The computation of nullity and rank of a linear transformation. Let $V$ be the linear space of all real functions continuous on $[a, b]$. If $f\in V, g=T(f)$ means that $$g(x)=\int_a^b f(t)\sin(x-t)\,dt\hspace{1 cm} for\ a\le x\le b$$
We can see this as a linear transformation $T: V\to V$ the linearity of which is easy to show.$\\$ Now how do you compute the nullity and rank of this transformation?
I'm rather new to linear algebra so I'd appreciate it if you include lots of detail and explanation in your answer. Thank you in advance.
P.S: the final answer is infinite nullity and 2nd rank (according to the back of the book).
From calculus II by Tom M. Apostol - page 36
 A: Start with definitions:
Nullility is the dimension of the null space, which for this problem is equivalent to asking for set of all non-zero, linearly independent functions $f$ such that $T(f)=0$. 
To get some intuition for why the answer is infinite nullility, suppose $b=-a$ and $f$ is an even function (i.e. $f(x)=f(-x)$), I think you can show that $T(f)=0$ (think about the integral from a to 0 and then from 0 to b -- these must have equal magnitude but opposite signs). Now the question becomes how many linearly independent functions span the space of all even functions? I think you can show that the Taylor series for an even function must be of the form $\sum_{n=0}^{\infty}\alpha_n\frac{x^{2n}}{(2n)!}$, which gives you the result (you've written all possible even functions as a linear combination of even powers of $x$, of which there are infinitely many)
To chase after rank, you want to apply a similar line of reasoning: you want to count how many linearly independent functions $f_{n}$ there are such that $T(f_n)\ne0$
