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I have a linear operator $L(p(x)) = -p(x+1) + p(x) + \frac6x \int_0^x p(y) dy$ with eigenvalues of it's representative matrix found to be 6, 3, and 2. From this the respective eigen vectors are (1,0,0) , (1,3,0) , (3,8,4).

I'm asked to find the eigenfunctions for this linear operator. I know that I need to find some function f s.t $L(f) = \lambda f$ but I don't know how to actually do this.

If someone could explain that would be great.

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  • $\begingroup$ Your statements are highly confused. You defined $L$ as function of the real variable $x$, but then talk about it as an operator on functions, not real numbers. You mention eigenfunctions, but also list 3 elements of $\Bbb R^3$ as being "eigenvectors", though there is nothing involving $\Bbb R^3$ in this problem. I can easily suppose you actually mean $L(p)(x)$ - that is, $L$ is an operator on the function $p$. But the three 3D "eigenvectors" are a complete non-sequitur here. They make no sense at all. $\endgroup$ – Paul Sinclair May 5 at 18:40
  • $\begingroup$ @PaulSinclair Then I'll edit it to make sense, I did in fact mean L(p)(x) as an operator, it was a typo, and the eigenvectors are the eigenvectors relating to the matrix that respresents L on the space of polynomials of degree 3. $\endgroup$ – mathPhys May 7 at 16:47
  • $\begingroup$ ?? That $L$ is meant to only be an operator on a finite dimensional space of polynomials is CRITICAL INFORMATION that should have been in your problem statement. Yet with your edit, it still isn't there. Also, the space of 3rd degree polynomials is 4-dimensional (coefficients of $\{x^3, x^2, x, 1\}$) so I assume you actually mean the space of quadratic (2nd degree) polynomials. $\endgroup$ – Paul Sinclair May 8 at 0:03
  • $\begingroup$ And now I am wondering what your difficulty is. You have the eigenfunctions expressed in vector form. You apparenly know the transformation between vectors and polynomials. Why have you not used it to convert your eigenvectors into eigenfunctions? $\endgroup$ – Paul Sinclair May 8 at 0:13
  • $\begingroup$ @PaulSinclair So the eigen functions here are just the polynomials represented by the eigenvectors? $\endgroup$ – mathPhys May 8 at 9:02

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