Let be a $ 2\times 2 $ real matrix $ A $ that has the property that $ \left| A^d-I_2 \right| =\left| A^d+I_2 \right| , $ for all $ d\in\{ 2014,2016 \} . $ Prove that $ \left| A^n-I_2 \right| =\left| A^n+I_2 \right| , $ for any natural number $ n. $
Problem
Source: Romanian National Olympiad 2016, grade xi, p.1
Tags: linear algebra, matrix
25.08.2019 21:53
$A=0$ so $ | A^n-I_2 | =1 = | A^n+I_2 | $
25.08.2019 22:02
dominicleejun wrote: $A=0$ so $ | A^n-I_2 | =1 = | A^n+I_2 | $ Why $ A=0 $ ?
25.08.2019 22:27
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.
26.08.2019 21:22
@dominic: this works if $A$ can be brought into diagonal form. This is not always true.
26.08.2019 21:26
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 $\lambda=\mu=0$.