Suppose that $T : V \rightarrow V$ is a linear transformation over R whose minimal polynomial has no multiple roots. Show that, relative to a suitable basis, $T$ can be represented by an $n$ by $n$ matrix with at most $2n$ non-zero entries, where $n = \dim(V)$.
My attempt at a solution is by noticing that because the minimal polynomial has no multiple roots, it has to be equal to the characteristic polynomial for $T$ as well. From there my idea kind of surrounds upon the idea that $T$ is almost diagonalizable, but not enough. So I thought we could pick up a basis of eigenvectors and then just extend it to span the entire space. But that didn't end up working anywhere.