Given a square matrix A (say with complex entries), which is the sparsest matrix which is similar to A?

I guess it has to be its Jordan normal form but I am not sure.


  • A matrix is sparser than other if it has less nonzero entries.

  • Two square $n \times n$ matrices $A,C$ are similar if there exists and invertible matrix $P$ such that $A = P^{-1}CP$

  • $\begingroup$ If the matrix is diagonalisable, clearly its Jordan form is the sparsest one in its similarity class because you cannot get fewer than $\operatorname{rank}(A)$ entries. $\endgroup$ – user1551 Oct 19 '18 at 11:32
  • $\begingroup$ My intuition says the same, but apart from trivial cases ($A$ is diagonaliseable) I do not see an immediate proof of that. $\endgroup$ – TZakrevskiy Oct 19 '18 at 11:32
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    $\begingroup$ Simulposted to MO, mathoverflow.net/questions/313224/sparsest-similar-matrix DON'T DO THAT! $\endgroup$ – Gerry Myerson Oct 19 '18 at 12:00

The companion matrix to the polynomial $(x^2-1)^2=1-2x^2+x^4$ is $$\pmatrix{0&0&0&-1\cr1&0&0&0\cr0&1&0&2\cr0&0&1&0\cr}$$ which has Jordan form $$\pmatrix{1&1&0&0\cr0&1&0&0\cr0&0&-1&1\cr0&0&0&-1\cr}$$ which has more nonzero entries.

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    $\begingroup$ The following paper claims that the companion matrix is the sparsest in a certain sense: doi.org/10.1016/j.laa.2012.08.017 $\endgroup$ – Julian Kuelshammer Oct 19 '18 at 13:24
  • $\begingroup$ That is an interesting example. But if we call the first matrix $A$ and the second one $C$, to me it seems (motivated by Jordan normal form definition) among all matrices that are NOT equal to $A$, $C$ is the sparsest similar matrix to $A$. Can your example be generalized to prove my statement is incorrect? $\endgroup$ – abolfazl Oct 19 '18 at 16:41
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    $\begingroup$ You can always get a different but equally sparse matrix by conjugating by some diagonal and/or permutation matrix . . . $\endgroup$ – Noam D. Elkies Oct 19 '18 at 18:34

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