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By the circulant matrix $C \in M_n(\mathbb{R})$, we mean that

$$ C = \left[\begin{array}{c|c|c|c} e_n & e_1 & \cdots & e_{n-1} \end{array}\right] $$

where $e_1,\dots,e_n$ are the standard basis vectors in $\mathbb{R}^n$. Let $U=(u_{ij})$ be a upper triangular matrix. We say it is super upper triangular if $u_{i, i+1}=0$.

Here, we just focus on the super upper triangular matrices $U$ whose entries $u_{i,j}$ are either $0$ or $1$.

Q. Can we characterize super upper triangular matrices $U$ such that $C+U$ are diagonalizable?

Remarks:

  1. In this MathOverflow post, an example of non-diagonalizable $C+U$ is given when the size $n=8$.

  2. We (checked by MATLAB) found out the following result by experience: If $n\leq6$, all matrices $C+U$ are diagonalizable. When $n=7$ among all sums, there are just $518$ cases such that $C+U$ are not diagonalizable.

  3. One may check that $C+U$ is diagonalizable if and only if the roots of the characteristic polynomial are all simple.

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  • $\begingroup$ No, they are different. Please take a lookt at the definition again. $\endgroup$
    – ABB
    Jun 2, 2023 at 13:52
  • $\begingroup$ Indeed an strictly upper triangular matrix is an upper triangular matrix having 0s along the diagonal as well as the lower portion. $\endgroup$
    – ABB
    Jun 2, 2023 at 13:52
  • $\begingroup$ I don't think 3. is true. When I did my search, I seem to remember coming across an example with a root of multiplicity two that was still diagonalisable, before I came across the example I gave you. $\endgroup$ Jun 12, 2023 at 12:54

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