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Notations:

$\mathcal{M}_n(\Bbb{R}) $: the set of all $n×n$ matrices over $\Bbb{R}$

$\chi_A(x)$: Characteristic polynomial of $A$

$m_A(x)$ : Minimal polynomial of $A$

$A\sim B$ : $\exists P\in Gl_n(\Bbb{R})$ such that $B=P^{-1}AP$

Conjecture: $(\forall A, B\in \mathcal{M}_n(\Bbb{R})$ with $\chi_A=\chi_B$ and $m_A=m_B$ implies $A\sim B) $implies $n\le 3$

Attempt:

Let's work with converse.

For $n\ge 4,\exists A, B\in\mathcal{M}_n(\Bbb{R}) $ with $\chi_A=\chi_B$ and $m_A=m_B$ such that $A\not\sim B\tag{1}$

Special cases :

$\boxed{n=4}:$

$A=\left(\begin{array}{cccc} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 \end{array}\right)$

$B= \left(\begin{array}{cccc} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right)$

$$\chi_A(x) =x^4=\chi_B(x) $$ and $$m_A(x) =x^2=m_B(x) $$

$\begin{align}\textrm{rank}(A)=2\neq 1=\textrm{rank}(B) &\implies A\not\sim B\\\end{align}$

$\boxed{n=5}$:

$A=\left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$

$B= \left(\begin{array}{ccccc} 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 & 0 \end{array}\right)$

$$\chi_A(x) =x^5=\chi_B(x) $$ and $$m_A(x) =x^2=m_B(x) $$

$\begin{align}\textrm{rank}(A)=2\neq 1=\textrm{rank}(B) &\implies A\not\sim B\\\end{align}$

$\boxed{\text{General cases}(n\ge 4)}$:

Strategy: Find two nilpotent matrices $A,B \in \mathcal{M}_n(\Bbb{R})$ with degree of nilpotency $2$ and $\textrm{rank}(A) \neq \textrm{rank}(B)$

$A=(a_{ij}) $ and $B=(b_{ij}) $

where $a_{ij}=\begin{cases} 1 &(i,j)\in\{(2,1),(n-1,n)\}\\0& \text{otherwise}\end{cases}$

$b_{ij}=\begin{cases} 1 &(i,j) =(2,1)\\0& \text{otherwise}\end{cases}$

Then $$\chi_A(x) =x^n=\chi_B(x) $$ and $$m_A(x) =x^2=m_B(x) $$

$\begin{align}\textrm{rank}(A)=2\neq 1=\textrm{rank}(B) &\implies A\not\sim B\\\end{align}$

For $n\le 3$ and $\forall A,B\in\mathcal{M}_n(\Bbb{R})$ with $\chi_A=\chi_B $ and $m_A =m_B $ implies $A\sim B $(here)$\tag{2}$

Conclusion: $(1) $ and $(2) $ together implies that the conjecture is true.

Question:

  1. Is my attempt correct?
  1. What about the direct proof $[$ starting with any two matrices having the said properties and then showing $\textrm{deg}(\chi_A) =\textrm{deg}(\chi_B)\le 3]$ ?

Note: Two matrices are similar iff both have the same Jordan normal form upto the permutation of Jordan blocks.

Sourav Ghosh
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    Your idea works, but your examples are not entirely correct, since your minimal polynomials are certainly not $X^2$, but $X^{n-2}$. However, they still do their job. – GreginGre Apr 14 '23 at 07:17
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    This is already known, e.g., see here, or posts on this site (see this post for your examples.) – Dietrich Burde Apr 14 '23 at 12:27
  • @DietrichBurde I believe the proof of "$(\forall A, B\in \mathcal{M}_n(\Bbb{R}) $ with $\chi_A=\chi_B$ and $m_A=m_B$ implies $A\sim B) $ implies $n\le 3$ can't be found on MSE? – Sourav Ghosh Apr 14 '23 at 13:02
  • $(1)$ and $(2)$ are on MSE, and as you say yourself - this immediately proves your conjecture. I didn't search yet, but I think it is also stated on MSE somewhere (but I did not downvote!) – Dietrich Burde Apr 14 '23 at 13:04
  • I only found this post, where DonAntonio states it as "a nice exercise", but does not prove it. I remember it was proved somewhere on MSE, but cannot find it right now - so perhaps you are right. – Dietrich Burde Apr 14 '23 at 13:14
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    Just take block-diagonal matrices whose first diagonal block are the matrices A and B of your first example, and all other entries are 0. – Christophe Leuridan Apr 19 '23 at 12:15

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