1
$\begingroup$

Question: Can we find two diagonalizable traceless matrices $A,B\in M(n,\Bbb C)$ with the same minimal polynomial but such that $A$ and $B$ are not similar?

There was a related question here. In that case the answer was positive, but they don't assume that $A$ and $B$ are diagonalizable and traceless.


Attempt: The fact that $A$ and $B$ are diagonalizable with the same minimal polynomial means that their set of eigenvalues are the same. Hence, they could only differ in the multiplicities of their eigenvalues. Let $\lambda_1,\ldots,\lambda_k\in\Bbb C$ be those distinct eigenvalues. Wlog, $A$ and $B$ are already diagonal, so $$A={\rm diag}(\lambda_1,\ldots,\lambda_1,\lambda_2,\ldots,\lambda_2,\ldots,\lambda_k,\ldots,\lambda_k),$$ where $\lambda_j$ is repeated $m_j$ times, and $$B={\rm diag}(\lambda_1,\ldots,\lambda_1,\lambda_2,\ldots,\lambda_2,\ldots,\lambda_k,\ldots,\lambda_k),$$ where $\lambda_j$ is repeated $n_j$ times. Now the condition that they are traceless means $$ \begin{align} m_1\lambda_1+\cdots+m_k\lambda_k&=0\\ n_1\lambda_1+\cdots+n_k\lambda_k&=0. \end{align} $$ Thus, the question can be reformulated as follows:

Reformulation: Can we find distinct $\lambda_1,\ldots,\lambda_k\in\Bbb C$ and two different vectors $(m_1,\ldots,m_k),(n_1,\ldots,n_k)\in(\Bbb Z_{\ge 1})^k$ such that $$ \begin{align} m_1\lambda_1+\cdots+m_k\lambda_k&=0\\ n_1\lambda_1+\cdots+n_k\lambda_k&=0 \\ m_1+\cdots+m_k&=n_1+\cdots+n_k? \end{align} $$

$\endgroup$

1 Answer 1

2
$\begingroup$

$1+1+1+1-1-1-1-1+2+2-2-2=1+1+1+1+1+1-1-1+2-2-2-2$

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .