TITLE: Why is there no simpler form of a matrix than the Jordan or Frobenius normal form?
QUESTION [8 upvotes]: The Jordan and Frobenius normal forms of a linear map $A:\Bbb R^n \rightarrow \Bbb R^n$ seem to be maximally simple representations of $A$ in the sense that one of them contains as few nonzero entries as possible. But how do you prove that? 
More precise, show that for every $A:\Bbb R^n \rightarrow \Bbb R^n$ and every Basis $B$ of $\Bbb R^n$, the transformation matrix $_B A _B$ has at least as many nonzero entries as the Jordan normal form or the Frobenius normal form of $A$ or, otherwise, give a counterexample.

REPLY [2 votes]: The Frobenius normal form of
$$\begin{pmatrix}1 & 0 \\ 0 & 2\end{pmatrix}$$
is
$$\begin{pmatrix}0 & 1 \\ -2 & 3\end{pmatrix}.$$
So the Frobenius normal form doesn't always give the minimum possible number of nonzero entries.
For the Jordan normal form, see the answer of Gerry Myerson.