{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

term_test_2_w10_soln

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

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

Math 136 Term Test 2 Solutions 1. Short Answer Problems [1] a) What is the deﬁnition of a basis B of a vector space V ? Solution: A basis is a linearly independent spanning set. [1] b) What is the deﬁnition 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) Find 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 ﬁnd 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 deﬁned by L ( ~x ) = 2 ~x - ( ~x · ~a ) ~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 ] · ~a ) ~a = 2 k~x + 2 ~ y - ( k [ ~x · ~a ] + [ ~ y · ~a ]) ~a = k [2 ~x - ( ~x · ~a ) ~a ] + 2 ~ y - ( ~ y · ~a ) ~a = kL ( ~x ) + 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 2 1 4 2 2 1 . 1

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

View Full Document
2 2. Let T : R 3 R 2 be deﬁned by T ( x 1 , x 2 , x 3 ) = ( x 1 + x 3 , 0). [2] a) Prove that T is linear. Solution: Let ~x = ( x 1 , x 2 , x 3 ), ~ y = ( y 1 , y 2 , y 3 ) and k R . Then T ( k~x + ~ y ) = T ( k ( x 1 , x 2 , x 3 ) + ( y 1 , y 2 , y 3 )) = T ( kx 1 + y 1 , kx 2 + y 2 , kx 3 + y 3 ) = ( kx 1 + y 1 + kx 3 + y 3 , 0) = k ( x 1 + x 3 , 0) + ( y 1 + y 3 , 0) = kT ( ~x ) + T ( ~ y ) [3] b) Find the standard matrix of 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 1 0 1 0 0 0 ´ . [2] c) Find a spanning set for the nullspace of T . Solution: The nullspace of T is the set of all ( x 1 , x 2 , x 3 ) such that T ( x 1 , x 2 , x 3 ) = (0 , 0). Hence, we must have
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### Page1 / 8

term_test_2_w10_soln - Math 136 1 Short Answer Problems...

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

View Full Document
Ask a homework question - tutors are online