TITLE: 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 [6 upvotes]: 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.

REPLY [3 votes]: While the answer provided by @cooper.hat is probably the one you want (as it finishes your otherwise correct proof), I would like to offer a different approach.
If $A^2=A-(1/2)I$, then multiplying by $A$ repeatedly and using induction, we see that $A^n=a_n A +b_nI$, and further that $A^{n+1}=a_n A^2 +b_n A=(a_n+b_n)A-(1/2)a_nI.$  Looking at the first few $(a_n,b_n)$ pairs, we have 
$$(1,0), (1,-1/2), (1/2,-1/2),(0,-1/4)$$
So $A^{4}=(-1/4)I$.  This makes it easy to compute larger powers of $A$ and see they are tending towards $0$.

REPLY [2 votes]: Let's do a few computations. We have:
$$A^2=A-{1\over 2}I$$
This means $A^3={A\over 2}-{I\over 2}$ and:
$$A^4=A^2-A+{1\over 4}I=-{1\over 4}I$$
We can prove by induction that
$$\begin{align}
A^{4k}&=\left({-1\over 4k}\right)^k\cdot I\\
A^{4k+1}&=\left({-1\over 4k}\right)^k\cdot A\\
A^{4k+2}&=\left({-1\over 4k}\right)^k\cdot\left(A-{1\over 2}I\right)\\
A^{4k+3}&=\left({-1\over 4k}\right)^k\cdot\left({A\over 2}-{1\over 2}I\right)
\end{align}$$
Let
$$M=\sup\{1,\|A\|,\|A-{I\over 2}\|,\|{A\over 2}-{I\over 2}\|\}$$
We have
$$\|A^n\|\leq {M\over n^{\lfloor {n\over 4}\rfloor}}\to 0$$