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Let us consider the backward-shift matrix $B=(b_{ij})\in M_n(\mathbb{R})$ whose entries are given by $b_{k,k+1}=1$ and the other entries are all 0. We also consider $X=(x_{ij})\in M_n(\mathbb{R})$ where $x_{n,1}=1$ and the other entries are all zero.

Obviously, the nilpotent matrix $B$ admits only 0 as the eigenvalue. The sum $B+X$ is the circulant matrix and it is well-known that its spectrum are just the nth roots of $1$.

Let us consider an upper triangular matrix $A=(a_{i,j})\in M_n(\mathbb{R})$ with

$$\begin{align*} a_{i,i+1}&=1& \\ a_{i,j}&=0 ~~ &~~ j\leq i \\ a_{i,j}&\in\{0,1\} &~~ i+2\leq j \end{align*} $$

Q. Are the eigenvalues of $A+X$ distinct?

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  • $\begingroup$ If $A$ is zero and $n \ge 2$, then $A+E$ has only the eigenvalue $0$ with geometric multiplicity $n-1$ and algebraic multiplicity $n$. Am I overlooking an assumption? $\endgroup$
    – Jochen Glueck
    Commented May 25, 2023 at 15:35
  • $\begingroup$ Yes, we need to make some modification. $\endgroup$
    – ABB
    Commented May 25, 2023 at 15:36
  • $\begingroup$ Instead of $i\le j+2$, did you mean $j\ge i+2$? $\endgroup$
    – Iosif Pinelis
    Commented May 25, 2023 at 15:54
  • $\begingroup$ @Iosif Pinelis, Yes. Thanks. $\endgroup$
    – ABB
    Commented May 25, 2023 at 16:00

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