Eigenvectors of derivative I'm trying to consider how linear algebra relates to calculus.  It seems to me that the only eigenvectors of the derivative operator on $\Bbb R$ are the functions $ce^{kx}$ for constants $c$ and $k$.  The eigenvalues associated with these eigenspaces are $k$.  I'm not sure how to prove uniqueness of this, but these seem to be the only ones.
OK.  What about derivatives on $\Bbb R^n$?  $e^{\mathbf x}$ is not a thing because a power series of vectors $\mathbf x$ doesn't make sense (I think).  So what are the eigenvalues and eigenvectors of the derivative in this case?  
Are there any other interesting results from linear algebra that apply to calculus?
 A: Denote the differential operator by $D$. The equation $Df=\lambda f$ is equivalent to $f'=\lambda f$. Multiplying by $e^{-\lambda x}$ we obtain:
$$f'(x)e^{-\lambda x}-e^{-\lambda x}\lambda f(x)=0$$
which is the same as
$$(f(x)e^{-\lambda x})'=0$$
Which means that there exists a constant $c$ such that
$$f(x)e^{-\lambda x}=c \ \text{for all $x$}$$
Hence $$f(x)=ce^{\lambda x}$$
as desired.
Regarding $\mathbb{R}^n$: Recall that a function $f:A \subset \mathbb{R}^n \to \mathbb{R}^m$ ($A$ open) is said to be differentiable at $x \in A$ if there exists a linear transformation, denoted by $Df(x)$ or $f'(x)$ such that: 
$$\lim_{h \to 0}{\left \| f(x+h)-f(x)-(Df(x))(h) \right \| \over \left \| h \right \|}=0$$
Note that $Df(x)$ is linear function between $\mathbb{R}^n$ and $\mathbb{R}^m$. Furthermore, $Df(x)$ is the only linear transformation which verifies the limit above.
As you can see the question $f=f'$ doesn't make as much sense in this context.
However, if we take a linear transformation $T: \mathbb{R}^n \to \mathbb{R}^m$ and any $x \in \mathbb{R}^n$ it is easy to verify, by using the limit above, that:
$$T=T'(x)$$
