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HW7sol

HW7sol - (Partial Solutions to Homework 7 Problem 20.3 Let...

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(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

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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 { u, v , u × v } . To get an orthonormal basis, we divide each vector by its magnitude: { u || u || , v || v || , u × v || u × v || } .
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HW7sol - (Partial Solutions to Homework 7 Problem 20.3 Let...

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