I'm working on the following linear algebra problem: Let $R[x]_n$ be the space of polynomials of degree less than or equal to $n$. Find the eigenvectors and eigenvalues of the operator $\frac{1}{x}\int_0^x f(t)dt$ on the space $R[x]_n$.

Here's my attempt at a solution:

Define the operator $T$ by $Tf(x) = \frac{1}{x}\int_0^x f(t)dt$ , where $f(x) = a_nx^n + a_{n-1}x^{n-1} + ... + a_1x + a_0 \in R[x]_n$. Then $Tf(x) = \frac{1}{x}(\frac{a_n}{n+1}x^{n+1} + \frac{a_{n-1}}{n}x^n + ... + \frac{a_1}{2}x^2 + \frac{a_0}{1}x^1)$

$= \frac{a_n}{n+1}x^n + \frac{a_{n-1}}{n}x^{n-1} + ... + \frac{a_1}{2}x + \frac{a_0}{1}x^0$.

We now find eigenvalues $\lambda$ of the operator $T$:

$Tf(x) = \lambda f(x)$ (where $f(x) \neq 0$)

$\Rightarrow$ $\frac{a_n}{n+1}x^n + \frac{a_{n-1}}{n}x^{n-1} + ... + \frac{a_1}{2}x + \frac{a_0}{1}x^0 = \lambda a_nx^n + \lambda a_{n-1}x^{n-1} + ... + \lambda a_1x + \lambda a_0$

$\Rightarrow$ by matching up coefficients, $\frac{a_n}{n+1} = \lambda a_n , \frac{a_{n-1}}{n} = \lambda a_{n-1}, ... , \frac{a_0}{1} = \lambda a_0$.

This is the part where I get tripped up. Based on the $n+1$ above equations, it seems that it is impossible to find such a $\lambda$ that will simultaneously solve all $n+1$ equations. Thus, it seems that there are no eigenvalues, and therefore, no corresponding eigenvectors.

Would this be correct? Thanks a lot in advance math friends!



1 Answer 1


Hint: there are eigen values. If you take $a_i=1$ and all other $a_j=0$ you see that $\lambda =\frac 1 j$ is an eigen value for $1\leq j \leq n+1$. Now it shouldn't be difficult for you to show that these are the only eigen values. It is also easy to write down eigen functions corresponding to these eigen values.

  • $\begingroup$ Hi Kavi! Thanks for the reply. I believe I understand what you're saying - the only way it's possible that lambda is an eigenvalue is if the original polynomial only has one term a_ix^i ! And in this case, the eigenvalue is just 1/(i+1)! Thus, we get n eigenvalues - one for each polynomial f(x) = a_ix^i with degree i, where $1 < = i < = n$? And those polynomials f(x) there are exactly the eigenfunctions? $\endgroup$ Oct 6, 2019 at 19:06
  • 1
    $\begingroup$ @MoBehzadKang Yes, what you are saying is right. $\endgroup$ Oct 6, 2019 at 23:12

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