Math 110 - Fall 05 - Lectures notes # 15 - Oct 3 (Monday) Our goal is to study inverses of linear transformations. Def: Let T: V -> W be linear. A function U: W -> V is called an inverse of U of T if (1) UT: V -> V is the identity function UT = I_V on V, i.e. I_V(v) = v for all v in V (2) TU: W -> W is the identity function TU = I_W on W, i.e. I_W(w) = w for all w in W If T has an inverse, we call T invertible, and write U = T^{-1} As noted in App B, inverse are unique. From now on, we will assume that V and W are finite dimensional. Recall our earlier theorem: Thm 1: T: V -> W is invertible if and only if rank(T) = dim(W) = dim(V). Lemma: (1) Let T: V -> W and S: W -> Z both be invertible. Then ST: V -> Z is invertible and (ST)^{-1} = T^{1} S^{1} (2) if it exists then T^{-1}: W -> V is linear (3) T^{-1}: W -> V is invertible too, and (T^{-1})^{-1} = T Proof: (1) If T is invertible it makes a 1-to-1 correspondence between V and W via T(v) = w. Similary S creates a 1-to-1 correspondence between W and Z via S(w) = z. So S(T(v)) = z is a 1-to-1 correspondence between V and Z so ST is invertible. If S(T(v)) = z, then we need to show v = T^{-1}(S^{-1}(z)): v = I_V (v) ... def of I_V = (T^{-1}T)(v) ... def of T^{-1} = T^{-1}(T(v)) ... def of function composition = T^{-1}(I_W(T(v))) ... def of I_W = T^{-1}(S^{-1}S)((T(v))) ... def of S^{-1} = (T^{-1}S^{-1})(S(T(v))) ... associativity of composition = (T^{-1}S^{-1})(z) ... def of S(T(v)) so T^{-1}S^{-1} is the inverse of ST (2) We need to show T^{-1}(c*x1+x2) = c*T^{-1}(x1) + T^{-1}(x2) Let T^{-1}(x1) = y1 and T^{-1}(x2) = y2, or T(y1) = x1 and T(y2) = x2 Then T^{-1}(c*x1 + x2) = T^{-1}(c*T(y1) + T(y2)) ... def of y1,y2 = T^{-1}(T(c*y1+y2)) ... T linear = c*y1+y2 ... T^{-1}T = I_V = c*T^{-1}(x1)+T^{-1}(x2) ... def of y1,y2 (3) Apply def of inverse to T^{-1} to see that its inverse is T Continuing our goal of connecting properties of linear transformations and matrices, we define: Def: Let A be an n-by-n matrix. Then A is invertible if there is an n-by-n matrix B such that AB = BA = I_n = n x n identify matrix Thm 2: Let V and W be finite dimensional vectors spaces with ordered bases beta and gamma, resp. Let T: V -> W be linear. Then T is invertible if and only if [T]_beta^gamma is invertible, in which case ([T]_beta^gamma)^{-1} = [T^{-1}]_gamma^beta Proof: Suppose T is invertible. Then by Thm 1 dim(V) = dim(W) = n and so [T]_beta^gamma is n x n. Now T^{-1}T = I_V and so I_n = [I_V]_beta^beta ... by def of I_V = [T^{-1}T]_beta^beta = [T^{-1}]_gamma^beta * [T]_beta^gamma ... by an earlier theorem and similarly I_n = [I_W]_gamma^gamma ... by def of I_W = [T T^{-1}]_gamma^gamma = [T]_beta^gamma * [T^{-1}]_gamma^beta So [T^{-1}]_gamma^beta satisfies the definition of the inverse of matrix [T]_beta^gamma. Now suppose A = [T]_beta^gamma is invertible; let B be the inverse. We need to show T is invertible and B=[T^{-1}]_gamma^beta Let beta = {v_1,...,v_n} and gamma = {w_1,...,w_n} and define U: W -> V by U(w_j) = sum_{i=1 to n} B_ij*v_i. We will show that U = T^{-1}: TU(w_j) = T(sum_{i=1 