Linear transformations, images for continuous functions

[member 428835]

Homework Statement

Let ##C## be the space of continuous real functions on ##[0,\pi]##. With any function ##f\in C##, associate another function ##g=T(f)## defined by $$g=T(f)\equiv \int_0^\pi \cos(t-\tau) f(\tau) \, d \tau$$
a) Show ##T## is a linear transformation from ##C## to ##C##.
b)What is the image of ##T##? Find a basis for it.
c) List a set of linearly independent vectors that are in the null space of ##T##. What is ##dimN(T)##?

Homework Equations

Linear algebra stuff. Too much to write I think.

The Attempt at a Solution

a)##T(af + bg) = \int\cos(t-\tau)(af + bg) = \int\cos(t-\tau)af + \int\cos(t-\tau)bg = a\int\cos(t-\tau)f + b\int\cos(t-\tau)g = aT(f)+bT(g)##.
b) I want to say the space of continuous real functions on ##[0,\pi]##. Does this even make sense though?
c)By inspection, ##a\sin(t-\tau)## but this is all I can see. Are there any others? In this case, I would say ##dimN(T) = 1## so far since I can only think of the one example. Obviously ##0## is as well but this is not linearly independent of what I have listed.

 

[Ray Vickson]

Homework Statement

Let ##C## be the space of continuous real functions on ##[0,\pi]##. With any function ##f\in C##, associate another function ##g=T(f)## defined by $$g=T(f)\equiv \int_0^\pi \cos(t-\tau) f(\tau) \, d \tau$$
a) Show ##T## is a linear transformation from ##C## to ##C##.
b)What is the image of ##T##? Find a basis for it.
c) List a set of linearly independent vectors that are in the null space of ##T##. What is ##dimN(T)##?

Homework Equations

Linear algebra stuff. Too much to write I think.

The Attempt at a Solution

a)##T(af + bg) = \int\cos(t-\tau)(af + bg) = \int\cos(t-\tau)af + \int\cos(t-\tau)bg = a\int\cos(t-\tau)f + b\int\cos(t-\tau)g = aT(f)+bT(g)##.
b) I want to say the space of continuous real functions on ##[0,\pi]##. Does this even make sense though?
c)By inspection, ##a\sin(t-\tau)## but this is all I can see. Are there any others? In this case, I would say ##dimN(T) = 1## so far since I can only think of the one example. Obviously ##0## is as well but this is not linearly independent of what I have listed.

Even if ##f(\tau)## is defined only on ##\tau \in [0,\pi]##, is ##g(t)## defined for ##t## outside the interval ##[0,\pi]##?

 

[member 428835]

Even if ##f(\tau)## is defined only on ##\tau \in [0,\pi]##, is ##g(t)## defined for ##t## outside the interval ##[0,\pi]##?

This was not specified, so I'm leaning towards no.

 

[pasmith]

Homework Statement

Let ##C## be the space of continuous real functions on ##[0,\pi]##. With any function ##f\in C##, associate another function ##g=T(f)## defined by $$g=T(f)\equiv \int_0^\pi \cos(t-\tau) f(\tau) \, d \tau$$
a) Show ##T## is a linear transformation from ##C## to ##C##.
b)What is the image of ##T##? Find a basis for it.
c) List a set of linearly independent vectors that are in the null space of ##T##. What is ##dimN(T)##?

Homework Equations

Linear algebra stuff. Too much to write I think.

[itex]\cos(t - \tau) = \cos t \cos \tau + \sin t \sin \tau[/itex] seems extremely relevant, as it should solve part (b) for you.

The Attempt at a Solution

a)##T(af + bg) = \int\cos(t-\tau)(af + bg) = \int\cos(t-\tau)af + \int\cos(t-\tau)bg = a\int\cos(t-\tau)f + b\int\cos(t-\tau)g = aT(f)+bT(g)##.
b) I want to say the space of continuous real functions on ##[0,\pi]##. Does this even make sense though?

