<s>[INST] Let $A$ be a $2 \times 2$ real matrix such that $A^2 - A + (1/2)I = 0$. Prove that $A^n \to 0$ as $n \to \infty$. Question: Let $A$ be a $2 \times 2$ matrix with real entries such that $A^2 - A + (1/2)I = 0$, where $I$ is the $2 \times 2$ identity matrix and $0$ is the $2 \times 2$ zero matrix. Prove that $A^n \to 0$ as $n \to \infty$.
My attempt: Here is my idea so far. Consider $A$ as a $2 \times 2$ matrix over the field of complex numbers. Now, the polynomial 
$$g(t) = t^2 - t + \frac{1}{2} $$
factors as 
$$g(t) = \left(\frac{1}{2} - \frac{i}{2}\right)\left(\frac{1}{2}+\frac{i}{2}\right)$$
over $C$. This means that the minimal polynomial of $A$ over $C$ is: 
$$m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right), m(t) = \left(t-(\frac{1}{2} + \frac{i}{2})\right), \ \ \ \text{ or } \ \ \ m(t) = \left(t-(\frac{1}{2} - \frac{i}{2})\right)\left(t-(\frac{1}{2} + \frac{i}{2})\right),$$
and thus we have $A = Q D Q^{-1}$, where
$$D =\left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 - i)}
\end{array}} \right],$$
$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 + i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right],\ \ \ \text{ or } $$
$$D = \left[ {\begin{array}{*{20}{c}}
{\frac{1}{2}(1 - i)}&0\\
0&{\frac{1}{2}(1 + i)}
\end{array}} \right].$$
Now, $A^n = QD^nQ^{-1}$. 
Here is where I get stuck. $D^n$ doesn't seem to be converging to $0$ as $n$ approaches infinity. In addition, I am concerned that my strategy is bad, since we are talking about a real matrix and I'm using a minimal polynomial over $C$. Is it still true that $A$ must have one of the three forms above, even if $A$ is supposed to be real? 
Thanks for any help/suggestions you may be able to provide.
 [/INST] You can get $A^n=A^{n-1}-1/2A^{n-2}=(A^{n-2}-1/2A^{n-3})-1/2A^{n-2}=1/2A^{n-2}-1/2A^{n-3}=-1/4A^{n-4}$. Then the conclusion is obvious by induction on k, while let n=4k+i.
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