If P=[ … ] and Q=PAP^T, then P^T Q^2015 P is :

$$P=[\begin{matrix}\frac{\sqrt{3}}{2} & \frac{1}{2} \\ -\frac{1}{2} & \frac{\sqrt{3}}{2}\end{matrix}] {P}^{T}=[\begin{matrix}\frac{\sqrt{3}}{2} & \frac{-1}{2} \\ \frac{1}{2} & \frac{\sqrt{3}}{2}\end{matrix}]$$P=[23​​−21​​21​23​​​]PT=[23​​21​​2−1​23​​​]

$$P{P}^{T}={P}^{T}P=I$$PPT=PTP=I

${Q}^{2015}=(PA{P}^{T}) (PA{P}^{T})$Q2015=(PAPT)(PAPT) ......($2015$2015 terms)

$=P{A}^{2015} {P}^{T}$=PA2015PT ...$(i)$(i)

$\therefore {P}^{T} {Q}^{2015} P={P}^{T}P {A}^{2015}{P}^{T}P={A}^{2015}$∴PTQ2015P=PTPA2015PTP=A2015 from..$(i)$(i)

Given $A=[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}]$A=[10​11​]

$\Rightarrow$⇒ ${A}^{2}=[\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}] [\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}]$A2=[10​11​][10​11​]=[10​21​]

$\Rightarrow$⇒ ${A}^{3}=[\begin{matrix}1 & 2 \\ 0 & 1\end{matrix}] [\begin{matrix}1 & 1 \\ 0 & 1\end{matrix}]=[\begin{matrix}1 & 3 \\ 0 & 1\end{matrix}]$A3=[10​21​][10​11​]=[10​31​]

$\therefore$∴ ${A}^{2015}=[\begin{matrix}1 & 2015 \\ 0 & 1\end{matrix}]$A2015=[10​20151​]