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General Linear TransformationsDenitionWe have seen that a linear transformation L from Rn to Rm is a function with domainRn, range a subset of Rm satisfying

1) L(u + v) = L(u) + L(v) 2) L(cu) = cL(u)for any vectors u and v and scalar c.We can use the analogous denition for a linear transformation of vector spaces.

DenitionLet V and W be vector spaces. Then a linear transformation from V to W isa function with domain V and range a subset of W satisfying 1) L(u + v) = L(u) + L(v) 2) L(cu) = cL(u)for any vectors u and v in V and scalar c.

ExamplesExampleLet V be the vector space of (innitely) dierentiable functions and dene D to be thefunction from V to V given by

D(f(t)) = f '(t)Then D is a linear transformation since D(f(t) + g(t)) = (f(t) + g(t))' = f '(t) + g'(t) = D(f(t)) + D(g(t))and

D(cf(t)) = (cf(t))' = c f '(t) = cD(f(t))

ExampleLet V be the space of continuous functions and dene I to be the function from V to Vgiven by

General Linear Transformations http://ltcconline.net/greenl/courses/203/MatrixOnVector...

1 of 5 Saturday 23 August 2014 07:09 PM

Then I is a linear transformation. (Check this)

ExampleLet V be M2x2 and W be P3 then define L to be the function from V to W with

Then L is a linear transformation since

= (a1 + a2)t3 + (b1 + b2)t2 + (c1 + c2)t + (d1 + d2) = a1t3 + b1t2 + c1t + d1 + a2t3 + b2t2 + c2t + d2

and

ExampleLet V = P2 and let W be the real numbers. Show that the function L from V to W defined by

L(at2 + bt + c) = abcis not a linear transformation.



Solution We can pick just about any example and show that either the first or second property does not hold. Forexample, let

v = 2t2 + 3t + 4 and c = 5then

L(cv) = L(10t2 + 15t + 20) = (10)(15)(20) = 3000and

cL(v) = 5L(2t2 + 3t + 4) = 5(2)(3)(4) = 120since these are not equal, L is not a linear transformation.

PropertiesWhen we looked at linear transformations from Rn to Rm, we stated and provedseveral properties. A close look at these proofs will show that they only used theproperties of vector spaces and linearity. We now state the properties. For each ofthe theorems below, assume that L is a linear transformation from a vector space V toa vector space W, and u, v, v1, v2, ... ,vn are vectors in V.Theorem1. L(0) = 02. L(u - v) = L(u) - L(v)3. L(c1v1 + c2v2 + ... + cnvn) = c1L(v1) + c2L(v2) + ... + cnL(vn) We will prove statement 3 and leave the rest for you. We prove the statement byinduction.For n = 1, the statement is just property 2 of a linear transformation.

L(c1v1) = c1L(v1)Now assume that the statement is true for n = k. Then

L(c1v1 + c2v2 + ... + ckvk) = c1L(v1) + c2L(v2) + ... + ckL(vk) We have



L(c1v1 + c2v2 + ... + ckvk + ck+1vk+1) = L((c1v1 + c2v2 + ... + ckvk) + ck+1vk+1) = L(c1v1 + c2v2 + ... + ckvk) + L(ck+1vk+1) = c1L(v1) + c2L(v2) + ... + ckL(vk) + ck+1L(vk+1)

So by mathematical induction the theorem is true.

We have seen that general linear transformations behave the same as lineartransformation from Rn to Rm. The next theorem solidies this fact.TheoremLet S = {v1, v2, ... ,vn} be a basis for V. And let L be a lineartransformation from V to a vector space W. Then L is completelydetermined by the image of the basis S.This means that if we know L(, L(v2), ... ,L(vn) then we know L(v) for any vector v.

ProofIf v is a vector in V, then since S is a basis, we can write

v = c1v1 + c2v2 + ... + ckvkso that

L(v) = L(c1v1 + c2v2 + ... + ckvk) = c1L(v1) + c2L(v2) + ... + ckL(vk)

ExampleLet L be the linear transformation from P1 to M2x2 such that

Find L(3 + t).



SolutionWe need to find the coordinates of v = 3 + t with respect to the basis S = {1 + t, 1 - t}. We have

so that v = 2(1 + t) + 1(1 - t)

and L(v) = L(2(1 + t) + 1(1 - t)) = 2L(1 + t) + L(1 - t)

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