# Eigenvalues and Eigenvectors of a Specific Integral Operator

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!

~Mo

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.
• 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? Oct 6, 2019 at 19:06