Let S=A=pmatrix-1&a0&bpmatrix: a,b1,2,,100 and let T_n=A S: A^n(n+1)=I.

Question

Let $S=\left\{A=\begin{pmatrix}-1&a\0&b\end{pmatrix}: a,b\in\{1,2,\dots,100\}ight\}$ and let $T_n=\{A\in S: A^{n(n+1)}=I\}$. Then the number of elements in $\displaystyle\bigcap_{n=1}^{100}T_n$ is

Answer: 100

Accepted Answer

The matrix $A=\begin{pmatrix}-1&a\0&b\end{pmatrix}$ has eigenvalues $-1$ and $b$ (upper triangular). For $A^{n(n+1)}=I$ to hold for every $n\ge1$, both eigenvalues must satisfy $\lambda^{n(n+1)}=1$ for all $n$, forcing $\lambda\in\{1,-1\}$; since one eigenvalue is fixed at $-1$, we need $b=1$ (so that $b^{n(n+1)}=1$ always) and the $(-1)^{n(n+1)}=1$ automatically since $n(n+1)$ is always even.

Also the off-diagonal entry $a$ must not obstruct $A^{n(n+1)}=I$: since $A$ is triangular with eigenvalues $-1,1$ (distinct), $A$ is diagonalizable, and the condition holds for any $a\in\{1,\dots,100\}$ once $b=1$.

So $b=1$ fixed, $a$ free over $100$ values: $|{	extstyle\bigcap_n}T_n|=100$.

Solutions · 1