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 $\operatorname{rank}(T) = 2$?

$T(f(x)) = 3x^3\int_0 ^1 yf(y) dy-5x^4\int_0^1 y^2f(y) dy$.
So all functions are mapped to $ax^3+bx^4$. Therefore the dimension of image space is 2.
But we know that rank of transformation is same as dimension of its image space.
