Problem: Let $V = C[0,1]$ be the real vector space of all continuous real valued functions on $[0,1]$, and let $T$ be the linear operator on $V$ given by
\[
(Tf)(x) = \int_{0}^{1} \sin (x + y) f(y) \,\mathrm{d}y,~ x \in [0,1].
\]
Then find the dimension of the range space of $T$.
Solution: Note that
\begin{align*}
(Tf)(x) & = \int_{0}^{1} \sin (x + y) f(y) \,\mathrm{d}y \\
& = \int_{0}^{1} \left( \sin x \cos y + \cos x \sin y \right) f(y) \,\mathrm{d}y \\
& = \sin x \int_{0}^{1} \cos y f(y) \,\mathrm{d}y + \cos x \int_{0}^{1} \sin y f(y) \,\mathrm{d}y \\
& = \alpha \sin x + \beta \cos x,
\end{align*}
where
\[
\alpha = \int_{0}^{1} \cos y f(y) \,\mathrm{d}y, ~ \text{ and } ~ \beta = \int_{0}^{1} \sin y f(y) \,\mathrm{d}y.
\]
Therefore, for any $f\in V$, $(Tf)$ is in the span of $\{ \sin x, \cos x \} $, therefore, a basis for range space of $T$ will be $\{ \sin x, \cos x \} $ and hence the dimension of range of $T$ will be $2$.