1 Linear Transformations We will study mainly nite-dimensional vector spaces over an arbitrary eld Fi.e. vector spaces with a basis. (Recall that the dimension of a vector space V (dimV ) is the number of elements in a basis of V .) DEFINITION 1.1 (Linear transformation) Given vector spaces U and V , T : U V is a linear transformation (LT) if T(u + v) = T(u) + T(v) for all , F, and u,v U. Then T(u+v) = T(u)+T(v), T(u) = T(u) and n T u = T(u ). k k k k k=1 k=1 EXAMPLES 1.1 Consider the linear transformation T = TA: Vn(F) m (F) where A = [ij] is m n, dened Ay T (X) = AX. x1 Note that V (F) = the set of all n-dimensional column ector of n . xn Fsometimes written F . Note that if Tn: V (F)m V (F) is a linear transformationA then T = T , where A = [T(E )T(E )] and 1 n 1 0 0 . E = ,...,E = . 1 . n 0 0 1 Note: x1 v V (F), v = . = x E + + x E n . 1 1 n n xn 1 www.notesolution.comIf V is a vector space of all innitely dierentiable functions on R, then T(f) = a D f + a D n1f + + a Df + a f 0 1 n1 n denes a linear transformation T : V V . The set of f such that T(f) = 0 (i.e. the kernel of T) is important. Let T : U V be a linear transformation. Then we have the following denition: DEFINITIONS 1.1 (Kernel of a linear transformation) KerT = {u U T(u) = 0} (Image of T) ImT = {v V u U such that T(u) = v} Note: KerT is a subspace of U. Recall that W is a subspace of U if 1. 0 W, 2. W is closed under addition, and 3. W is closed under scalar multiplication. PROOF. that KerT is a subspace of U: 1. T(0) + 0 = T(0) = T(0 + 0) = T(0) + T(0). Thus T(0) = 0, so 0 KerT. 2. Let u,v KerT; then T(u) = 0 and T(v) = 0. So T(u + v) = T(u) + T(v) = 0 + 0 = 0 and u + v KerT. 3. Let u KerT and F. Then T(u) = T(u) = 0 = 0. So u KerT. EXAMPLE 1.1 KerT A = N(A), the null space of A = {X V (n) AX = 0} and ImT = C(A), the column space of A A = A ,1.,A n 2 www.notesolution.com

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