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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? $$T(f(x))=\int_{0}^{1}(3x^3y-5x^4y^2)f(y)dy$$

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    $\begingroup$ Can you see an obvious upper bound? $\endgroup$ – Arnaud Mortier Mar 23 '18 at 22:51
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$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 $T(f(x))$?

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  • $\begingroup$ thus we can conclude rank T=4? $\endgroup$ – user519338 Mar 24 '18 at 0:12
  • $\begingroup$ @J.D No... for any $f, T(f)$ will be of the form Ax^4 + Bx^3$ The rank of $T$ is $2.$ $\endgroup$ – Doug M Mar 26 '18 at 16:19
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Hint: $T(f)$ is a actually a polynomial...

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