Basis of $\ker T$ for $\mathcal{A}=\left\{ e^{n\cdot x}\mid n\in\mathbb{Z}\right\}$? I am given the linear transformation $T:V\rightarrow V$ defined by $f\mapsto f''+2f'-3f$, where $V\subset\mathcal{C}^\infty$, the linear space of all real value smooth functions defined over $\mathbb{R}$.
I am asked to find a basis of $\text{ker }T$, which I assumes involves writing out some change of basis matrix.
What values of $x$ should I be using to calculate the coefficients? I understand that $f''+2f'-3f=n^2e^{nx}+2ne^{nx}-3e^{nx}$ for $f=e^{nx}.$ $\,\mathcal{B}=\left\{1,x,x^2\right\}$ seems relevant, but so does $P^n$. Where do I start?
 A: If $V$ means the subspace generated by $\mathcal{A}$, you should look at the (finite) linear combinations of the basis elements that are mapped to $0$ by $T$. It's quite clear that those functions form a linearly independent set (you get a Vandermonde determinant).
Let's start with looking at the action of $T$ on one basis element. Set $f_n(x)=e^{nx}$. Then
$$
T(f_n)=f_n''+2f_n'-3f_n
$$
and
$$
f_n''(x)+2f_n'(x)-3f_n(x)=(n^2+2n-3)f_n(x)
$$
so $T(f_n)=(n^2+2n-3)f_n$. If we have a linear combination, say
$$
f=\sum_{k=1}^r \alpha_kf_{n_k}
$$
then
$$
T(f)=\sum_{k=1}^r \alpha_k(n_k^2+2n_k-3)f_{n_k}
$$
and $T(f)=0$ implies
$$
\alpha_k(n_k^2+2n_k-3)=0\qquad(k=1,2,\dots,r)
$$
Now, $t^2+2t-3=0$ implies $t=-3$ or $t=1$. Thus, if $n_k\ne1$ and $n_k\ne-3$ (for $k=1,\dots,r$), the equation tells $f=0$. Hence the condition $f\in\ker T$ is equivalent to $f=\alpha_1f_1+\alpha_{-3}f_{-3}$.

If you know how to solve a linear differential equation with constant coefficient, you have that any $\mathcal{C}^\infty$ function defined on $\mathbb{R}$ satisfying $f''+2f'-3f=0$ must be of the form $f(x)=ae^x+be^{-3x}$, so the result above is confirmed.
A: Note that
$v \in ker(T) \iff T(v)=0$
if $v = e^{n x}$, $T(v) = n^2e^{nx}+2ne^{nx}−3 = 0 \iff$ $n=3$ or $n=-1$
So $\mathcal{C}=\{e^{3x},e^{-x}\} \in ker(T)$. If you show that $\mathcal{A}$ generates $V$, you are done, that is, $\mathcal{C}$ is a basis of $ker(T)$.
