Prelim 2 [Solutions]

Prelim 2 [Solutions] - Math 2940 Prelim 2 Solutions You are...

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Math 2940, Prelim 2 Solutions October 27, 2011 You are NOT allowed calculators or the text. SHOW ALL WORK! 1) (a) (10 points) Find a basis for the subspace of all vectors in R 4 orthogonal to both (1 , 0 , 2 , 3) and (1 , 0 , 0 , 1). (b) (8 points) Let W R 4 be the subspace spanned by { (0 , 2 , 0 , 7) , (1 , 1 , 1 , 4) , ( 1 , 1 , 1 , 3) , (3 , 1 , 3 , 5) } . What is the dimension of W ? Find a matrix A whose left null space is exactly W . Solution: (a) Row reducing the matrix ° 100 1 102 3 ± we get the row reduced echelon form ° 1 001 2 ± so the null space of our original matrix has basis { (0 , 1 , 0 , 0) , (1 , 0 , 2 , 1) } . The null space is necessarily the orthogonal complement of the row space. (b) Row reducing the matrix 11 14 11 13 02 07 31 35 we get the row reduced echelon form ° 1011 / 2 0107 / 2 ± . A basis for the null space of our matrix is then { ( 1 / 2 , 7 / 2 , 0 , 1) , ( 1 , 0 , 1 , 0) } . Writing these as columns gives us ourt matrix 1 / 2 1 7 / 20 01 10 . 2) (8 points) a) Find the matrix P that projects vectors °v R 3 onto the plane x +2 y z = 0. b) (8 points) Express the vector ( 2 , 2 , 1) as the sum of a vector in the plane from part (a), and a vector normal to that plane. Solution: (a) The plane is a two dimensional subspace of R 3 . The vectors (1 , 0 , 1) and (0 , 1 , 2) are independent and in the plane so they form a basis. Set A = 12 so the projection matrix is P = A ( A T A ) 1 A T = ° 101 012 ± 1 ° ± = 1 6 5
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Prelim 2 [Solutions] - Math 2940 Prelim 2 Solutions You are...

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