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 $p(x)$?
Dimension of null space of a given problem
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linear-algebra
2 Answers
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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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0Don'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 $T(p)=0$?