HW7sol

# HW7sol - (Partial) Solutions to Homework 7 Problem 20.3:...

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

(Partial) Solutions to Homework 7 Problem 20.3: Let P n ( t ) be the vector space of degree n polynomials, and d dt : P n ( t ) 7→ P n - 1 ( t ) be the derivative operator. Find the matrix of d dt in the bases { 1 ,t,. ..,t n } for P n ( t ) and { 1 ,t,. ..,t n - 1 } for P n - 1 ( t ). Solution: In order to ﬁnd the matrix of a given linear transformation with respect to some bases, we calculate the linear transformation on each element of the basis of the domain, and express the answer in terms of the basis of the range. d dt (1) = 0 = 0 · 1 + 0 · t + 0 · t 2 + ··· d dt ( t ) = 1 = 1 · 1 + 0 · t + 0 · t 2 + ··· d dt ( t 2 ) = 2 t = 0 · 1 + 2 · t + 0 · t 2 + ··· d dt ( t 3 ) = 3 t 2 = 0 · 1 + 0 · t + 3 · t 2 + ··· . . . Then each list of coeﬃcients becomes a column vector in the matrix of the linear transformation with respect to those two bases: M = 0 1 0 0 ··· 0 0 2 0 ··· 0 0 0 3 ··· . . . . . . Not part of the problem, but note that this matrix is not invertible (its deter- minant is zero). This makes sense, since diﬀerentiation is not invertible (if I tell you d dt ( f ( t )) = 3, you don’t know if f ( t ) = 3 t , or f ( t ) = 3 t + 2, etc.). Problem 21.3: Let u,v be independent vectors in R 3 , and P = span { u,v } be the plane spanned by u and v . i. Is the vector v = v - u · v u · u u in the plane P ? ii. What is the angle between v and u ? iii. Given your solution to the above, how can you ﬁnd a third vector perpen- dicular to both u and v ? iv. Construct an orthonormal basis for R 3 from u and v . v. Test your abstract formulae starting with u = ( 1 2 0 ) and v = ( 0 1 1 ) . 1

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

View Full Document
Solution: i. Yes, this vector is in the plane P because it is a linear combination of u and v . ii. The angle between v and u is π/ 2 because we can calculate the dot product: v · u = ( v - u · v u · u u ) · u = v · u - ( u · v )( u · u ) u · u = 0 (remember, the dot product is symmetric, i.e., u · v = v · u .) iii. We could get a third vector orthogonal to the both u and v by taking the cross product of the two. iv. An orthogonal basis is given by
This is the end of the preview. Sign up to access the rest of the document.

## This note was uploaded on 03/23/2010 for the course MAT 022A taught by Professor Pon during the Spring '10 term at UC Davis.

### Page1 / 5

HW7sol - (Partial) Solutions to Homework 7 Problem 20.3:...

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

View Full Document
Ask a homework question - tutors are online