<s>[INST] Finding the smallest power of $A$ such that $A^n = I$ Let $A=\begin{bmatrix}0 & 1\\-1 & 1\end{bmatrix}$ then the smallest positive positive integer $n$ such that $A^n = I$ is :
(a) $1$ (b) $2$ (c) $4$ (d) $6$
proof: option (d) 6.
The characteristic polynomial of $A$ is $\lambda^2 - \lambda + 1$. So the eigenvalues of $A$ are $\omega, \omega^2$ where $\omega$ is a cube root of $-1$ and not equal to $-1$. So the eigenvalues of $A^n$ are $\omega^n$ and $\omega^{2n}$ and the eigenvalue of $I$ is $1$, so $\omega^{2n} = 1$ and $\omega^n = 1$ which implies $n$ as a multiple of $3$ and $2$ which means $6$ is the only option we have.
Is my reasoning correct?? And can we solve this without using eigenvalues?
 [/INST] Since $\lambda^2-\lambda +1$ is characteristic, it follows that $A^2 - A + I = 0$, i.e. \begin{align}
A^2 &= A - I\\
A^3 &= A^2 - A = -I\\
A^4 &= -A \\
A^5 &= A^2\cdot A^3 = I - A\\
A^6 &= A^3\cdot A^3 = I.
\end{align}
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