2 20 points Consider the matrix A 1 3 1 2 1 1 1 1 2 1 1 5 2 3 1 4 1 v 3 1 w 1 1

# 2 20 points consider the matrix a 1 3 1 2 1 1 1 1 2 1

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(2) (20 points) Consider the matrix A = 1 0 3 1 2 1 1 0 1 1 2 1 0 1 5 2 3 1 4 1 , v = 0 0 0 3 1 , w = 1 1 a 1 1 a + 1 1) Find an orthogonal bases of Col A ; 2) Find the orthogonal projection of v on Col A ; 3) Determine a for which w is in Col A ; 4) Find the orthogonal projection of 2 v on Nul A T ; 5) What is the maximal number of linearly independent rows of A ? Explain; 6) What is the dimension of Nul A T ? Explain. (3) (15 points) With as little computation as possible, determine which of the following is diagonalizable. A = 3 7 9 5 7 1 1 2 9 1 2 1 5 2 1 4 , B = 3 0 0 0 7 1 0 0 9 1 2 0 5 2 1 4 , C = 3 7 0 0 7 1 0 0 0 0 2 1 0 0 1 4 , D = 3 7 0 0 7 1 0 0 0 0 2 1 0 0 1 4 . (4) (15 points) Consider four polynomials p 1 = 1 + t, p 2 = 1 t + t 2 + 2 t 3 , p 3 = 1 t + t 2 + 3 t 3 , p 4 = t t 2 3 t 3 in P 3 and the subspace H of P 3 spanned by p 1 , p 2 , p 3 . 1) Show that { p 1 , p 2 , p 3 , p 4 } is a basis of P 3 ; 2) Which of q = t 2 + t 3 and r = 1 + t + t 2 + t 3 is in H . For the one in H , find the coordinates with respect to the basis { p 1 , p 2 , p 3 } of H ; 3) Show that the restriction D ( p ) = p : H P 2 of the derivative transformation on H is invertible. (5) (17 points) Circle the right answers (no reasons needed, multiple answers possible) 1) Suppose a linear transformation T : M (2 , 3) P 3 is onto. Then the dimension of the kernel of T is 0 1 2 3 4 5 6 2) Suppose u , v and u + v are nonzero. Then Proj u + v ( x + y ) is equal to Proj u ( x + y ) + Proj v ( x + y ) Proj u x + Proj v y Proj u + v x + Proj u + v y Proj u x + Proj v x + Proj u y + Proj v y 3) Suppose A is a symmetric 3 × 3 matrix, with det( A λI ) = ( λ 1)( λ + 2) 2 . Suppose (1 , 2 , 3) and (1 , 1 , 1) are eigenvectors with eigenvalue 2. Then the following are also eigenvectors of A ( 1 , 2 , 1) (3 , 2 , 1) ( 1 , 2 , 3) (1 , 2 , 1) (2 , 1 , 2) (0 , 1 , 2) 4) Suppose the solution of the equation A 5 × 7 x = 0 contains two free variables. Then the following are true rank A = 2 dimNul A = 2 dimNul A T = 2
rank A = 3 dimNul A = 3 dimNul A T = 3 rank A = 5 dimNul A = 5 dimNul A T = 5 (6) (23 points) True or False (no reason needed) 1) If Ax = 0 has only trivial solution, then the columns of A is a basis of Col A 2) If Ax = 0 has only trivial solution, then the rank of A is the number of columns of A 3) If Ax = 0 has only trivial solution, then the rank of A is the number of rows of A 4) If Ax = b has solution for all b , then the columns of A is a basis of Col A 5) If Ax = b has solution for all b , then the rank of A is the number of columns of A 6) If Ax = b has solution for all b , then the rank of A is the number of rows of A 7) If dim V = n and B spans V , then B contains at least n vectors 8) If B spans V and contains n vectors, then dim V n 9) If a linear transformation T : V W is onto, then dim V dim W 10) If a linear transformation T : V W is one-to-one, then dim V dim W
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