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For any $n\in \mathbb{N}$, let $P_{n}$ denote the vector space of all polynomials with real coefficients and of degree at most $n$. Define linear transformation $T \colon P_n \rightarrow P_{n+1}$ by $T(p)(x) = p'(x)-\int _0^xp(t)dt$. How to find out the dimension of the null space of $T$, where $p'(x)$ is the derivative of $p(x)$?

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The first step is to try to figure out what the kernel/image are.

A basis for $P_n$ is given by $1$, $x$, $x^2,\ldots,x^n$. We have: $$\begin{align*} T(1) &= (1)' - \int_0^x 1\,dt\\ &= -t\Bigm|_0^x = -x.\\ T(x) &= (x)' - \int_0^x t\,dt\\ &= 1 - \frac{1}{2}x^2\\ T(x^2) &= (x^2)' - \int_0^x t^2\,dt\\ &= 2x - \frac{1}{3}x^3\\ &\vdots\\ T(x^n) &= (x^n)' - \int_0^x t^n\,dt\\ &= nx^{n-1} - \frac{1}{n+1}x^{n+1}. \end{align*}$$ If $p(x) = a_0+a_1x+\cdots+a_nx^n$, under what conditions will $T(p(x))=0$?

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    i need little more explanation sir. tried to do by taking n = 3 then i got dim of kernal zero. but i am not sure whether i am right or not?2012-05-08
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    There is nothing more to be explained; the only further thing I could do is solve the problem for you. At this point, what you need to do is be sure of your computations. *How* did you figure out that the kernel was trivial in the $n=3$ case? Does the argument apply to any other value of $n$?2012-05-08
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    I made a matrix for T , taking n = 2 and found that dimension of range space of the matrix is coming out to be 3 which means that dim of null space must be trivial. but i am not sure about the general case.2012-05-08
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    Don't do the matrix; look at the linear transformation. Look at the images. Look at what you are getting. Or take a polynomial $p(x) = a_0 +a_1x+\cdots+a_rx^r$ with $a_r\neq 0$ and **compute** the value of $T(p)$. What do you get?2012-05-08
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Hint: What is the leading term of $T(p)$ in terms of the leading term of $p$? What does this tell you about $p$ such that $T(p)=0$?