MATH 304Linear AlgebraLecture 20:Change of coordinates (continued).Linear transformations.Basis and coordinatesIf {v1, v2, . . . , vn} is a basis for a vector space V ,then any vector v ∈ V has a unique representationv = x1v1+ x2v2+ · · · + xnvn,where xi∈ R. The coefficients x1, x2, . . . , xnarecalled the coordinates of v with respect to theordered basis v1, v2, . . . , vn.The mappingvector v 7→ its coordinates (x1, x2, . . . , xn)is a one-to-one correspondence between V and Rn.This correspondence respects linear operations in Vand in Rn.Examples. • Coordinates of a vectorv = (x1, x2, . . . , xn) ∈ Rnrelative to the standardbasis e1= (1, 0, . . . , 0, 0), e2= (0, 1, . . . , 0, 0),. . . ,en= (0, 0, . . . , 0, 1) are (x1, x2, . . . , xn).• Coordinates of a matrixa bc d∈ M2,2(R)relative to the basis1 00 0,0 10 0,0 01 0,0 00 1are (a, b, c, d).• Coordinates of a polynomialp(x) = a0+ a1x + · · · + an−1xn−1∈ Pnrelative tothe basis 1, x, x2, . . . , xn−1are (a0, a1, . . . , an−1).Change of coordinates: general caseLet V be a vector space of dimension n.Let v1, v2, . . . , vnbe a basis for V and g1: V → Rnbe thecoordinate mapping corresponding to this basis.Let u1, u2, . . . , unbe another basis for V and g2: V → Rnbe the coordinate mapping corresponding to this basis.Vg1ւg2ցRn−→ RnThe composition g2◦g−11is a transformation of Rn.It has the form x 7→ Ux, where U is an n×n matrix.U is called the transition matrix from v1, v2. . . , vntou1, u2. . . , un. Columns of U are coordinates of the vectorsv1, v2, . . . , vnwith respect to the basis u1, u2, . . . , un.Problem. Find the transition matrix from thebasis p1(x) = 1, p2(x) = x + 1, p3(x) = (x + 1)2to the basis q1(x) = 1, q2(x) = x, q3(x) = x2forthe vector space P3.We have to find coordinates of the polynomialsp1, p2, p3with respect to the basis q1, q2, q3:p1(x) = 1 = q1(x),p2(x) = x + 1 = q1(x) + q2(x),p3(x) = (x+1)2= x2+2x+1 = q1(x)+2q2(x)+q3(x).Hence the transition matrix is1 1 10 1 20 0 1.Thus the polynomial identitya1+ a2(x + 1) + a3(x + 1)2= b1+ b2x + b3x2is equivalent to the relationb1b2b3=1 1 10 1 20 0 1a1a2a3.Problem. Find the transition matrix from thebasis v1= (1, 2, 3), v2= (1, 0, 1), v3= (1, 2, 1) tothe basis u1= (1, 1, 0), u2= (0, 1, 1), u3= (1, 1, 1).It is convenient to make a two-step transition:first from v1, v2, v3to e1, e2, e3, and then frome1, e2, e3to u1, u2, u3.Let U1be the transition matrix from v1, v2, v3toe1, e2, e3and U2be the transition matrix fromu1, u2, u3to e1, e2, e3:U1=1 1 12 0 23 1 1, U2=1 0 11 1 10 1 1.Basis v1, v2, v3=⇒ coordinates xBasis e1, e2, e3=⇒ coordinates U1xBasis u1, u2, u3=⇒ coordinates U−12(U1x)=(U−12U1)xThus the transition matrix from v1, v2, v3tou1, u2, u3is U−12U1.U−12U1=1 0 11 1 10 1 1−11 1 12 0 23 1 1=0 1 −1−1 1 01 −1 11 1 12 0 23 1 1=−1 −1 11 −1 12 2 0.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.A linear mapping ℓ : V → R is called a linearfunctional on V .If V1= V2(or if both V1and V2are functionalspaces) then a linear mapping L : V1→ V2is calleda linear operator.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.Remark. A function f : R → R given byf (x) = ax + b is a linear transformation of thevector space R if and only if b = 0.Properties of linear mappingsLet L : V1→ V2be a linear mapping.• L(r1v1+ · · · + rkvk) = r1L(v1) + · · · + rkL(vk)for all k ≥ 1, v1, . . . , vk∈ V1, and r1, . . . , rk∈ R.L(r1v1+ r2v2) = L(r1v1) + L(r2v2) = r1L(v1) + r2L(v2),L(r1v1+ r2v2+ r3v3) = L(r1v1+ r2v2) + L(r3v3) == r1L(v1) + r2L(v2) + r3L(v3), and so on.• L(01) = 02, where 01and 02are zero vectors inV1and V2, respectively.L(01) = L(001) = 0L(01) = 02.• L(−v) = −L(v) for any v ∈ V1.L(−v) = L((−1)v) = (−1)L(v) = −L(v).Examples of linear mappings• Scaling L : V → V , L(v) = sv, where s ∈ R.L(x + y) = s(x + y) = sx + sy = L(x) + L(y),L(rx) = s(r x) = r (sx) = rL(x).• Dot product with a fixed vectorℓ : Rn→ R, ℓ(v) = v · v0, where v0∈ Rn.ℓ(x + y) = (x + y) · v0= x · v0+ y · v0= ℓ(x) + ℓ(y),ℓ(rx) = (r x) · v0= r (x · v0) = rℓ(x).• Cross product with a fixed vectorL : R3→ R3, L(v) = v × v0, where v0∈ R3.• Multiplication by a fixed matrixL : Rn→ Rm, L(v) = Av, where A is an m×nmatrix and all vectors are column vectors.Linear mappings of functional vector spaces• Evaluation at a fixed pointℓ : F (R) → R, ℓ(f ) = f (a), where a ∈ R.• Multiplication by a fixed functionL : F (R) → F (R), L(f ) = gf , where g ∈ F (R).• Differentiation D : C1(R) → C (R), L(f ) = f′.D(f + g) = (f + g )′= f′+ g′= D(f ) + D(g),D(rf ) = (rf )′= rf′= rD(f ).• Integration over a finite intervalℓ : C (R) → R, ℓ(f ) =Zbaf (x) dx, wherea, b ∈ R, a < b.Properties of linear mappings• If a linear mapping L : V → W is invertible thenthe inverse mapping L−1: W → V is also linear.• If L : V → W and M : W → X are linearmappings then the composition M ◦ L : V → X isalso linear.• If L1: V → W and L2: V → W are linearmappings then the sum L1+ L2is also linear.Linear differential operators• an ordinary differential operatorL : C∞(R) → C∞(R), L = g0d2dx2+ g1ddx+ g2,where g0, g1, g2are smooth functions on R.That is, L(f ) = g0f′′+ g1f′+ g2f .• Laplace’s operator ∆ : C∞(R2) → C∞(R2),∆f =∂2f∂x2+∂2f∂y2(a.k.a. the Laplacian; also denoted by ∇2).Mm,n(R): the space of m×n matrices.• α : Mm,n(R) → Mn,m(R), α(A) = AT.α(A + B) = α(A) + α(B) ⇐⇒ (A + B)T= AT+ BT.α(rA) = r α(A) ⇐⇒ (rA)T= rAT.Hence α is linear.• β : M2,2(R) → R, β(A) = det A.Let A =1 00 0and B =0 00 1.Then A + B =1 00 1.We have det(A) = det(B) = 0 while det(A + B) = 1.Hence β(A + …
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