If matrix $X$ is sparse, does it mean that $e^X$ is also sparse?

Question

Let's say, that we have a sparse matrix $X$. Then, is $e^X$ also sparse?

At first glance I would say, that it is not always true, because we could expand this function in Taylor series and I think, that consecutive powers of matrix $X$ do not have to be sparse.

Answer 1

Here is a counter example : \begin{align*} M= \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \end{align*}

We can compute \begin{align*} M^2= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \end{align*}

and so \begin{align*} M^p=M^{p \text{ mod }2} \end{align*}

This gives \begin{align*} e^M &= \begin{bmatrix} 1 & 0\\ 0 & 1 \end{bmatrix} \sum_{k=0}^\infty \frac{1}{(2k)!} + \begin{bmatrix} 0 & 1\\ 1 & 0 \end{bmatrix} \sum_{k=0}^\infty \frac{1}{(2k+1)!}\\ &=\begin{bmatrix} \cosh(1) & \sinh(1)\\ \sinh(1) & \cosh(1) \end{bmatrix} \end{align*} Which is not sparse.