dx2= 2x =⇒ AD=0 1 00 0 2• L : P3→ P3, (Lp) (x) = p(x + 1).Let ALbe the matrix of L w.r. t. the basis 1, x, x2.L1 = 1, Lx = 1 + x, Lx2= (x + 1)2= 1 + 2x + x2.=⇒ AL=1 1 10 1 20 0 1Problem. Consider a linear operator L : R2→ R2,Lxy=1 10 1xy.Find the matrix of L with respect to the basisv1= (3, 1), v2= (2, 1) .Let N be the desired matrix. Columns of N are coordinates ofthe vectors L(v1) and L( v2) w.r.t. the basis v1, v2.L(v1) =1 10 131=41, L(v2) =1 10 121=31.Clearly, L(v2) = v1= 1v1+ 0v2.L(v1) = αv1+ βv2⇐⇒3α + 2β = 4α + β = 1⇐⇒α = 2β = −1Thus N =2 1−1 0.Problem. Find the matrix of the identity mappingL : R2→ R2, L(x) = x with respect to the bases:(i) e1= (1, 0), e2= (0, 1) and e1, e2;(ii) v1= (3, 1), v2= (2, 1) and v1, v2;(iii) v1, v2and e1, e2;(iv) e1, e2and v1, v2.The desired matrix is the transition matrix fromthe first basis to the second one:A1= A2=1 00 1, A3=3 21 1,A4= A−13=1 −2−1 3.Change of basisConsider a linear operator L : Rn→ Rn. I t i s givenby L(x) = Ax, x ∈ Rn, where A is an n×n matrix.Let u1, u2, . . . , unbe a nonstandard basis for Rn.If we change the standard coordinates in Rnto thecoordinates relati ve to u1, u2, . . . , un, then theoperator L is given by L(y) = By, y ∈ Rn, where Bis another n×n matrix.A is the matrix of L relative to the standar d basis.B is the matrix of L relative to the basis u1, . . . , un.Columns of A ar e vectors L(e1), . . . , L(en).Columns of B are coordinates of vectorsL(u1), . . . , L(un) relative to the basis u1, . . . , un.Problem 1. Given the matrix A, find the matrix B.Problem 2. Given the matrix B, find the matrix A.Let uj= (u1j, u2j, . . . , unj), j = 1, 2, . . . , n.Consider the transition matrix U = (uij) from thebasis u1, . . . , unto the standard basi s .Theorem A = UBU−1, B = U−1AU.Proof: It is enough to prove the 2nd formula asB=U−1AU =⇒ UBU−1=U(U−1AU)U−1=(UU−1)A(UU−1)=A.Take any x ∈ Rnand let y be the (vector of) nonstandardcoordinates of x.Then Uy are standard coordinates of x=⇒ AUy are standard coordinates of L(x)=⇒ U−1AUy are nonstandard coordinates of L(x).Problem. Consider a linear operator L : R2→ R2,Lxy=1 10 1xy.Find the matrix of L with respect to the basisv1= (3, 1), v2= (2, 1) .Let S be the matrix of L with respect to the standard basis,N be the matrix of L with respect to the basis v1, v2, and U bethe transition matrix from v1, v2to e1, e2. Then N = U−1SU.S =1 10 1, U =3 21 1,N = U−1SU =1 −2−1 31 10 13 21 1=1 −1−1 23 21 1=2 1−1
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