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 1 0 0 1 1 0 2 3 we get the row reduced echelon form 1 0 0 1 0 0 1 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 1 1 1 4 1 1 1 3 0 2 0 7 3 1 3 5 we get the row reduced echelon form 1 0 1 1 / 2 0 1 0 7 / 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 / 2 0 0 1 1 0 . 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 = 1 0 0 1 1 2 so the projection matrix is P = A ( A T A ) 1 A T = 1 0 0 1 1 2 1 0 1 0 1 2 1 0 0 1 1 2 1 1 0 1 0 1 2 = 1 6
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