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Unformatted text preview: SOLUTION TO ASSIGNMENT 2  MATH 251, WINTER 2007 1. Let B = { (1 , 1) , (1 , 5) } and C = { (2 , 1) , (1 , 1) } be bases of R 2 . Find the change of basis matrices B M C and C M B between the bases B and C . Let v = ( 8 28 ) with respect to the standard basis. Find [ v ] B and [ v ] C . Proof. We have St M B = 1 1 1 5 , St M C = 2 1 1 1 . Thus, B M St = 1 4 5 1 1 1 . Therefore, B M C = 1 4 5 1 1 1 2 1 1 1 = 1 4 9 6 1 2 , and C M B = ( B M C ) 1 = 1 3 2 6 1 9 . We have 8 28 B = B M St 8 28 = 1 4 5 1 1 1 8 28 = 3 5 and 8 28 C = C M B 8 28 B = 1 3 2 6 1 9 3 5 = 12 16 . 2. Which of the following functions is a linear map? (provide proof): (1) T : R 2 → R 2 , T ( x,y ) = (3 x 2 y,x + y ). (2) T : R 2 → R 2 , T ( x,y ) = ( x 2 y,x + y + 1). (3) T : R [ x ] 3 → R 2 , T ( f ( x )) = ( f (1) ,f (1)). (4) T : R [ x ] 3 → R [ x ] 4 , T ( f ( x )) = xf ( x ) + f (1) . In each case where T is a linear map, find its kernel. Proof. (1) T is a linear map. Its kernel are the vectors ( x,y ) s.t. 3 x = 2 y,x = y , namely, just (0 , 0). (2) T is not a linear map. T (2 , 0) = (4 , 3), while 2 T (1 , 0) = 2(1 , 2) = (2 , 4) 6 = T (2 , 0). (3) T is a linear map. Its kernel are the polynomials that vanish at 1 and whose derivatives vanish at 1. Writing a polynomial as a 3 x 3 + a 2 x 2 + a 1 x + a , the conditions are a 3 + a 2 + a 1 + a = 0 , 3 a 3 + 2 a 2 + a 1 = 0 . (4) Again, T is a linear map. Its kernel is just 0, because if f is not zero then xf ( x ) is a nonzero polynomial with no constant term, so for every scalar r , xf ( x ) + r is not zero, in particular, xf ( x ) + f (1) is not zero. 1 2 SOLUTION TO ASSIGNMENT 2  MATH 251, WINTER 2007 3. Let T : V → W be a surjective linear map between vector spaces of finite dimension. Prove that if dim( V ) = dim( W ) then T is an isomorphism. Proof. We have dim(Ker( T )) = dim( V ) dim(Im( T )) = dim( W ) dim( W ) = 0 and so Ker( T ) can only be the zero subspace. Thus, T is injective. 4. Prove the following Proposition. Proposition 0.1. Let V and W be vector spaces over F . Let B = { b 1 ,...,b n } be a basis for V and let t 1 ,...,t n be any elements of W . There is a unique linear map T : V → W, such that T ( b i ) = t i , i = 1 ,...,n. Proof. Define T ( n X i =1 α i b i ) = n X i =1 α i t i . Since any vector in V can be written uniquely as ∑ n i =1 α i b i this gives a well defined function T : V → W , which clearly satisfies T ( b i ) = t i . It only remains to check that T is linear: T ( α n X i =1 α i b i ) = T ( n X i =1 αα i b i ) = n X i =1 αα i t i = α n X i =1 α i t i = αT ( n X i =1 α i b i ) , and T ( n X i =1 α i b i + n X i =1 β i b i ) = T ( n X i =1 ( α i + β i ) b i ) = n X i =1 ( α i + β i ) t i = n X i =1 α i t i + n X i =1 β i t i = T ( n X i =1 α i b i )+ T ( n X i =1 β i b i ) ....
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This note was uploaded on 08/26/2008 for the course MATH 251 taught by Professor Goren during the Spring '08 term at McGill.
 Spring '08
 Goren
 Matrices

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