test3-review_Fall2010

test3-review_Fall2010 - Solutions to practice problems for...

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Solutions to practice problems for Test 3 1. Label the following statements as true or false . true (a) Every vector space that is spanned by a finite set has a finite basis. false (b) A vector space cannot have more than one basis. true (c) If a vector space has a finite basis, then the number of vectors in every basis is the same. false (d) The dimension of P n is n . true (e) There exists an isomorphism from P 3 to R 4 . true (f) There exists an isomorphism from R n to R m if and only if n = m . true (g) If F is a linear transformation from R 5 to R 3 and the dimension of the kernel of F is 2, then F is onto. false (h) Suppose that V is a finite-dimensional vector space, that S 1 is a linearly independent subset of V , and that S 2 is a subset of V that spans V . Then S 1 can contain more vectors than S 2 . true (i) If S is a basis for the vector space V , then every vector in V can be written as a linear combination of vectors in S in only one way. true (j) If V is a vector space having dimension n , and if S is a subset of V with n vectors, then S is linearly independent if and only if S spans V . true (k) If X and Y are two bases for a vector space V , then the change of basis matrix Y I X is invertible and ( Y I X ) 1 = X I Y . true (l) If F is an isomorphism from a vector space V to a vector space W , then X is a basis of V if and only if F ( X ) is a basis of W . false (m) For any scalar k and square matrix A , det( kA ) = k det( A ). true (n) For any n × n matrices, A and B , (det A )(det B ) = det( AB ). true (o) For any n × n matrices, A and B , det( AB ) = det( BA ). 2. Use a matrix representation to find bases for the kernel and the image of the linear transformation T : P 2 P 3 defined by T ( a + bx + cx 2 ) = ( a +2 b 2 c )+(2 a +2 b ) x +( a + b 4 c ) x 2 +(3 a +2 b +2 c ) x 3 Solution. Let X = (1 ,x,x 2 ) be the standard basis of P 2 and let Y = (1 ,x,x 2 ,x 3 ) be the standard basis on P 3 . Then Y T X = [ K Y ( T (1)) K Y ( T ( x )) K Y ( T ( x 2 )) ] = [ K Y (1 + 2 x x 2 + 3 x 3 ) K Y (2 + 2 x + x 2 + 2 x 3 ) K Y ( 2 4 x 2 + 2 x 3 ) ] = 1 2 2 2 2 0 1 1 4 3 2 2
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The reduced row echelon form of Y T X is 1 0 2 0 1 2 0 0 0 0 0 0 A basis for the column space of Y T X is B = 1 2 1 3 , 2 2 1 2 So a basis for the image of T is K 1 Y ( B ) = K 1 Y 1 2 1 3 ,K 1 Y 2 2 1 2 = (1 + 2 x x 2 + 3 x 3 , 2 + 2 x + x 2 + 2 x 3 ) The general solution to the equation Y T X ⃗x = 0 is x 1 = 2 t, x 2 = 2 t, x 3 = t.
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test3-review_Fall2010 - Solutions to practice problems for...

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