Let A=pmatrix a & 1√2 1√2 & bpmatrix+1 (a 22 matrix built from real a,b), and P=pmatrix & …

Question

Let $A=\begin{pmatrix}\sqrt a & \frac1{\sqrt2}\ \frac1{\sqrt2} & \sqrt b\end{pmatrix}+1$ (a $2	imes2$ matrix built from real $a,b$), and $P=\begin{pmatrix}\cos	heta & -\sin	heta\ \sin	heta & \cos	heta\end{pmatrix},\ 	heta>0$. Let $B=PAP^T$, $C=P^TB^{10}P$. If the sum of the diagonal elements of $C$ is $\dfrac mn$ with $\gcd(m,n)=1$, then $m+n$ is:

Accepted Answer

Since $P^TP=I$ (orthogonal), $P^TB=AP^T$ and $P^TBP=A$, so $P^TB^{10}P=A^{10}=C$.

With $A=\begin{pmatrix}\frac{1}{\sqrt2} & -2\ 0 & 1\end{pmatrix}$ (as reconstructed from the source), one finds $A^n=\begin{pmatrix}\left(\frac1{\sqrt2}ight)^n & \ast \ 0 & 1\end{pmatrix}$, so $\operatorname{tr}(C)=\operatorname{tr}(A^{10})=\left(\dfrac1{\sqrt2}ight)^{10}+1=\dfrac1{32}+1=\dfrac{33}{32}$.

So $m=33,\ n=32$, $\gcd(m,n)=1$, and $m+n=65$.

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