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I came across a similar question here, which should guarantee that $B$ is nilpotent, but I wonder if knowing that the matrices are similar (not just having the same eigenvalues) is enough to conclude that $B = 0$. I'm particularly interested in the case where $A$ is diagonalizable, but I would not be surprised if that isn't relevant here.

I considered using the Jordan-Chevalley decomposition to show that $B = 0$ since if $A = P(A + tB)P^{-1}$ for $t \neq 0$, we have $A = A + 0$ as well as $A = PAP^{-1} + tPBP^{-1}$, where $A$ and $PAP^{-1}$ are diagonalizable and $0$ and $tPBP^{-1}$ are nilpotent. However, I don't know that $A$ and $B$ commute, so this decomposition is not necessarily unique. Any help would be appreciated.

Joey
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    No. E.g. $A+tB=\pmatrix{1&t\ 0&0}$ is similar to $A=\pmatrix{1&0\ 0&0}$ for all $t$ and the latter is diagonalisable (because it has two distinct eigenvalues). – user1551 Aug 05 '22 at 17:04

1 Answers1

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For the non-diagonalizable case, take $$ A = \begin{bmatrix} 1 & 1 \\0 & 1 \end{bmatrix}, \qquad B = \begin{bmatrix} 0 & 1 \\0 & 0 \end{bmatrix}. $$ For all $t$ except $t=-1$, we have $A+tB$ similar to $A$.

GEdgar
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