I have the differential operator: $$\frac{d}{dx} +2x \cdot$$ operating on $P$, the polynomials. I want to know the dimension of the kernel of $\frac{d}{dx} +2x\cdot$.
My argument is that the dimesnion is zero because solving $$\frac{dy}{dx} +2xy=0$$ produces $$y(x) = Ae^{-x^2}$$ which is clearly not in $P$. But, my friend argues that since we can write $$e^{-x^2} = \sum_{n=0}^{\infty}\frac{(-x^2)^n}{n!}$$ the solution is in $P$, and hence dimension is 1. I'm not comfortable with this argument, but I can't explain why. Is my friend wrong? Why or why not?
If you apply the operator to any non-trivial polynomial with degree $n$, then the result will have degree $n+1$ by the second term.
Only the zero polynomial thus results in the zero polynomial.