Midterm 1-Fall 2005-Solutions

Midterm 1-Fall 2005-Solutions - AMS 510.01 Fall 2005,...

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Unformatted text preview: AMS 510.01 Fall 2005, Midterm #1 Solutions Name: Student ID: Score: /100 ( + /15 bonus) 1. Indicate whether each of the following statements is true or false. m and n should be taken as any constant positive integer. (2 points each). (a) ~u + ~v = ~v + ~u for all ~u,~v V , where V is any vector space. True: addition is commutative in all vector spaces. (b) The set of all square matrices forms a field. False: The multiplicative inverse ( A- 1 A = AA- 1 = I ) does not exist for all square matrices; matrix multiplication is not generally commutative ( AB 6 = BA ). (c) ~u ~v = ~v ~u for all ~u,~v C n . False: For complex vectors, ~u ~v = ( ~v ~u ) * , where * is the complex conjugate. (d) The set of all vectors ~x R n forms an algebra. False: An algebra requires closure under addition, scalar multiplication, and multiplication; there is no general product of two vectors giving another vec- tor. (e) The expression f ( X ) = 2 X 2 + X + 5 is defined when X is any n n (square) matrix. True: The set of n-square matrices form an algebra, meaning any polynomial is defined for n-square matrices (with scalars replaced by scalar matrices). (f) The set of all m n matrices forms a vector space. True: Addition and scalar multiplication are closed and are appropriately commutative, associative, and distributive; the additive identity and inverse (and the scalar identity) also exist. (g) If f ( a~u + b~v ) = af ( ~u ) + bf ( ~v ) for all a,b K and all ~u,~v V , then f ( ~x ) is a linear map acting on V. True: The definition of a linear map is one that preserves linear combinations of vectors. (h) If AB = I then A and B must be square matrices. False: AB may equal the n n identity matrix when A is n m and B is m n , even if m 6 = n . (i) { (1 ,- 1) , (1 , 0) , (1 , 1) } is a basis for R 2 . False: R 2 is a 2-dimensional vector space, and thus any basis must contain two vectors; (1 , 0) = (1 ,- 1) + (1 , 1), and thus (1 , 0) is linearly dependant on { (1 ,- 1) , (1 , 1) } . (j) ( ~u ~v ) ~u = ( ~u ~v ) ~v = 1 for all ~u,~v R 3 . False: ( ~u ~v ) ~u = ( ~u ~v ) ~v = 0 6 = 1; ~u ~v is orthogonal to both ~u and ~v . 2. Let ~u = (1 , 1 , 1) and ~v = (0 , 1 , 0). (a) Find a non-zero vector, ~w , in the plane of ~u and ~v such that ~w is orthogonal to ~u . (4 points) If ~w is in the plane of ~u and ~v , then ~w = a~u + b~w ( i.e. ~w is a linear combination of ~u and ~v . Thus: ~w = a~u + b~v = a (1 , 1 , 1) + b (0 , 1 , 0) = ( a,a + b,a ) If ~w is orthogonal to ~u , then ~w ~u = 0: ~w ~u = 0 ( a,a + b,a ) (1 , 1 , 1) = 0 a + ( a + b ) + a = 0 3 a + b = 0 Let a be any non-zero number (say a = 1), then b =- 3 a =- 3, and: ~w = ~u- 3 ~v = (1 , 1 , 1)- 3(0 , 1 , 0) = (1 ,- 2 , 1) (b) Find a vector ~x that is orthogonal to both ~u and ~v . (4 points) The cross-product of two vectors in R 3 is orthogonal to both vectors. Thus, let ~x = ~u ~v :...
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This note was uploaded on 02/09/2010 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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Midterm 1-Fall 2005-Solutions - AMS 510.01 Fall 2005,...

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