math 304 Test2sample solution

math 304 Test2sample solution - MATH 304505 Spring 2011...

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MATH 304–505 Spring 2011 Sample problems for Test 2: Solutions Any problem may be altered or replaced by a different one! Problem 1 (15 pts.) Let M 2 , 2 ( R ) denote the vector space of 2 × 2 matrices with real entries. Consider a linear operator L : M 2 , 2 ( R ) → M 2 , 2 ( R ) given by L parenleftbigg x y z w parenrightbigg = parenleftbigg x y z w parenrightbigg parenleftbigg 1 2 3 4 parenrightbigg . Find the matrix of the operator L with respect to the basis E 1 = parenleftbigg 1 0 0 0 parenrightbigg , E 2 = parenleftbigg 0 1 0 0 parenrightbigg , E 3 = parenleftbigg 0 0 1 0 parenrightbigg , E 4 = parenleftbigg 0 0 0 1 parenrightbigg . Let M L denote the desired matrix. By definition, M L is a 4 × 4 matrix whose columns are coordinates of the matrices L ( E 1 ) , L ( E 2 ) , L ( E 3 ) , L ( E 4 ) with respect to the basis E 1 , E 2 , E 3 , E 4 . We have that L ( E 1 ) = parenleftbigg 1 0 0 0 parenrightbigg parenleftbigg 1 2 3 4 parenrightbigg = parenleftbigg 1 2 0 0 parenrightbigg = 1 E 1 + 2 E 2 + 0 E 3 + 0 E 4 , L ( E 2 ) = parenleftbigg 0 1 0 0 parenrightbigg parenleftbigg 1 2 3 4 parenrightbigg = parenleftbigg 3 4 0 0 parenrightbigg = 3 E 1 + 4 E 2 + 0 E 3 + 0 E 4 , L ( E 3 ) = parenleftbigg 0 0 1 0 parenrightbigg parenleftbigg 1 2 3 4 parenrightbigg = parenleftbigg 0 0 1 2 parenrightbigg = 0 E 1 + 0 E 2 + 1 E 3 + 2 E 4 , L ( E 4 ) = parenleftbigg 0 0 0 1 parenrightbigg parenleftbigg 1 2 3 4 parenrightbigg = parenleftbigg 0 0 3 4 parenrightbigg = 0 E 1 + 0 E 2 + 3 E 3 + 4 E 4 . It follows that M L = 1 3 0 0 2 4 0 0 0 0 1 3 0 0 2 4 . Problem 2 (20 pts.) Find a linear polynomial which is the best least squares fit to the following data: x 2 1 0 1 2 f ( x ) 3 2 1 2 5 We are looking for a function f ( x ) = c 1 + c 2 x , where c 1 , c 2 are unknown coefficients. The data of the problem give rise to an overdetermined system of linear equations in variables c 1 and c 2 : c 1 2 c 2 = 3 , c 1 c 2 = 2 , c 1 = 1 , c 1 + c 2 = 2 , c 1 + 2 c 2 = 5 . 1
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This system is inconsistent. We can represent it as a matrix equation A c = y , where A = 1 2 1 1 1 0 1 1 1 2 , c = parenleftbigg c 1 c 2 parenrightbigg , y = 3 2 1 2 5 . The least squares solution c of the above system is a solution of the system A T A c = A T y : parenleftbigg 1 1 1 1 1 2 1 0 1 2 parenrightbigg 1 2 1 1 1 0 1 1 1 2 parenleftbigg c 1 c 2 parenrightbigg = parenleftbigg 1 1 1 1 1 2 1 0 1 2 parenrightbigg 3 2 1 2 5 ⇐⇒ parenleftbigg 5 0 0 10 parenrightbigg parenleftbigg c 1 c 2 parenrightbigg = parenleftbigg 3 20 parenrightbigg ⇐⇒ braceleftbigg c 1 = 3 / 5 c 2 = 2 Thus the function f ( x ) = 3 5 +2 x is the best least squares fit to the above data among linear polynomials. Problem 3 (25 pts.) Let V be a subspace of R 4 spanned by the vectors x 1 = (1 , 1 , 1 , 1) and x 2 = (1 , 0 , 3 , 0).
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