Dimension of null space of linear map? Suppose V is the real vector space of real-valued functions with a derivative. What is the dimension of the null space of the linear map
$$Tf = x\frac{df}{dx} - 4f\;\;?$$
The basis I found for the null space is $\{x^4\}$, which gives dimension 1, but the hint for problem was that the dimension is not 1. Why is that so?
 A: You can solve the equation $x{dy\over dx}=4y$ using separation of variables:
$$\tag{1}
{1\over y}\,dy={4\over x}\,dx.
$$
We want to avoid division by zero; so, in solving $(1)$, we consider the cases where $x<0$ and $x>0$.  Integration of $(1)$ shows that for $k$ a constant,  $y=k x^4$ is a solution for both the domain $x>0$ and the domain $x<0$.
One may then verify that the  solutions of $(1)$ over $\Bbb R$ have the form 
$$\tag{2}
f(x)=\cases{cx^4,&$x\ge0$\cr d x^4,&$x\le 0$ }
$$
where $c$ and $d$ are constants (in particular, one can  (and should)  show such an $f$ is differentiable at $0$ with $f'(0)=0$).
Note that any such $f$ is a linear combination of the independent functions
obtained by taking $(c=1, d=0)$ and $(c=0,d=1)$ in $(2)$.
A: We have a simple linear homogeneneous differential equation of order $\,1\,$ to solve
$$Tf=0\Longrightarrow x\frac{df}{dx}=4f\Longrightarrow \int\frac{df}{f}=4\int\frac{dx}{x}\Longrightarrow$$
$$\log|f|=4\log|x|+C\Longrightarrow f=kx^4\,\,,\,k=\,\text{a constant}$$
and yes: you're answer is correct.
Whose hint was that  $\,\dim\ker T\neq 1\,$ ?
