{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Assignment 4 Solution

# Assignment 4 Solution - D-20 = 0 This implies 4 C 8 D = 36...

This preview shows pages 1–2. Sign up to view the full content.

CS3251, Homework 4 Solution 1. Problem Set 4.1, p. 191, # 5 (a) If Ax = b and A T y = 0 then y b because b C ( A ) and y N ( A T ) C ( A ). (b) If A T y = c and Ax = 0 then x c because c C ( A T ) and x N ( A ) C ( A T ). 2. Problem Set 4.2, p. 203, # 7 We have a 1 = ( - 1 , 2 , 2) T , a 2 = (2 , 2 , - 1) T , a 3 = (2 , - 1 , 2) T and P = aa T a 2 2 , is the projection on a vector a . Then P 1 = 1 9 1 - 2 - 2 - 2 4 4 - 2 4 4 , P 2 = 1 9 4 4 - 2 4 4 - 2 - 2 - 2 1 , P 3 = 1 9 4 - 2 4 - 2 1 - 2 4 - 2 4 . It’s easy to check that P 1 + P 2 + P 3 = I . 3. Problem Set 4.2, p. 204, # 17 ( I - P ) 2 = ( I - P )( I - P ) = I - 2 P + P 2 and since P 2 = P we have ( I - P ) 2 = I - P . When P projects on C ( A ), I - P projects on N ( A T ). 4. Problem Set 4.3, p. 205, # 29 B is a m × n matrix with full row rank, i.e., r = r ( B ) = m n . Therefore, B has m linearly independent rows. Recall that A has linearly independent columns if and only if A T A is non-singular, see 4G p. 200. The n × m matrix A = B T has m linearly independent columns. Thus N ( A ) = 0 = N ( A T A ). This implies that A T A = BB T is non-singular. 5. Problem Set 4.3, p. 215, # 4 t = 0 , 1 , 3 , 4, b = (0 , 8 , 8 , 20) T . E = Ax - b 2 2 = 4 i =1 ( C + Dt i - b i ) 2 = C 2 + ( C + D - 8) 2 + ( C + 3 D - 8) 2 + ( C + 4 D - 20) 2 . ∂F ∂C = 2 C + 2( C + D - 8) + 2( c + 3 D - 8) + 2( C + 4 D - 20) = 0 ∂F ∂D = 2(

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

View Full Document
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: D-20) = 0 This implies 4 C + 8 D = 36 8 C + 26 D = 112 . 1 A T Ax = A t b is given by ± 4 8 8 26 ²± C D ² = ± 36 112 ² . 6. Problem Set 4.5, p. 217, # 12 b = ( b 1 , . . . , b m ) and a = (1 , . . . , 1). Observe that there is a change in notation and b, a denote a row vectors. Thus b T , a T will denote column vectors. (a) a T a ˆ x = a T implies that m ˆ x = ∑ m i =1 b i and ˆ x = 1 m ∑ m i =1 b i . (b) e = b-a ˆ x = ( b 1-ˆ x, . . . , b m-ˆ x ). Then k e k 2 2 = ∑ m i =1 ( b i-ˆ x ) 2 and k e k 2 = { ∑ m i =1 ( b i-ˆ x ) 2 } 1 / 2 . (c) b = (1 , 2 , 6), a = (1 , 1 , 1), ˆ b = 3 and p = (3 , 3 , 3). Then for e = b-p = (-2 ,-1 , 3) we have pe T = 0 and P = a T a k a T k 2 2 = 1 3 1 1 1 1 1 1 1 1 1 . 2...
View Full Document

{[ snackBarMessage ]}

### Page1 / 2

Assignment 4 Solution - D-20 = 0 This implies 4 C 8 D = 36...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document
Ask a homework question - tutors are online