

Study with the several resources on Docsity
Earn points by helping other students or get them with a premium plan
Prepare for your exams
Study with the several resources on Docsity
Earn points to download
Earn points by helping other students or get them with a premium plan
Homework problems related to linear transformations and vector spaces for math 121a. Topics include finite rank linear transformations, finding the matrix of a linear transformation in a given basis and its dual, the relationship between determinants of a linear transformation and its adjoint, and weak formulations in functional analysis and control theory.
Typology: Assignments
1 / 3
This page cannot be seen from the preview
Don't miss anything!


(a) Prove that ~x is a solution of (1) if and only if
[~x, f ] = [~y, A′f ] for all f ∈ V ′, (2)
where A′^ is the adjoint of A.
(b) The requirement that ~x satisfy (2) is sometimes too strong to be readily satisfied. In such cases, it is weakened by restricting the choice of the functional f to a certain subset T of V ′. A vector ~x ∈ V is called a T -weak solution of (1) if [A~x, f ] = [~y, f ] for all f ∈ T ; (3) the elements of T are then called test functionals. Suppose A is invertible. Under what conditions on T is a T -weak solution of (1) unique? (c) Equation (3) can be written in the form [~x, A′f ] = [~y, f ] for all f ∈ T However, finding A′^ is sometimes much less practical than replacing A′^ by a “weaker version.” An operator A′ T ∈ L(V ) is called the T -weak adjoint of A if [~x, f ] = [~y, A′ T f ] for all f ∈ T ; (4) Is the weak T -adjoint generally unique? (d) We now give an example in which equation (1) fails to have a solution, yet has a T -weak solution for a suitable choice of T. Let
[x, fφ] =
− 1
x(t)φ(t)dt
Take the set T to consist of all such fφ (with φ ∈ U ). Let
y(t) =
− 1 if − 1 ≤ t ≤ 0 1 if 0 < t ≤ 1
thus, y(t) is an element of V. Let A = (^) dtd , i. Prove that with this choice of A and ~y = y(t), equation (1) has no solution. ii. Calculate A′ T. iii. Prove that a T -weak solution x(t) of (4) exists, by finding at least one such solution. If x(t) is such a solution, then y(t) is called the weak derivative of x(t). iv. Prove that if x(t) is differentiable and if y(t) is its weak derivative y(t), then ∫ (^1)
− 1
[x′(t) − y(t)φ(t)dt = 0
Does this mean that x′(t) = y(t) at every t?