Why is there no simpler form of a matrix than the Jordan or Frobenius normal form? 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. 
 A: This might be a counterexample for the Jordan form. The companion matrix of the polynomial $$(x^2-1)^3=x^6-3x^4+3x^2-1$$ is $$A=\pmatrix{0&0&0&0&0&1\cr1&0&0&0&0&0\cr0&1&0&0&0&-3\cr0&0&1&0&0&0\cr0&0&0&1&0&3\cr0&0&0&0&1&0\cr}$$ with $8$ nonzero entries, and I think the Jordan form for this matrix is $$\pmatrix{1&1&0&0&0&0\cr0&1&1&0&0&0\cr0&0&1&0&0&0\cr0&0&0&-1&1&0\cr0&0&0&0&-1&1\cr0&0&0&0&0&-1\cr}$$ with $10$ nonzero entries. I'm not so familiar with the Frobenius normal form. (EDIT:) I think it's another name for "rational canonical form," and I think the rational canonical form for $$B=\pmatrix{2&0\cr0&3\cr}$$ is $$\pmatrix{0&-6\cr1&5\cr}$$ So it seems to me that the matrix $$A\oplus B=\pmatrix{A&0\cr0&B\cr}$$ has $10$ nonzero entries, its Frobenius normal form has $11$, and its Jordan form has $12$. 
A: 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.
