Let $A=(a_{ij})$ be an $n\times n$ invertible matrix over $\mathbb{C}$ and $D=diag(a_{11},a_{22},\dots,a_{nn})$ be the diagonal matrix whoes diagonal entries are same as $A$. Suppose $A-D$ is nilpotent. Is it true that $D$ is invertible?

For $n=2$, since every nilpotent matrix with zero diagonal entries is either upper- or lower-triangular, I already know this is true for $n=2$.

Thank you!

Thank you for user1551 for giving a counter example for $n\geq 3$.

I have modified the question a bit. I would like to assume the matrix $A$ having the property that $$a_{ij}\neq 0\Rightarrow a_{ji}=0.$$ Will it be true that $D$ is invertible under this assumption?


No. Here is a counterexample for every $n\ge3$: $$ A=\pmatrix{0&0&1\\ 0&1&-1\\ 1&1&1\\ &&&I_{n-3}} =\pmatrix{0&0&1\\ 0&0&-1\\ 1&1&0\\ &&&0_{n-3}}+\pmatrix{0\\ &1\\ &&1\\ &&&I_{n-3}}. $$ When $n=3$, we have $$ (A-D)^2=\pmatrix{0&0&1\\ 0&0&-1\\ 1&1&0}^2=\pmatrix{1&1&0\\ -1&-1&0\\ 0&0&0} \text{ and } (A-D)^3=0. $$


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