the rank of a linear transformation Let $V$ be vector space consisting of all continuous real-valued functions defined on the closed interval $[0,1]$ (over the field of real numbers) and $T$ be linear transformation from $V$ to $V$ defined by
$$(Tf)(x) = \int_0 ^1 (3x^3 y - 5x^4 y^2) f(y)\,\mathrm dy$$
Why is $\operatorname{rank}(T) = 2$?

Note that $(Tf)(x) = 3x^3 \int_0^1 y f(x)dy - 5 x^4 \int_0^1 y^2 f(y)dy$, hence
$Tf \in \operatorname{sp} \{ x \mapsto x^3, x \mapsto x^4 \}$, so $\dim {\cal R} T \le 2$.
If we choose $f(x) = {2\over 3} -x$, we see $(Tf)(x) = {5 \over 36} x^4$, and if we choose $f(x) = {4\over 3} -x$, we see that $(Tf)(x) ={1 \over 24} x^3$, hence
$\dim {\cal R} T = 2$.
