# Rank of a linear transformation.

Let $V$ be the vector space of all continuous function from $[0,1]$ to $\mathbb{R}$ on the field $\mathbb{R}$. What is the rank of the linear transformation $T:V\rightarrow V$ which was defined as below? $$T(f(x))=\int_{0}^{1}(3x^3y-5x^4y^2)f(y)dy$$

• Can you see an obvious upper bound? – Arnaud Mortier Mar 23 '18 at 22:51

$T(f(x)) =$$\displaystyle\int_{0}^{1}(3x^3y-5x^4y^2)f(y)dy\\ \displaystyle3x^3\int_{0}^{1}yf(y)\ dy-5x^4\int_0^1 y^2 f(y)\ dy$
$\int_{0}^{1}yf(y)\ dy$ is a constant as is $\int_0^1 y^2 f(y)\ dy$
So what does that say about $T(f(x))$?
• @J.D No... for any $f, T(f)$ will be of the form Ax^4 + Bx^3$The rank of$T$is$2.$– Doug M Mar 26 '18 at 16:19 Hint:$T(f)\$ is a actually a polynomial...