MA2940F09SamplePrelim2Sol - Practice Prelim 2 Solutions 1....

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Unformatted text preview: Practice Prelim 2 Solutions 1. Note that P 4 is a 5-dimensional linear space with basis B = { 1 ,x,x 2 ,x 3 ,x 4 } . (a) In terms of the basis B , T a b c = a- 2 b + 3 c 3 a + 2 b + c a + 2 b- c a + c B = 1- 2 3 3 2 1 1 2- 1 1 1 a b c To find the image (and kernel) of T , row-reduce the matrix of T : 1- 2 3 3 2 1 1 2- 1 1 1 - 3 R 1- R 1- R 1- 1- 2 3 8- 8 4- 4 2- 2 8 4 2- 1- 2 3 1- 1 1- 1 1- 1 - R 2- R 2- 1- 2 3 1- 1 The pivots in the first and second columns indicate these columns are independent in the original matrix, so im( T ) = span 1 3 1 1 B , - 2 2 2 B = span 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 A basis for im( T ) is given by 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 . (b) Set p 1 ( x ) = 1 + 3 x + x 2 + x 4 and p 2 ( x ) =- 2 + 2 x + 2 x 2 . Then p ( x ) = ( a + c ) p 1 ( x ) + ( b- c ) p 2 ( x ). So the coordinates with respect to the basis 1 + 3 x + x 2 + x 4 ,- 2 + 2 x + 2 x 2 are a + c b- c . (c) For the kernel, the row-reduced matrix gives us the relations x 1- 2 x 2 + 3 x 3 = 0 and x 2- x 3 = 0. x 3 is a free variable, so set x 3 = t . Then x 2 = t and x 1 = 2 t- 3 t =- t . Thus, ker( T ) consists of all vectors - t t t , where t is any real number. So ker( T ) = span - 1 1 1 . Alternatively, this problem can be solved by inspection: if we notice the relationship among the columns: ~v 1- ~v 2 = ~v 3 , where ~v i is the i th column of the matrix for T , we obtain the same answers. (Since the rank of the matrix is 2, the rank-nullity theorem tells us the nullity must be 1, so we have found eveything in the kernel) 2. (a) W is not a subspace of V . It is not closed under scalar multiplication: I = 1 0 0 1 is in W since I 2 = I . But 2 I is not in W since (2 I ) 2 = 4 I 6 = 2 I . (b) W is a subspace of V : 0 (ie. the function f ( t ) = 0) is in W : f ( t ) = 0 for all t . In particular, f (3) = 0. W is closed under scalar multiplication: If f ( t ) is in W and k is a scalar, then d ( kf ) dt = k df dt for all t . So d ( kf ) dt t =3 = k df dt t =3 = k 0 = 0. Hence, kf ( t ) is in W ....
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MA2940F09SamplePrelim2Sol - Practice Prelim 2 Solutions 1....

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