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.

  • $\begingroup$ I was able to get it, the proof is by induction over the decomopsition of the vector space by the primary decomposition theorem. $\endgroup$ – Nick R Feb 10 '14 at 12:32

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