Given the invertible matrix $A$ so that $A+A^{-1}=2I_n$,

which of the following equalities stand true?






I know the formula for $A^{-1}$, but I'm not sure if and how should I use it here or what else should I apply.

Could I have some hints on how to approach this? Thank you

  • $\begingroup$ try multiply A in both sides in some equations. A * A^-1 = I. $\endgroup$ – J Tg Sep 1 '17 at 7:49
  • $\begingroup$ Hint: Consider that $I_n$ = $AA^{-1}$=$A^{-1}A$.Now get to work on the algebra. $\endgroup$ – Mathemagician1234 Sep 1 '17 at 7:51
  • $\begingroup$ multiplying $A$ both sides one gets characteristic equation $A^2-2A+I=0$ $\endgroup$ – Arun Sep 1 '17 at 7:54
  • 1
    $\begingroup$ $$(x+\dfrac1x)^2=x^2+2+\dfrac1{x^2}.$$ $$(x+\dfrac1x)^3=?$$ $\endgroup$ – Jyrki Lahtonen Sep 1 '17 at 7:59
  • $\begingroup$ It is easy to eliminate three of these by simple substitution into the original equation $\endgroup$ – Mark Bennet Sep 1 '17 at 9:02


First move would be to consider let $A=I$, that would eliminate a few options.

Also, try to cube both sides.


Hint :$$A+A^{-1}=2I \to (A+A^{-1}=2I)^3$$simplify then to what you want $$A^3+(A^{-1})^3+AAA^{-1}+AA^{-1}A+A^{-1}AA+AA^{-1}A^{-1}+A^{-1}AA^{-1}+A^{-1}A^{-1}A\\=A^3+A^{-3}+3A+3A^{-1}=8I\\\to \\A^3+A^{-3}+3\underbrace{(A+A^{-1})}_{2I}=8I$$

  • $\begingroup$ @Alexander :It is almost a real solution , $\endgroup$ – Khosrotash Sep 1 '17 at 8:06

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