Part (a) required you also to show that [itex]T(f) \in C[/itex] for every [itex]f \in C[/itex]. Part (b) requires you to go further.

c)By inspection, ##a\sin(t-\tau)## but this is all I can see. Are there any others? In this case, I would say ##dimN(T) = 1## so far since I can only think of the one example. Obviously ##0## is as well but this is not linearly independent of what I have listed.

Doing part (b) properly will suggest further examples of vectors in [itex]\ker T[/itex].

 

[member 428835]

[itex]\cos(t - \tau) = \cos t \cos \tau + \sin t \sin \tau[/itex] seems extremely relevant, as it should solve part (b) for you.

I don't follow you. I can't see how this identity helps us find the image of ##T##. Could you elaborate?

Part (a) required you also to show that [itex]T(f) \in C[/itex] for every [itex]f \in C[/itex]. Part (b) requires you to go further.

How to you go about showing this part of a)? I would try but I don't know where to start.

Thanks for replying!

 

[pasmith]

I don't follow you. I can't see how this identity helps us find the image of ##T##. Could you elaborate?

Try expressing [itex]T(f)[/itex] in the form [tex]
T(f) =\int_0^\pi \left(\sum_{n=1}^N A_n(t)B_n(\tau)\right)f(\tau)\,d\tau = \sum_{n=1}^N A_n(t) \int_0^\pi B_n(\tau) f(\tau)\,d\tau.[/tex]

 

[Ray Vickson]

I don't follow you. I can't see how this identity helps us find the image of ##T##. Could you elaborate?
How to you go about showing this part of a)? I would try but I don't know where to start.

Thanks for replying!

I think that PF rules forbid us from answering that question.

 

[member 428835]

Try expressing [itex]T(f)[/itex] in the form [tex]
T(f) =\int_0^\pi \left(\sum_{n=1}^N A_n(t)B_n(\tau)\right)f(\tau)\,d\tau = \sum_{n=1}^N A_n(t) \int_0^\pi B_n(\tau) f(\tau)\,d\tau.[/tex]

So something like this $$\int_0^\pi \cos(t-\tau)f(\tau) \, d\tau = \int_0^\pi (\cos t\cos\tau +\sin t \sin \tau)f(\tau) \, d\tau\\ = \cos t\int_0^\pi\cos\tau f(\tau) \, d \tau + \sin t \int_0^\pi\sin \tau f(\tau) \, d\tau$$ But I still don't see how this helps us find a basis. If you feel uncomfortable helping me with this one since I clearly can't do it well, perhaps we could make a simpler example and you could help me through it so I can see the strategy? I don't want to violate the rules on PF, I'm just confused how all of this is related.

 

[pasmith]

[itex]\int_0^\pi \cos \tau f(\tau)\,d\tau[/itex] is a definite integral of a real-valued function, so it's just a real number!

 

[member 428835]

How does this look?

 

[pasmith]

You need to rethink the kernel.

I sense some confusion. The input of [itex]T[/itex] is a function of one real variable, which we are labelling [itex]\tau[/itex]. The output of [itex]T[/itex] is another function of one real variable, which we are labelling [itex]t[/itex]. Setting [itex]f(\tau) = \sin(t - \tau)[/itex] therefore is nonsense; trying [itex]f(\tau) = \sin (\alpha - \tau)[/itex] for real [itex]\alpha[/itex] makes sense, but it turns out that that is not in the kernel for any [itex]\alpha[/itex].

Finding functions in the kernel is actually straightforward: You can readily show that if [itex]h(\tau) = a \cos \tau + b \sin \tau[/itex] then [itex]T(h) = \frac{\pi}{2}h[/itex], and having established that the image of [itex]T[/itex] is the space spanned by [itex]\cos[/itex] and [itex]\sin[/itex] it then follows that [tex]
T^2(f) = \tfrac{\pi}2T(f)[/tex] for all [itex]f \in C[/itex], and hence by linearity of [itex]T[/itex] that [tex]
T(f) - \tfrac{\pi}{2}f \in \ker T[/tex] for every [itex]f \in C[/itex]. This gives you a way to easily find non-trivial members of the kernel, provided you are careful with your choice of [itex]f[/itex].

 

[member 428835]

Thanks! I really appreciate your help!