235_tt1_w10_soln

235_tt1_w10_soln - Math 235 Term Test 1 Solutions 1. Short...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: Math 235 Term Test 1 Solutions 1. Short Answer Problems [3] a) Let A = 1 0 0- 1 0 0 1 1 0 0 0 . Write a basis for the rowspace, columnspace and nullspace of A . Solution: A basis for the row space is { (1 , , ,- 1) , (0 , , 1 , 1) } . A basis for the column space is 1 , 1 . A basis for the nullspace is 1 , 1- 1 1 [2] b) Let B = { ~v 1 ,...,~v n } be orthonormal in an inner product space V and let ~v = a 1 ~v 1 + + a n ~v n . Prove that a i = < ~v,~v i > . Solution: Taking the inner product of both sides with ~v i to get < ~v,~v i > = < a 1 ~v 1 + + a n ~v n ,~v i > = a 1 < ~v 1 ,~v i > + + a n < ~v n ,~v i > = a i since B is orthonormal. [1] c) Determine if < p,q > = p (0) q (0) + p (1) q (1) is an inner product for P 2 . Solution: It is not an inner product since if p ( x ) = x 2- x then < p,p > = 0(0) + 0(0) = 0 but p ( x ) 6 = ~ 0. [2] d) State the Rank-Nullity Theorem. Solution: Suppose that V is an n-dimensional vector space and that L : V W is a linear mapping into a vector space W . Then rank( L ) + nul( L ) = n. [2] e) Find the rank and nullity of the linear mapping T : P 2 M (2 , 2) defined by T ( a + bx + cx 2 ) = c b c . Solution: The range of T is given by { c b c | b,c arbitrary scalars } . A basis for that is { 1 0 0 1 , 0 1 0 0 } . So the rank of L is two. Using (a), the nullity is equal to 1, noting that the domain P 2 has dimension three. 1 2 [3] 2. Let L : M (2 , 2) M (2 , 2) be given by L ( A ) = 1 2 3 4 A T . Find the matrix for L relative to the standard basis B of M (2 , 2) , where B = 1 0 0 0 , 0 1 0 0 , 0 0 1 0 , 0 0 0 1 . Solution: L ( 1 0 0 0 ) = 1 2 3 4 1 0 0 0 = 1 0 3 0 . By inspection, its coordinate vector relative to the basis B is 1 3 which will form the first column of [ L ] B . Similar computation according to the definition of L yield3 L ( 0 1 0 0 ) = 1 2 3 4 0 0 1 0 = 2 0 4 0 , L ( 0 0 1 0 ) = 1 2 3 4 0 1 0 0 = 0 1 0 3 , L ( 0 0 0 1 ) = 1 2 3 4 0 0 0 1 = 0 2 0 4 ....
View Full Document

Page1 / 7

235_tt1_w10_soln - Math 235 Term Test 1 Solutions 1. Short...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online