Finding all eigenvectors and eigenvalues of a linear operation on a function Here is the question I am stuck on: 
Consider $T \in \mathrm{Hom}(\Bbb{R}[x]_{\le 2} ,\Bbb{R}[x]_{\le 2} )$ given by
$$ (Tf)(x)=\int_{-1}^1(x-y)^2f(y)dy-2f(0)x^2$$
for all $f \in \Bbb{R}[x]_{\le 2}$. Find all eigenvalues and eigenvectors for T.
So far, I'm trying to apply T to the monomial basis of $\Bbb{R}[x]_{\le 2}$.
I got that $T(1)=-2x$, but where I'm stuck is how to evaluate $T(x)$ for the $f(y)$ and $f(0)$ terms. I thought that if I applied $T(x)$ to $f(y)$, it would equal $y$ because the function is $x$, and $f(y)$ means $x=y$. When I evaluate this with the $f(y)$ term equal to $y$, I get back that $T(x)=-(4/3)x-2x^3$. The $2x^3$ term's degree is too high for the bounds, so I'm not sure how to proceed from here. 
Is it wrong to say that $x$ evaluated at $y$ equals $y$? 
Any help would be greatly appreciated.

Note that we should integrate with respect to $y$, not $x$. Write $f(x) = x$. Then
$$ T(f)(x) = \int_{-1}^1 (x - y)^2 f(y) \, dy - 2f(0)x^2 = \int_{-1}^1 (x - y)^2 y \, dy = \int_{-1}^1 (x^2y - 2xy^2 + y^3) \, dy =\left[ \frac{x^2y^2}{2} - \frac{2xy^3}{3} + \frac{y^4}{4} \right]^{y = 1}_{y = -1} = -\frac{4}{3} x. $$
