{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

2.22.3 - Homework 4 Solutions 2.2 Problem 1 1 TRUE See...

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

View Full Document Right Arrow Icon
Homework 4 - Solutions 2.2 Problem 1 1. TRUE. See Theorem 2.7 (a). 2. TRUE. See last paragraph on page 80. 3. FALSE. It will be a n × m matrix. 4. TRUE. See Theorem 2.8 (a). 5. TRUE. See Theorem 2.7. 6. FALSE. L ( V, W ) is a collection of functions from V to W , and L ( W, V ) is a collection of functions from W to V . These are not the same unless V and W are the same. 2.2 Problem 4. T 1 0 0 0 = 1 T 0 1 0 0 = 1 + x 2 T 0 0 1 0 = 0 T 0 0 0 1 = 2 x The coordinates of these four polynomials with respect to the basis β form the four columns of the matrix [ T ] γ β : [ T ] γ β = 1 1 0 0 0 0 0 2 0 1 0 0 2.2 Problem 9. Let z 1 = a 1 + ib 1 , z 2 = a 2 + ib 2 , and α R . . Then T ( z 1 + αz 2 ) = T (( a 1 + ib 1 ) + α ( a 2 + ib 2 )) = T (( a 1 + αa 2 ) + i ( b 1 + αb 2 )) = ( a 1 + αa 2 ) i ( b 1 + αb 2 ) = ( a 1 ib 1 ) + α ( a 2 ib 2 ) = z 1 + α z 2 = T ( z 1 ) + αT ( z 2 ) This proves T is linear. T (1) = 1, and T ( i ) = i . In the basis { 1 , i } , the coordinates for T (1) and T ( i ) are (1 , 0) and (0 , 1) (written as columns). Therefore, [ T ] β = 1 0 0 1
Background image of page 1

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

View Full Document Right Arrow Icon
2.2 Problem 10. The j th column of [ T ] β is obtained by expressing T ( v j ) in the basis β . Since T ( v j ) = v j + v j - 1 , we get T ( v j ) = 0 v 1 + · · · + 1 v j - 1 + 1 v j · · · 0 v n The coordinates for T ( v j ) in the basis β are (0 , . . . , 1 , 1 , . . ., 0), where the only nonzero entries are at the j 1 and j positions (written as a column vector). For the special case j = 1, the only nonzero entry is the first entry (since v j - 1 is taken to be 0). Therefore, [ T ] β = 1 1 0 0 · · · 0 0 1 1 0 · · · 0 0 0 1 1 · · · 0 . . . 0 0 · · · 0 1 1 0 · · · 0 1 This is a matrix which has (a) 1 on each diaginal entry and each entry above a diagonal entry (b) zero in all other entries. 2.2 Problem 16. Let the dimension of V and W be n . Assume that the rank of T is k . Let S = { v 1 , . . . , v n - k } be a basis of N ( T ). Extend S to a basis β = { v 1 , . . . , v n - k , v n - k +1 , . . . , v n } of V (by Corollary on Page 51). Define w j = T ( v j ) for j = n k + 1 , . . . , n . Let R = { w n - k +1 , . . . , w n } .
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}