to n} B_ij*v_i) ... by def of U(w_j) = sum_{i=1 to n} B_ij*T(v_i) ... since T linear = sum_{i=1 to n} B_ij*(sum_{k=1 to n} A_ki*w_k) ... by def of T(v_i) = sum_{i=1 to n} sum_{k=1 to n} B_ij*A_ki*w_k ... move B_ij into summation = sum_{k=1 to n} sum_{i=1 to n} B_ij*A_ki*w_k ... reverse order of summation = sum_{k=1 to n} w_k * (sum_{i=1 to n} B_ij*A_ki) ... move w_k out of inner summation = sum_{k=1 to n} w_k * (AB)_kj ... def of matrix product AB = sum_{k=1 to n} w_k * (I_n)_kj ... since AB = I_n = w_j ... def of I_n Since I_W(w_j) = w_j for all j too, by uniqueness we have TU = I_W One can similarly show UT(v_i) = v_i and so UT = I_V Ex: T: R^2 -> R^2 where T(x1,x2) = (x1 - 2*x2, -2*x1 + x2) beta = gamma = {e_1,e_2} standard ordered basis. T is invertible because we can solve T(x1,x2) = (y1,y2) for (x1,x2) given any (y1,y2) as follows: y1 = x1 - 2*x2 => y2+2*y1 = -3*x2 => x2 = (-2/3)*y1 + (-1/3)*y2 y2 = -2*x1 + x2 => x1 = y1 + 2*x2 = (-1/3)*y1 + (-2/3)*y2 and T^{-1}(y1,y2) = ( (-1/3)*y1 + (-2/3)*y2, (-2/3)*y1 + (-1/3)*y2 ) so [T^{-1}]_gamma^beta = [ -1/3 -2/3 ; -2/3 -1/3 ] Also [T]_beta^gamma = [ [T*e1]_gamma, [T*e2]_gamma ] = [ [1; -2]_gamma, [-2 ; 1]_gamma ] = [ 1 -2 ; -2 1 ] and so ([T]_beta^gamma)^{-1} = [ 1 2 ; 2 1 ]/det = [1 2 ; 2 1]/(-3) = [T^{-1}]_gamma^beta ... as expected Corollary: If V = W and beta = gamma, and T : V -> W is invertible, then ([T]_beta)^{-1} = [T^{-1}]_beta Proof: Follows from Thm 2. Corollary: Let A be an n x n matrix. Then A is invertible if and only if L_A is invertible Proof: Homework! One-to-one correspondences have a special name when they are linear: Def: Let V and W be vector spaces. If there is an invertible linear transformation T: V -> W, then we say V and W are isomorphic vector spaces, and T is called an isomorphism. Ex: V = R^2, W = P_1(R), T((a1,a2)) = a1 + a2*x is an isomorphism Thm: Let V and W be finite dimensional vector spaces. Then they are isomorphic if and only if dim(V) = dim(W). Proof: If V and W are isomorphic, let T: V -> W be an isomorphism. Then we know T is invertible if and only if rank(T) = dim(V) = dim(W). Now suppose dim(V) = dim(W) = n. Let beta = {v_1,..,v_n} and gamma = {w_1,...,w_n} be ordered bases for V and W, resp. Define T by T(v_i) = w_i for i=1,...,n. Then R(T) = span(T(v_1),..,T(v_n)) = span(w_1,..,w_n) = W so rank(T) = n = dim(W) = dim(V), so T is an isomorphism. Corollary: Let V be a vector space over F. Then V is isomorphic to F^n if and only if n = dim(V). Thm: Let V and W have finite dimensions n and m, resp., with ordered bases beta and gamma, resp. Then Phi: L(V,W) -> M_{m x n}(F) defined by Phi(T) = [T]_beta^gamma is an isomorphism Proof: We have proved most of this already: We know Phi is linear. It is one-to-one because the property of a basis implies that sum_i a_i*w_i = 0 if and only if all a_i = 0, so the only way [T]_beta^gamma can be the zero matrix is if T(v_i) = 0_W for all i, implying T(v) = 0 for all v. It is onto because given a matrix A, if we define T(v_i) = sum_{j=1 to n} A_ji*w_j then T is linear and [T]_beta^gamma = A.