General Linear Transformations Definition We have seen that a linear transformation L from Rn to Rm is a function with domain Rn, 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 definition for a linear transformation of vector spaces. Definition Let V and W be vector spaces.  Then a linear transformation from V to W is a 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. Examples Example Let V be the vector space of (infinitely) differentiable functions and define D to be the function 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)) Example Let V be the space of continuous functions and define I to be the function from V to V given by         Then I is a linear transformation.  (Check this) Example Let 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         Example Let V  =  P2 and let W be the real numbers.  Show that the function L from V to W defined by         L(at2 + bt + c)  =  abc is not a linear transformation.   Solution  We can pick just about any example and show that either the first or second property does not hold.  For example, let         v  =  2t2 + 3t + 4        and        c  =  5 then         L(cv)  =  L(10t2 + 15t + 20)  =  (10)(15)(20)  =  3000 and         cL(v)  =  5L(2t2 + 3t + 4)  =  5(2)(3)(4)  =  120 since these are not equal, L is not a linear transformation. Properties When we looked at linear transformations from Rn to Rm, we stated and proved several properties.  A close look at these proofs will show that they only used the properties of vector spaces and linearity.  We now state the properties.  For each of the theorems below, assume that L is a linear transformation from a vector space V to a vector space W, and u, v, v1, v2, ... ,vn are vectors in V. Theorem 1.  L(0)  =  0 2.  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 by induction. 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 linear transformation from Rn to Rm.  The next theorem solidifies this fact. Theorem Let S  =  {v1, v2, ... ,vn} be a basis for V.  And let L be a linear transformation from V to a vector space W.  Then L is completely determined 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.   Proof If v is a vector in V, then since S is a basis, we can write         v  =  c1v1 + c2v2 + ... + ckvk  so that         L(v)  =  L(c1v1 + c2v2 + ... + ckvk)           =  c1L(v1) + c2L(v2) + ... + ckL(vk)    Example Let L be the linear transformation from P1 to M2x2 such that          Find L(3 + t).   Solution We 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)           Back to the Linear Algebra Home Page