$A=0$ so $ | A^n-I_2 | =1 = | A^n+I_2 | $

dominicleejun wrote: $A=0$ so $ | A^n-I_2 | =1 = | A^n+I_2 | $ Why $ A=0 $ ?

let A have 2 eigenvalues $\lambda$ and $\mu$ $(\lambda^{2014}-1)(\mu^{2014}-1)=(\lambda^{2014}+1)(\mu^{2014}+1)$ $(\lambda^{2016}-1)(\mu^{2016}-1)=(\lambda^{2016}+1)(\mu^{2016}+1)$ $\lambda^{2014}+\mu^{2014}=\lambda^{2016}+\mu^{2016}=0$ $\lambda=\mu=0$ i think A doesnt have to be 0 but A is nilpotent. $ | A^n-I_2 | =1 = | A^n+I_2 | $ still holds.

@dominic: this works if $A$ can be brought into diagonal form. This is not always true.

alexheinis wrote: @dominic: this works if $A$ can be brought into diagonal form. This is not always true. Weel, actually into triangular form, if you require $\lambda,\mu$ to be real. But you can take $\lambda,\mu$ to be complex and elaborate a little more the proof above to get that indeed $\lambda=\mu=0$.