Timeline for Prove that $A+I$ is invertible if $A$ is nilpotent
Current License: CC BY-SA 3.0
3 events
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| Dec 1, 2016 at 15:06 | comment | added | Manos | @ArturoMagidin: Agreed. I was thinking about the algebraic closure. | |
| May 3, 2012 at 21:20 | comment | added | Arturo Magidin | You want to be careful: you can have a matrix for which all eigenvalues are $0$, but the matrix is not nilpotent (because it doesn't have "enough" eigenvalues); e.g., the $3\times 3$ matrix over $\mathbb{R}$ given by $$\left(\begin{array}{rrr}0 & -1 & 0\\1 & 0 & 0\\0 & 0 & 0\end{array}\right).$$ The characteristic polynomial is $t(t^2+1)$, so the only (real) eigenvalue is $0$, but the matrix is not nilpotent. So it depends on how you interpret "all its eigenvalues" (true if that means "in an algebraic closure of the ground field"). | |
| May 3, 2012 at 21:13 | history | answered | Manos | CC BY-SA 3.0 |