MATH 304Linear AlgebraLecture 21:General linear equations.Matrix transformations.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 properties 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.Range and kernelLet V , W be vector spaces and L : V → W be alinear mapping.Definition. The range (or image) of L is th e setof all vectors w ∈ W such that w = L(v) for somev ∈ V . The range of L is denoted L(V ).The kernel of L, denoted ker L, is the set of allvectors v ∈ V such that L(v) = 0.Theorem (i) The range of L is a subspace of W .(ii) T he kernel of L is a subspace of V .General linear equationsDefinition. A linear equation is an equation of the formL(x) = b,where L : V → W is a linear mapping, b is a given vectorfrom W , and x is an unknown vector from V .The range of L is the set of all vectors b ∈ W such that theequation L(x) = b has a solution.The kernel of L is the solution set of the homogeneous linearequation L(x) = 0.Theorem If the linear equation L(x) = b is solvable then thegeneral solution isx0+ t1v1+ ··· + tkvk,where x0is a particular solution, v1, . . . , vkis a basis for thekernel of L, and t1, . . . , tkare arbitrary scalars.Example.x + y + z = 4,x + 2y = 3.L : R3→ R2, Lxyz=1 1 11 2 0xyz.Linear equation: L(x) = b, where b =43.1 1 141 2 03→1 1 1 401 −1 −1→1 0 2 501 −1 −1x + 2z = 5y − z = −1⇐⇒x = 5 − 2zy = −1 + z(x, y, z) = (5 − 2t, −1 + t, t) = (5, −1, 0) + t(−2, 1, 1).Example. u′′(x) + u(x) = e2x.Linear operator L : C2(R) → C (R), Lu = u′′+ u.Linear equation: Lu = b, where b(x) = e2x.It can be shown that the range of L is the entirespace C (R) while the kernel of L is spanned by thefunctions sin x and cos x.Observe that(Lb)(x) = b′′(x)+b(x) = 4e2x+e2x= 5e2x= 5b(x).By linearity, u0=15b is a particular solution.Thus the general solution isu(x) =15e2x+ t1sin x + t2cos x.Matrix transformationsAny m×n matrix A gives rise to a transformationL : Rn→ Rmgiven by L(x) = Ax, wh ere x ∈ Rnand L(x) ∈ Rmare regarded as column vectors.This transformation is linear.Indeed, L(x + y) = A(x + y) = Ax + Ay = L(x) + L(y),L(rx) = A(r x) = r (Ax) = rL(x).Example. Lxyz=1 0 23 4 70 5 8xyz.Let e1= (1, 0, 0), e2= (0, 1, 0), e3= (0, 0, 1) be thestandard bas is for R3. We ha ve that L(e1) = (1, 3, 0),L(e2) = (0, 4, 5), L(e3) = (2, 7, 8). Thus L(e1), L(e2), L(e3)are columns of the matrix.Problem. Find a linear mapping L : R3→ R2such that L(e1) = (1, 1), L(e2) = (0, −2),L(e3) = (3, 0), where e1, e2, e3is the standardbasis for R3.L(x, y , z) = L(xe1+ ye2+ ze3)= xL(e1) + yL(e2) + zL(e3)= x(1, 1) + y (0, −2) + z(3, 0) = (x + 3z, x − 2y)L(x, y , z) =x + 3zx − 2y=1 0 31 −2 0xyzColumns of the matrix are vectors L(e1), L(e2), L(e3).Theorem Supp ose L : Rn→ Rmis a linear map. Thenthere exists an m×n matrix A such that L(x) = Ax for allx ∈ Rn. Columns of A are vectors L(e1), L(e2), . . . , L(en),where e1, e2, . . . , enis the standard basis for Rn.y = Ax ⇐⇒y1y2...ym=a11a12. . . a1na21a22. . . a2n............am1am2. . . amnx1x2...xn⇐⇒y1y2...ym= x1a11a21...am1+ x2a12a22...am2+ ··· + xna1na2n...amnLinear transformations of R2Any linear mapping f : R2→ R2is represented asmultiplication of a 2-dimensional column vector by a2×2 matrix: f (x) = Ax orfxy=a bc dxy.Linear transformations corresponding to particularmatrices can have various geometric properties.A =0 −11 0Rotation by 90oA = 1√2−1√21√21√2!Rotation by 45oA =−1 00 1Reflection inthe vertical axisA =0 11 0Reflection inthe line x − y = 0A =1 1/20 1Horizontal shearA =1/2 00 1/2ScalingA =3 00 1/3SqueezeA =1 00 0Vertical projection onthe horizontal axisA =0 −10 1Horizontal projectionon the line x + y = 0A =1 00
View Full Document
Unlocking...