midterm_2_sol - 2.035 Midterm Exam Part 2 Spring 2007...

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2.035: Midterm Exam - Part 2 Spring 2007 SOLUTION Problem 1: Consider the set V of all 2 × 2 skew-symmetric matrices, i.e. matrices of the form 0 x x = , −∞ < x < . (1.1) x 0 a) Define the addition of two vectors by 0 x + y 0 x 0 y x + y = where x = , y = , (1.2) x y 0 x 0 y 0 and the multiplication of a vector x by the scalar α as 0 αx α x = . (1.3) αx 0 Then one can readily verify that x + y V whenever x and y V , and that α x V for every scalar α whenever x V . Thus V is a vector space. b) Since 0 x 0 y 0 0 y x = (1.4) x 0 y 0 0 0 it follows that y x + ( x ) y = o for any two vectors x and y . Thus every pair of vectors is linearly dependent. There is no pair of vectors that is linearly independent. c) It follows from item (b) that the dimension of the vector space is 1. d) If x and y are two vectors in V , and we define x y = xy + ( x )( y ) = 2 xy, (1.5) · then we can verify that x y has the properties · a) x y = y x , · · b) x ( α y + β z ) = α x y + β x z , · · · c) x x 0 with x x = 0 if and only if x = o , · · for all x , y , z V and all scalars α, β . Thus this is a proper definition of a scalar product.
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e) Since the dimension of V is 1, any one (non null) vector provides a basis for it. Pick 0 1 e = (1.6) 1 0 Note that the length of e is | e | = e · e = 1 + 1 = 2 . (1.7) Thus a basis composed of a unit vector is { e } where 0 1 / 2 e = . (1.8) 1 / 2 0 f) Let A be the transformation defined by 0 2 x Ax = for all vectors x V . (1.9) 2 x 0 We can readily verify that ( i ) the vector Ax V for every x V , and ( ii ) that A ( α x + β y ) = α Ax + β Ay for all vectors x , y and all scalars α, β . Therefore A is a linear transformation (tensor). g) From item (d), a b = 2( a )( b ) for any two vectors a and b . Thus it follows that · Ax y = 2(2 x )( y ) = 4 xy · while x Ay = 2( x )(2 y ) = 4 xy. · Thus Ax y = x Ay and so A is symmetric. · · h) If Ax = o then 2 x = 0 and so x = 0 from which it follows that x = o . Thus Ax = o if and only if x = o . Thus A is non-singular.
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