# Dimension of null space of a given problem

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 $p(x)$?

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 $T(p(x))=0$?
• 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 $n$? May 8, 2012 at 19:36
• 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? May 8, 2012 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 $T(p)=0$?