2
$\begingroup$

Let $$A= \begin{pmatrix} 0 & 1 & 0 \\ 1 & 0 & 1 \\ 1 & 0 & 0 \\ \end{pmatrix} $$ And I should calculate $A^2$ and $A^{12}$ by Cayley Hamilton theorem.

I found that the characteristic polynomial is $f_A(x)=x^3-x-1$ and thus by Cayley Hamilton: $A^3-A=I_{3}$ .

I tried to multiply by $A^9$ but it didn't lead to something simple to express $A^2$ and $A^{12}$ by.

Any suggestions?

$\endgroup$
2
$\begingroup$

First calculate $A^2$, then calculate $A^6=(A+I_3)^2=A^2 + 2A + I_3$ and finally $A^{12}=(A^6)^2$

$\endgroup$
  • $\begingroup$ You mean that I need to calculate clearly $A^2$ and than use it for the others? Thank You $\endgroup$ – A-H May 22 '16 at 18:48

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.