Exam2MA262fall2011

2 4 which of the following subsets s of the given

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4. Which of the following subsets S of the given vector space V is a subspace? (10 points) (I) V = M 2 ( R ), S = { A M 2 ( R ) : A = A T } ; (II) V = R 3 , S = { ( x, y, z ) : x - y + 5 z = 1 } ; (III) V = P 3 , S = { p ( x ) P 3 : p 0 (1) = p (0) } . (a) (I) and (II). (b) (I) only. (c) (I) and (III). (THIS IS THE ANSWER!) (d) (II) and (III). (e) None of the above. Solution. For (I) we see that if A, B M 2 ( R ) are such that A T = A and B T = B , then given any real number c , we have ( A + B ) T = A T + B T = A + B A + B S ; ( cA ) T = cA T = cA cA S. Thus, S is a subspace of M 2 ( R ) . For (II), we see that the vector (0 , 0 , 0) is NOT in S , so S is NOT a subspace. For (III), we have that if p 1 , p 2 P 3 are such that p 0 j (1) = p j (0) , j = 1 , 2 , then given any real number c ( p 1 + p 2 ) 0 (1) = p 0 1 (1) + p 0 2 (1) = p 1 (0) + p 2 (0) = ( p 1 + p 2 )(0) p 1 , p 2 S ; ( cp 1 ) 0 (1) = cp 0 1 (1) = cp 1 (0) = ( cp 1 )(0) cp 1 S. Thus, S is a subspace of P 3 . 5. Find all values of k such that the vectors (2 , - 4 , 0), (1 , k - 1 , 4) and (10 points) (0 , - 2 , 1 - k ) are linearly DEPENDENT. Solution. This is the same as finding all k such that 2 - 4 0 1 k - 1 4 0 - 2 1 - k = 0 . Expanding the first column 2 - 4 0 1 k - 1 4 0 - 2 1 - k = 2 k - 1 4 - 2 1 - k - - 4 0 - 2 1 - k = 2[( k - 1)(1 - k )+8] - 4( k - 1) = 0 . So we want to solve - 2 k 2 + 4 k + 14 - 4 k + 4 = 0 ⇒ - 2 k 2 + 18 = 0 k = ± 3 . 3
6. Find a basis for the subspace S = { ( x, y, z, t ) : x + 2 y + z - 2 t = 0 } of R 4 . (10 points) Solution. If we solve x + 2 y + z - 2 t = 0 for x , we get x = 2 t - 2 y - z . So S = { (2 t - 2 y - z, y, z, t ) } .

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