204ps42006sols

# 204ps42006sols - Problem Set 4 Solutions Economics 204...

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Problem Set 4 - Solutions Economics 204 - August 2006 1. Let Θ be the set of all continuous functions whose domain is the unit interval [0 , 1] and range is R . Let Φ be the subset consisting of all real polynomials (whose domain is restricted to the unit interval) of degree at most two: Φ ≡ { a + bx + cx 2 | a, b, c R } Note that the set Θ is a vector space over the field of real numbers and the subset Φ is a proper subspace. (a) Are the vectors { x, ( x 2 1) , ( x 2 + 2 x + 1) } linearly independent over R ? Apply the usual test for independence of vectors. Solve for A, B, and C such that Ax + B ( x 2 1) + C ( x 2 + 2 x + 1) = 0 Equating like powers of x we obtain the following system in the three unknowns: B + C = 0 A + 2 C = 0 C B = 0 C = B = B C = B = 0 A = 0. Thus, the three vectors are linearly independent over R . (b) Find a Hamel basis for the subspace Φ. Clearly, { 1 , x, x 2 } is linearly independent and spans Φ, so it is a Hamel basis and dim Φ = 3. Also, since the set of vectors in ( a ) is linearly independent and contains three elements, it is a basis. (c) What is the dimension of Φ ? What is the dimension of Θ ? Three and , respectively. To see that the dimension of Θ is infinite note that the set of vectors { 1 , x, x 2 , x 3 , ... } form a linearly independent set over R . Since Θ contains an infinite linearly indpendent set of vectors and the number of lin- early independent elements of a vector space cannot exceed the dimension, the dimension of Θ must be infinite. 2. Recall that a reﬂection across the x -axis can be achieved with the transformation ( x, y ) ( x, y ). Derive a transformation, T , which reﬂects a point across the line y = 3 x . 1

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(a) First, calculate the action of T on the points (1 , 3) and ( 3 , 1). Since (1 , 3) lies on the line y = 3 x it is unchanged by T so we know that T (1 , 3) = (1 , 3). The slope of the line y = 3 x is 3 and the slope of the vector ( 3 , 1) is 1 / 3. Because the vector is perpendicular with the line, reﬂecting it across the line takes is to (3 , 1), so T ( 3 , 1) = (3 , 1). (b) Next, write the matrix representation of T using these two vectors as bases. Let V = { v 1 , v 2 } = 1 3 , 3 1 . From (2a) we know that T ( v 1 ) = v 1 = 1 · v 1 +0 · v 2 and T ( v 2 ) = v 2 = 0 · v 1 1 · v 2 . So we write P = Mtx V ( T ) = 1 0 0 1 . (c) Find S and S - 1 , the matrices that changes coordinates under this basis to stan- dard coordinates and back again. We can easily find S , the matrix that changes coordinates in V to coordinates un- der the standard basis, E , because we already expressed the basis vectors v 1 and v 2 in terms of standard basis coordinates. We have S = Mtx E,V ( id ) = ( v 1 v 2 ) = 1 3 3 1 . The matrix that changes coordinates from E to V is simply the inverse of S . So S - 1 = Mtx V,E ( id ) = 1 10 1 3 3 1 = 1 / 10 3 / 10 3 / 10 1 / 10 .
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