0
$\begingroup$

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 $k\in\{0,1,...,n\}$)

|cite|improve this question
$\endgroup$
  • $\begingroup$ 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 '17 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 '17 at 19:49
1
$\begingroup$

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

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '17 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. $\endgroup$ – Robert Israel Oct 26 '17 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 '17 at 12:32
  • $\begingroup$ Hint: what does $A-aI$ do to a vector that has only one nonzero entry? $\endgroup$ – Robert Israel Oct 27 '17 at 18:50
1
$\begingroup$

Show that $$(A-aI)e_k=e_{k-1}$$ where $e_i$ is the canonical basis.

|cite|improve this answer
$\endgroup$
  • $\begingroup$ 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 '17 at 12:50

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.