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For which $n$ is it true that two $n \times n$ matrices are similar if they have the same minimal polynomial and the same characteristic polynomial?

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  • $\begingroup$ What do you believe? $\endgroup$ – I am not Paul Erdos Oct 2 '17 at 9:40
  • $\begingroup$ I have no idea, unfortunately. $\endgroup$ – Ludwwwig Oct 2 '17 at 9:43
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    $\begingroup$ I’ll think about it a bit more, it’s not very easy I feel. The tip is to use the Jordan form $\endgroup$ – Ludwwwig Oct 2 '17 at 9:53
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    $\begingroup$ I think that over the complex numbers, for $n=2,3$ you need only check the characteristic and minimal polynomials, but for $n=4$ they are not a sufficient criterion for similarity. $\endgroup$ – Joppy Oct 2 '17 at 10:06
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    $\begingroup$ The tag-info for characteristic-function explicitly says: "Do not use this tag if you are asking about ... the characteristic polynomial in linear algebra." (You might have noticed that tag-excerpt is displayed when you are adding a tag - in order to help with correct tagging.) $\endgroup$ – Martin Sleziak Oct 3 '17 at 5:07
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Counterexample for $n = 4$: $$\begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 1 \\ 0 & 0 & 0 & 2 \end{pmatrix}, \begin{pmatrix} 2 & 1 & 0 & 0 \\ 0 & 2 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 2 \end{pmatrix}.$$ From reading down the diagonal, both of these have only $\lambda = 2$ as an eigenvalue with multiplicity $4$. Computation reveals that $(z - 2)^2$ is the minimal polynomial (can be seen by the fact that $2$ is the size of the largest Jordan Block).

But, they are not similar, as the former has an eigenspace of dimension $2$, whereas the latter has an eigenspace of dimension $3$.

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If $n \ge 4$, we can get a counterexample as follows . . .

Choose a basis $e_1,...,e_n$ (for example, the standard basis).

Let $A$ be the $n{\,\times\,}n$ matrix such that

  • $Ae_1=0$
  • $Ae_k=e_1,\;$for $k>1$

and let $B$ be the $n{\,\times\,}n$ matrix such that

  • $Be_1=0$
  • $Be_2=0$
  • $Be_3=e_1$
  • $Be_k=e_2,\;$for $k \ge 4$

Then $A,B$ are nonzero, but $A^2$ and $B^2$ are both zero.

It follows that both of $A,B$ have minimal polynomial $x^2$, and characteristic polynomial $x^n$.

But $A$ has rank $1$, and $B$ has rank $2$, hence $A,B$ are not similar.

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Here is my attempt at a solution for $n = 2,3$. If the Jordan forms of two matrices $A, B$ are the same, then they are obviously similar. Hence, we consider all combinations of non-equal Jordan forms and try to show that the matrices still have to be similar.

$n = 2$

$A, B$ both have at most 2 eigenvalues. If both are same, then the Jordan forms could be

$ J_A = \begin{bmatrix}\lambda_1 & 1 \\0 & \lambda_1 \end{bmatrix}$

and

$ J_B = \begin{bmatrix}\lambda_1 & 0 \\ 0 & \lambda_1 \end{bmatrix}$

However, in this case, the minimal polynomial is no longer the same since $J_B-\lambda_1 = 0 \neq J_A - \lambda_1$. Hence, the matrices must have the same Jordan form.

If the two matrices have two eigenvalues $\lambda_1, \lambda_2$ then the Jordan form of the two matrices will be exactly the same, $\texttt{diag(lambda_1, lambda_2})$.

n = 3

Again, if we have three eigenvalues then the Jordan form will be the same for $A, B$ which implies that $A,B$ are similar.

I am not really sure on how to continue on from here. It feels like there are many cases to consider. Someone have any suggestion?

What happens for instance when we have $ J_A = \begin{bmatrix}\lambda_1 & 1 & 0 \\ 0 & \lambda_1 & 0 \\ 0 & 0 & \lambda_2 \end{bmatrix}$

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