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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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2 Answers 2

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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? –  srijan May 8 '12 at 19:35
    
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 $n$? –  Arturo Magidin May 8 '12 at 19:36
    
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. –  srijan May 8 '12 at 19:48
    
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? –  Arturo Magidin May 8 '12 at 20:04

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 $T(p)=0$?

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