# 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.

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Is your question about the Frobenius normal form, or the Jordan normal form? – Chris Eagle Dec 4 '12 at 22:45
At first I thought that the Jordan normal form is not as good as the Frobenius normal form when the minimalpolynomial consists of many zero coefficients, but now I'm not so sure because it seems that the less zeros your polynomial has, the less number of multiple roots, which give you the 1s in your Jordan form, are possible so they might be more or less equally good. – Dominik Dec 4 '12 at 22:59

## 2 Answers

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 $12$.

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The Frobenius normal form of a companion matrix is the companion matrix itself. – Chris Godsil Dec 5 '12 at 3:08

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.

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