Kernel of a linear transformation $D : \mathbb{C}^{\infty}(\mathbb{R}) \to \mathbb{C}^{\infty}(\mathbb{R})$

Question

Let $D : \mathbb{C}^{\infty}(\mathbb{R}) \to \mathbb{C}^{\infty}(\mathbb{R})$ be the function given by differentiation: $D(f) = f'$.

I've shown that $D$ is a linear transformation by using rules for derivative of a sum of two functions and derivative of a scalar (constant) multiplied by a function.

I now want to find the kernel of $D$ (if it is finite, find a basis $\mathbb{B}$ for $D$) aswell as the eigenvalues of $D$.

I interpreted $\mathbb{C}^{\infty}(\mathbb{R})$ as the smooth functions from $D$ to $\mathbb{R}$.

I've concluded that the only thing $D$ sends to $0$ are constants. Therefore: $\ker(D) = \{ a \in \mathbb{R}\}$. $\ker(D)$ is not finite and we cannot find a basis for $D$. Is my conclusion correct?

For the eigenvalues I would appreciate a hint.

Answer 1

You are right about what the kernel is, however, is it not finite? What's wrong with the basis consisting of the constant function $f(x)=1$?

Finding the eigenvalues works the same way. The ODE $$ Df=\lambda f\iff f'=\lambda f $$ might be of interest.

edit 1: Given the confusion below, the $\lambda$ I have written is meant to represent an eigenvalue in $\mathbb{R}$, not a matrix representing the linear transformation $D$.

edit 2: Solution. We are looking for functions $f$ which satisfy $$ Df=\lambda f $$ i.e. eigenvectors with eigenvalue $\lambda$. Recalling ode, we separate and integrate or ansatz solutions $$ f(x)=ce^{\lambda x} $$ Where $c$ is a constant. One may check that $Df=\lambda ce^{\lambda x}=\lambda f$ by the chain rule. This will be a valid eigenvector for any $\lambda\in \mathbb{R}$, an uncountable set of eigenvectors which is pairwise linearly independent.

Answer 2

Yes, the kernel is the set of constant functions, and yes, there are infinitely may such functions, but they are not linearly independent.

Any nonzero constant function generates all the other constant functions, so take any one of them, and you get a basis.

As regards the eigenvalues of $D$, note that if $\lambda\in\mathbb{R}$, and $f(x)=e^{\lambda x}$, then $D(f)=\lambda f$.