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Math 105 Semester Review - <strong>Solutions</strong> 3. u⊕( v⊕ w) = ( u1, u2) ⊕ ( v1+ w1+ 1, v2 + w2 + 1) = ( u1+ ( v1+ w1+ 1) + 1, u2 + ( v2 + w2 + 1) + 1) = (( u1+ v1+ 1) + w1+ 1, ( u2 + v2 + 1) + w2 + 1) = ( u1+ v1+ 1, u2 + v2 + 1 ) ⊕( w1, w2) = ( u⊕v) ⊕w 4. 0= 0 ( u1, u2) = ( −1, −1) u⊕ 0 = ( u ) ( ) ( ) ( ) u 1, u2 ⊕ −1, − 1 = u1− 1+ 1, u2 − 1+ 1 = u1, u2 = 5. − u = ( − 1 ) ( u1, u2) = ( −1−u1−1, −1−u2 − 1) = ( −2 −u1, −2−u1) u⊕− u = ( u1, u2) ⊕( −2 −u1, −2−u1) = ( u −2− u + 1, u −2− u + 1) = ( −1, − 1) = 0 1 1 2 2 k u1, u2 = k+ ku1− 1, k+ ku2 −1 ∈ k+ l u = k+ l u , u = k+ l + k+ l u − 1, k+ 1 + k+ l u −1 6. ( ) ( ) 2 ( ) ) 7. ( ) ( ) ( ) ( ) ( ) ( ) ( ) = (( k+ ku1− ) + ( l+ lu1− ) + ( k+ ku2 − ) + ( l+ lu2 − ) + = ( k+ ku1− 1, k+ ku2 −1) ⊕ ( l+ lu1− 1, l+ lu2 −1) = ( ku) ⊕( lu) 8. k( u⊕ v) = k( u1+ v1+ 1, u2 + v2 + 1) = ( k+ k( u1+ v1+ 1) − 1, l+ l( u2 + v2 + 1) −1) (( k ku1 1) ( k kv1 1) 1, ( k ku2 1) ( k kv2 1) = ( k+ ku1− 1, k+ ku2 −1) ⊕ ( k+ kv1− 1, k+ kv2 −1) = ( ku) ⊕( kv) 9. ( kl ) u = ( kl + klu1− 1, kl + klu2 −1) = ( k+ k( l+ lu1−1) − 1, k+ k( l+ lu2 −1) −1) = k ( l+ lu1− 1, l+ lu2 −1) = k( lu) 10. 1u = 1 ( u , u ) = ( 1+ u − 1,1+ u − 1 ) = ( u , u ) = u 1 2 1 2 1 1 1, 1 1 1 = + − + + − + + − + + − + 1 2 1 2 1 2 1) Winter 2006 Martin Huard 18
Math 105 Semester Review - <strong>Solutions</strong> 21. Is the set W a subspace of V Support your answer. ⎧ ⎡ a a+ b⎤ ⎫ a) W = ⎨⎢ : a, b∈ ⎬ a− b b ⎥ V = M2,2 ⎩⎣ ⎦ ⎭ ⎡0 0⎤ Yes W is nonempty since ⎢ ∈W 0 0 ⎥ . ⎣ ⎦ ⎡ a a+ b⎤ Let A = ⎢ a− b b ⎥ ⎣ ⎦ and ⎡ r r+ s⎤ B = ⎢ ∈W r− s s ⎥ . ⎣ ⎦ ⎡ a+ r a+ b+ r+ s⎤ ⎡ a+ r ( a+ r) + ( b+ s) ⎤ 1. A + B = ⎢ a b r s b s ⎥ = ⎢ ∈ ( a r) ( b s) ⎥ W ⎣ − + − + ⎦ ⎣ + − + b+ s ⎦ 2. ( + ) ⎡ ka k a b ⎤ ⎡ ka ka + kb⎤ kA = ⎢ ⎥ = W k( a− b) kb ⎢ ka − kb kb ⎥∈ ⎣ ⎦ ⎣ ⎦ b) W { A: A is nilpotent, A M2,2} 2 = ∈ V = M2,2 (A is nilpotent if A = 0 ) 0 1 No. Let A ⎡ ⎤ 0 0 = ⎢⎣ and 0 0 ⎥ B = ⎡ ⎤ ⎢ ⎦ 1 0 ⎥ ⎣ ⎦ . Then 2 2 A = 0 and B = 0 , so A, B∈ V but 0 1 A B ⎡ ⎤ + = ⎢ ∉V since 1 0 ⎥ ( ) 2 ⎡1 0⎤ A+ B = ⎢ ≠0 ⎣ ⎦ 0 1 ⎥ ⎣ ⎦ W = ax 3 −b: a, b∈ V = P3 c) { } Yes W is nonempty since 0∈ W . 3 p x = ax −b and q( x) = cx −d∈ W Let ( ) 3 p x + q x = ax − b+ cx − d = a+ c x − b+ d ∈ W 3 3 3 1. ( ) ( ) ( ) ( ) 3 3 2. ( ) ( ) { : 1 2 , ( ) 2 } kp x = k ax − b = akx −bk ∈W d) W = p( x) p( ) = p( ) p x ∈ P V = P2 Yes V is nonempty since p( x) = 0∈ V since p( 1) = 0= p( 2). Let p( x) , q( x) ∈ V . Then p( 1) = p( 2 ) and q( 1) = q( 2) . 1. ( p + q)( x) ∈ V since ( p+ q)( 1) = p( 1) + q( 1) = p( 2) + q( 2) = ( p+ q)( 2 ) 2. ( kp)( x) ∈ V since ( kp)( 1) = kp ( 1) = kp ( 2) = ( kp)( 2 ) 3 e) W = ( x, y, z) :2x− 3y+ z = 0, x, y, z∈ V = { } Yes W is nonempty since 0∈W Let u = x , y , z ∈W and v = x , y , z ∈W . ( ) 1 1 1 ( ) 2 2 2 Then 2x1− 3y1+ z1 =0 and 2x2 − 3y2 + z2 = 0 u+ v = x + x , y + y , z + z ∈W 1. ( 1 2 1 2 1 2) since 2( x + x ) − 3( y + y ) + ( z + z ) = ( 2x − 3y + z ) + ( 2x − 3y + z ) 1 2 1 2 1 2 1 1 1 2 2 2 = 0+ 0= 0 Winter 2006 Martin Huard 19
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