This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Math 311 second practice exam answers 1. ( D 2 I ) u ( x ) = u 00 ( x ) u ( x ) = 4 cos 2 x cos 2 x = 5 cos 2 x. 2. To show that these polynomials span the whole vector space, we note that for any a , a 1 , a 2 , a + a 1 x + a 2 x 2 = a 1 + a 1 ( x + x 2 ) + ( a 2 a 1 ) x 2 . To show that these polynomials are linearly independent, suppose that for some a, b, c , a 1 + b ( x + x 2 ) + cx 2 = 0 . Then a + bx + ( b + c ) x 2 = 0 , from which is follows that a = 0 , b = 0 , b + c = 0 , so also c = 0 . 3. Row reduction brings the matrix 1 1 1 2 4 2 3 5 1 to the form 1 1 1 0 1 2 0 0 . The first two columns are lead columns, so the first two of the original vectors Span 1 2 3 , 1 4 5 form a basis for this subspace. In particular, that subspace has dimension 2 . 4. 2 2 1 1 3 0 1 . Since the matrix is uppertriangular, we see right away that the eigenvalues are = 1 , 2 . For = 1 , the...
View
Full
Document
This note was uploaded on 08/28/2008 for the course MATH 311 taught by Professor Anshelvich during the Spring '08 term at Texas A&M.
 Spring '08
 Anshelvich
 Math, Polynomials, Vector Space

Click to edit the document details