term_test_2_w10_soln

term_test_2_w10_soln - Math 136 1. Short Answer Problems...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
Math 136 Term Test 2 Solutions 1. Short Answer Problems [1] a) What is the defnition oF a basis B oF a vector space V ? Solution: A basis is a linearly independent spanning set. [1] b) What is the defnition oF the dimension oF a vector space V ? Solution: The dimension oF a vector space is the number oF vectors in any basis For V . [2] c) ±ind a basis For R 3 that includes the vectors 1 2 3 , 3 2 1 . JustiFy. Solution: Since 1 2 3 , 3 2 1 is clearly linearly independent and the dimension oF R 3 is 3, we just need to fnd one vector in R 3 which does not lie in the span oF the vectors. We know that (1 , 2 , 3) × (3 , 2 , 1) = ( - 4 , - 8 , - 4) does not lie in the plane spanned by (1 , 2 , 3) and (3 , 2 , 1), hence a basis For R 3 that includes these vectors is 1 2 3 , 3 2 1 , - 4 - 8 - 4 [2] d) Let !a = (1 , 2) R 2 , and L : R 2 R 2 be defned by L ( !x ) = 2 - ( · ) !a. Determine iF L is linear. Solution: Let !x, !y R 2 and k R , then L ( k!x + ! y ) = 2( k!x + ! y ) - ([ k!x + ! y ] · ) =2 k!x +2 ! y - ( k [ · ]+[ ! y · ]) = k [2 - ( · ) ]+2 ! y - ( ! y · ) = kL ( )+ L ( ! y ) Hence L is linear. [2] e) Determine the standard matrix oF a linear mapping L : R 2 R 3 whose range is span { (1 , 2 , 1) } and whose nullspace is span { ( - 1 , 2) } . Solution: Since the range is span { (1 , 2 , 1) } all columns must be multiples oF (1 , 2 , 1). Since the nullspace is span { ( - 1 , 2) } , all columns must be multiples oF (2 , 1). Hence, the standard matrix could be any scalar multiple oF 21 42 . 1
Background image of page 1

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

View Full DocumentRight Arrow Icon
2 2. Let T : R 3 R 2 be defned by T ( x 1 ,x 2 3 ) = ( x 1 + x 3 , 0). [2] a) Prove that T is linear. Solution: Let !x =( x 1 2 3 ), ! y y 1 ,y 2 3 ) and k R . Then T ( k!x + ! y )= T ( k ( x 1 2 3 )+( y 1 2 3 )) = T ( kx 1 + y 1 , kx 2 + y 2 , kx 3 + y 3 ) 1 + y 1 + 3 + y 3 , 0) = k ( x 1 + x 3 , 0) + ( y 1 + y 3 , 0) = kT ( )+ T ( ! y ) [3] b) Find the standard matrix o± T . Solution: We have T (1 , 0 , 0) = (1 , 0) T (0 , 1 , 0) = (0 , 0) T (0 , 0 , 1) = (1 , 0) Hence [ T ]= ± T (1 , 0 , 0) T (0 , 1 , 0) T (0 , 0 , 1) ² = ³ 101 000 ´ . [2] c) Find a spanning set ±or the nullspace o± T . Solution: The nullspace o± T is the set o± all ( x 1 2 3 ) such that T ( x 1 2 3 ) = (0 , 0). Hence, we must have x 1 + x 3 = 0 or x 3 = - x 1 . There±or a vector in the nullspace has the ±orm ( x 1 2 , - x 1 x 1 (1 , 0 , - 1) + x 2 (0 , 1 , 0) . Thus null( T ) = span { (1 , 0 , - 1) , (0 , 1 , 0) } . Moreover, since the vectors are not scalar multiples o± each other, they are linearly inde- pendent and so { (1 , 0 , - 1) , (0 , 1 , 0) } is a basis ±or the nullspace o± T .
Background image of page 2
3 [5] 3. Determine the standard matrix of a reFection in R 2 in the line x 1 +2 x 2 =0 followed by a rotation by π 2 radians counterclockwise.
Background image of page 3

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

View Full DocumentRight Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 04/30/2010 for the course MATH 136 taught by Professor All during the Winter '08 term at Waterloo.

Page1 / 8

term_test_2_w10_soln - Math 136 1. Short Answer Problems...

This preview shows document pages 1 - 4. Sign up to view the full document.

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