A non-zero matrix $A$ is said to be nilpotent for some positive integer $k\geq2$. If $A$ is nilpotent then is $I+A$ invertible?? Where $I$ is the identity matrix.
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2This is probably a tenplicate. As a hint think about the eigenvalues of $A$ and $A+I$, by that specific order. – Git Gud Sep 03 '13 at 18:00
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http://math.stackexchange.com/questions/325318/prove-that-if-matrix-a-is-nilpotent-then-ia-is-invertible – njguliyev Sep 03 '13 at 18:00
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3You might also want to read the first sentence again. As currently written, it does not really make sense. – Tobias Kildetoft Sep 03 '13 at 18:04
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@GitGud It looks like if we got each of five voters to report a single different duplicate, we would have gotten all of them listed in the closure template... – rschwieb Sep 03 '13 at 18:53
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@rschwieb Gotta improve the synergy. – Git Gud Sep 03 '13 at 18:54
2 Answers
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Short answer: yes.
Long answer:
If $A^k=0$, consider the product $$A^{2n+1}+I=(A+I)(A^{2n}-A^{2n-1}+\cdots+I)$$ For a sufficiently large integer $n$.
Ben Grossmann
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Hint: Suppose $\,A^k=0\;$ , then:
$$(A-I)(A^{k-1}+A^{k-2}+...+A+I)=A^K-1\ldots$$
...but you have $\,A\color{red}+I\;$ , so what to do...? :)
DonAntonio
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