MATH 311-504Topics in Applied MathematicsLecture 2-5:Image and null-space (continued).General linear equations.Image and null-spaceLet V1, V2be vector spaces and f : V1→ V2be alinear mapping.V1: the domain of fV2: the range of fDefinition. The image of f (denoted Im f ) is theset of all vectors y ∈ V2such that y = f (x) forsome x ∈ V1. The null-space of f (denotedNull f ) is the set of all vectors x ∈ V1such thatf (x) = 0.Theorem The image of f is a subspace of therange. The null-space of f is a subspace of thedomain.More examples• M : P → P, (Mp)(x) = xp(x).p(x) = a0+ a1x + · · · + anxn=⇒ (Mp)(x) = a0x + a1x2+ · · · + anxn+1Null M = {0}, Im M = {p(x) ∈ P : p(0) = 0}.• I : P → P, (Ip)(x) =Zx0p(s) ds.p(x) = a0+ a1x + · · · + anxn=⇒ (Ip)(x) = a0x +12a1x2+ · · · +1n+1anxn+1Null I = {0}, Im I = {p(x) ∈ P : p(0) = 0}.General linear equationsDefinition. A linear equation is an equation of the formL(x) = b,where L : V → W i s a linear mapping, b is a given vectorfrom W , and x is an unknown vector f ro m V .The image of L is the set of all vectors b ∈ W such that theequation L(x) = b has a solution.The null-space of L is the solution set of the homogeneouslinear equation 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 spanning setfor the null-space o f L, and t1, . . . , tkare arbitrary scalars.Theorem If the linear equation L(x) = b issolvable then the general solution isx0+ t1v1+ · · · + tkvk,where x0is a particular s olution, v1, . . . , vkis aspanning for the null-space of L, and t1, . . . , tkarearbitrary scalars.Proof: Let x = x0+ t1v1+ · · · + tkvk. ThenL(x) = L(x0) + t1L(v1) + · · · + tkL(vk) = b.Conversely, if L(x) = b thenL(x − x0) = L(x) − L(x0) = b − b = 0.Hence x − x0belongs to Null L. It follows thatx − x0= t1v1+ · · · + tkvkfor some t1, . . . , tk∈ R.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 − 2 t, −1 + t, t) = (5, −1, 0) + t(−2, 1, 1).Example. u′′(x) + u(x) = e2x.Linear oper ator L : C2(R) → C (R), Lu = u′′+ u.Linear equation: Lu = b, where b(x) = e2x.It can be shown that the image of L is the entirespace C (R) while the null-space of L is spanned bythe functions sin x and cos x.Observe that(Lb)(x) = b′′(x)+b(x) = 4e2x+e2x= 5e2x= 5b(x).By linearity, u0=15b is a particul ar solution.Thus the general solution isu(x) =15e2x+ t1sin x + t2cos x.Let V1, V2be vector spaces and f : V1→ V2be alinear mapping.Definition. The map f is one-to-one if it mapsdifferent vectors from V1to different vectors in V2.That is, for any x, y ∈ V1we have thatx 6= y =⇒ f (x) 6= f (y).Theorem A linear mapping f is one-to-one if andonly if Null f = {0}.Proof: If a vector x 6= 01belongs to Null f , thenf (x) = 02= f (01) =⇒ f is not one-to-one.On the other hand, if Null f is trivial thenx 6= y =⇒ x − y 6= 0 =⇒ f (x − y) 6= 0=⇒ f (x) − f (y) 6= 0 =⇒ f (x) 6= f (y).Let f : V1→ V2be a linear mapping.Definition. The map f is onto if any vector fromV2is the image under f of some vector from V1.That is, if Im f = V2.If the mapping f is both one-to-o ne and onto, thenany y ∈ V2is uniquely represented as f (x), wherex ∈ V1. In this case, we define the inversemapping f−1by f−1(y) = x ⇐⇒ f ( x) = y.If the mapping f is only one-to-one, we can stilldefine the inverse mapping f−1: Im f → V1.Theorem The inverse of a linear mapping is alsolinear.Examples. • f : R2→ R3, f (x, y ) = (x, y , x).Null f = {0}, Im f is the plane x = z.The inverse mapping f−1: Im f → R2is given by(x, y , z) 7→ (x, y ).• g : R2→ R2, g(x) = Ax, where A =1 21 3.det A = 1 =⇒ A is invertibl e.g is one-to-one sinceAx = 0 =⇒ x = A−1(Ax) = A−10 = 0.g is onto since y = A(A−1y) for any y ∈ R2.The inverse mapping is given by g−1(y) =
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