I believe in my case it's the 2nd degree polynomial.

The "set of all 2nd degree polynomials" does not form a vector space. As dwsmith said, \(\displaystyle P_2\) is the set of all polynomials of degree 2

**or less**.

Random Variable showed how to represent this transformation as a matrix- worth knowing for itself.

But for this problem, you don't have to do that. As dwsmith said, if \(\displaystyle p= ax^2+ bx+ c\), then \(\displaystyle T(p)= xp'= x(2ax+ b)= 2ax^2+ bx\). If \(\displaystyle \lambda\) is an eigenvalue for T, we must have some a, b, c, not all 0, such that \(\displaystyle T(p)= 2ax^2+ bx= \lambda(ax^2+ bx+ c)\). Since that must be true for all x, we can set corresponding coefficients equal: \(\displaystyle 2a= \lambda a\), \(\displaystyle b= \lambda b\), and \(\displaystyle 0= \lambda c\).

If a is not 0, then \(\displaystyle \lambda= 2\), if b is not 0, then \(\displaystyle \lambda= 1\), and if c is not 0, then \(\displaystyle \lambda= 0\).

That is, 0 is an eigenvalue with eigenvectors any multiple of 1, 1 is an eigenvalue with eigenvectors any multiple of x, and 2 is an eigenvalue with eigenvectors any multiple of \(\displaystyle x^2\), exactly what Random Variable got.