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So I've got an $n\times n$ Matrix, $A$, such that it's $a$ on the diagonal and $1$ on the diagonal above the actual diagonal. I've shown the characteristic polynomial is $(x-a)^n$. I am asked to find its characteristic polynomial, playing around with smaller matrices I've got the feeling that the minimal polynomial is the same but can't prove it. Have tried to use induction but get stuck! Any help will be appreciated! (I get the minimal polynomial will be of the form $(x-a)^k$, where $k\in\{0,1,...,n\}$)

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  • $\begingroup$ Hint: do a 2 by 2 example by hand, then a 3 by 3 example by hand. There is a pattern for the number you call $k.$ It's amazing!! $\endgroup$
    – Will Jagy
    Oct 26, 2017 at 19:46
  • $\begingroup$ This is how I got the intuition that the two polynomials should be the same but I couldn't spot a pattern! $\endgroup$
    – Deke
    Oct 26, 2017 at 19:49

2 Answers 2

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Hint: find a vector $v$ such that $(A-aI)^{n-1} v \ne 0$.

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  • $\begingroup$ Would I be looking for a single vector for all n? Also, this would show that the powers less than $n-1$ are not zero would it? $\endgroup$
    – Deke
    Oct 26, 2017 at 19:42
  • $\begingroup$ Has to be a different vector for each $n$, because it's in $\mathbb R^n$. But they all have a similar form. $\endgroup$ Oct 26, 2017 at 23:13
  • $\begingroup$ I'm currently trying this method, but I am not sure what form $(A-aI)^{n-1}v $ takes so I'm finding it hard to find a $v$ for each n, any help? $\endgroup$
    – Deke
    Oct 27, 2017 at 12:32
  • $\begingroup$ Hint: what does $A-aI$ do to a vector that has only one nonzero entry? $\endgroup$ Oct 27, 2017 at 18:50
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Show that $$(A-aI)e_k=e_{k-1}$$ where $e_i$ is the canonical basis.

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  • $\begingroup$ I have been able to show this, but not sure how this leads to a proof of my question, unless I'm missing something? $\endgroup$
    – Deke
    Oct 27, 2017 at 12:50

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