MTH5102 Spring 2017 HW Assignment 4: Sec. 2.2, #3, 5; Sec. 2.3, #17; Sec. 2.4, #14, 17 The due date for this assignment is 2/22/
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1 MTH50 Spring 07 HW Assignment : Sec.. # 5; Sec.. #7; Sec.. # 7 The due date for this assignment is //7. Sec.. #. Let T : R R be defined by T (a a = (a a a a + a. Let β be the standard ordered basis for R and γ = {( 0 (0 ( }. Compute [T ] γ β. If α = {( ( } compute [T ]α β. Solution. First we have ( T ( 0 = ( = ( 0 + [ ( 0 ( ] Next = ( ( 0 + ( ( T (0 = ( 0 = ( ( 0 + ( (0 [T ] γ β [T ] γ β Sec.. #5. Let α = ( [T ] γ β = [T ( 0] γ [T (0 ] γ =. 0 ( = ( = ( = 0 ( = 0 ( [T ] γ α = [T ( ] γ [T ( ] γ = 7. {( 0 ( β = { x x } and γ = {}. ( 0 7 ( } (a Define T : M (F M (F by T (A = A t. Compute [T ] α. (b Define ( f T : P (R R by T (f (x = (0 f ( 0 f (
2 where denotes differentiation. Compute [T ] γ α. (c Define T : M (F F by tr (A. Compute [T ] γ α. (d Define T : P (R R by T (f (x = f (. Compute [T ] γ β. (e If ( A = 0 compute [A] α. (f If f (x = 6x + x compute [f (x] β. (g For a F compute [a] γ. Solution. (a We have 0 T = T = 0 T = T 0 ( = ( (b We have 0 [T ] α = 0. 0 T ( = = ( T (x = = ( + T ( x ( 0 = = [T ] γ α = 0.
3 (c We have ( 0 T = ( T = 0 ( T = 0 0 ( T = (d We have (e We have A = [T ] γ α = (. T ( = T (x = T ( x = ( = ( 0 [T ] γ β = (. ( 0 + ( [A] α = 0. ( + ( (f We have f (x = 6x + x = ( + ( 6 x + ( x [f (x] β = 6. (g For a F we have a = (a [a] γ = (a. Sec.. #7. Let V be a vector space. Determine all linear transformations T : V V such that T = T. Solution. Denote by L (V the set of all linear transformations from V into V. Define by S L (V the set S = { T L (V : T = T }.
4 Define by S L (V the set S = {P L (V : there exists subspaces M N V with V = M N such that P (x = x P (y = 0 for all x M y N}. We will prove that S = S. First if P S then there exists subspaces M and N of V such that V = M N with P (x = x for all x M and P (y = 0 for all y N. Then P = P since for any z V there exists x M and y N such that z = x + y P (z = P (x + y = P (x + P (y = x = P (x = P (P (x = P (P (z = P (z. Thus S S. Now we will prove the reverse inclusion. Let T S then define M by M = {x V : T (x = x} and define N = N (T = {y V : T (y = 0} i.e. the nullspace of T. It follows immediately that M and N are subspaces of V. We will show next that V = M N. Suppose that w M N then 0 = T (w = w M N = {0}. This proves that for the W = M + N V is a subspace of V and that W = M N. Now we will prove that W = V. If z V then z = T (z + (z T (z and since T (T (z = T (z = T (z T [z T (z] = T (z T (z = 0 this implies that z W. Thus W = M N = V. This implies that S S. Therefore S = S as desired. Finally for a characterization of S we construct" each element of S. Let M and N be subspaces of V such that M N = V. Define a map P : V V by P (x = x for all x M P (y = y for all y N and P (u + v = u for all u M v N. Then we will prove now that P S. First we must show that P is a well-defined function. To do this notice that P is a well-defined function on M and N separately. Suppose that x + y = x + y for some x x M and some y y N. Then since V = M N this implies that M N = {0} and hence x x = y y M N = {0} which means x = x P (x + y = x = x = P (x + y. This proves P is a well-defined function on V into V. Next we will prove that P is linear. Let z w V and c a scalar. Then since M N = V there exists x u M and y v N such that z = x + y w = u + v implying cx + u M and cy + v N and P (cz + w = P ((cx + u + (cy + v = cx + u = cp (z + P (w
5 which proves that P L (V. This also proves by definition of P that P S and hence P S. This completes the characterization. Sec.. #. Let {( } a a + b V = : a b c F. Construct an isomorphism from V to F. Solution. Define a map Φ : V F by ( a a + b Φ = (a b c. We will prove that Φ is a well-defined function which is linear and bijective (hence an an isomorphism from V to F. First if for some a b c d e f F we had a a + b d d + e = 0 f then this implies that a = d a+b = d+e c = f implying that a = d b = e c = f and hence (a b c = (d e f F. This proves Φ : V F is a well-defined function. Next if z w V and s F then there exists a b c d e f F such that a a + b d d + e z = w = 0 f and hence [ Φ (sz + w = Φ s = Φ ( a a + b + ( sa + d sa + d + sb + e 0 sc + f = (sa + d sb + e sc + f ( ] d d + e 0 f = s (a b c + (d e f a a + b d d + e = sφ + Φ. 0 f This proves that Φ : V F is a linear transformation. Moreover its obvious that Φ is onto. Finally as V has as a basis { ( } γ = then dim (V = dim ( F = and so by Theorem.5 Sec.. and the fact that Φ is onto implies that Φ is an isomorphism. This completes the construction. 5
6 Sec.. #7. Let V and W be finite-dimensional vector spaces and T : V W be an isomorphism. Let V 0 be a subspace of V. (a Prove that T (V 0 is a subspace of W. (b Prove that dim (V 0 = dim (T (V 0. Proof. (a We will use Theorem. Sec.. to prove this result. First 0 V 0 since V 0 is a subspace of V and since T : V W is linear then 0 = T (0 W. Second if x y T (V 0 then there exists a u v V 0 such that x = T (u y = T (v and since V 0 is a subspace then u + v V 0 and since T : V W is linear then x + y = T (u + T (v = T (u + v T (V 0. Finally if c F and x T (V 0 then there exists a u V 0 such that x = T (u and since V 0 is a subspace then cu V 0 and since T : V W is linear then cx = ct (u = T (cu T (V 0. Therefore by Theorem. Sec.. T (V 0 is a subspace of W. (b As T : V W be an isomorphism then its restriction T V0 to the subspace V 0 i.e. T V0 : V 0 T (V 0 defined by T V0 (u = T (u for all u V 0 is also an isomorphism implying by the Dimension Theorem Theorem. of Sec.. This completes the proof. dim (V 0 = nullity (T V0 + rank (T V0 = rank (T V0 = dim R (T V0 = dim (T (V 0. 6
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