# Linear transformations, images for continuous functions

1. Sep 11, 2016

### joshmccraney

1. The problem statement, all variables and given/known data
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 $dimN(T)$?

2. Relevant equations
Linear algebra stuff. Too much to write I think.

3. 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.

2. Sep 11, 2016

### Ray Vickson

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

3. Sep 11, 2016

### joshmccraney

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

4. Sep 11, 2016

### pasmith

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

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

Doing part (b) properly will suggest further examples of vectors in $\ker T$.

5. Sep 11, 2016

### joshmccraney

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.

6. Sep 11, 2016

### pasmith

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

7. Sep 11, 2016

### Ray Vickson

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

8. Sep 11, 2016

### joshmccraney

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.

9. Sep 12, 2016

### pasmith

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

10. Sep 12, 2016

### joshmccraney

How does this look?

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11. Sep 13, 2016

### pasmith

You need to rethink the kernel.

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

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

12. Sep 14, 2016

### joshmccraney

Thanks! I really appreciate your help!