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.
 A: The eigenvalues of $A$ (or $D$) satisfy $|\lambda | = { 1\over \sqrt{2}} <1$. Hence
$|\lambda|^n \to 0$.
A: 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 $0$.  
A: 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
$$\|A^n\|\leq {M\over n^{\lfloor {n\over 4}\rfloor}}\to 0$$
A: 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.
A: For generalisation, assume $$
        A=\begin{bmatrix}
        a & b \\
        c & d \\
        \end{bmatrix}
$$ Now, for every $2X2$ matrix, this equation is always valid $A^2-trace(A)A+det(A)I=0$ (prove this by youself, this is very easy). Now you can compare this equation with your original equation, you will get $a+d=-1$ and $ad-bc=\frac {1}{2}$. Assume any values of $a,b,c,d$ satistfying these equations which I had done earlier.
Assume the values $a=d=-0.5, -b=c=0.5$, then matrix will look like $$2A=
        \begin{bmatrix}
        -1 & -1  \\
         1 & -1  \\
        \end{bmatrix}
$$ This matrix satisfy all the given condition.Now you can proceed. You can get very easily that $A^n=0$ as $n \to \infty$ . And also you can also use the eigen value logic to prove that $A^n=0$ as $n \to \infty$
Eigen Value Logic can also be applied. Let the eigen value of this matrix be $e_1$ and $e_2$. From given equation it is clear that $det(A)=0.5=e_1e_2$ and both eigen values are non real. So $e_1e_2=e^2=0.5$ where $|e_1|=|e_2|=e$
Now $\alpha=|A^n|=|A|^n=2^{-n}$ and as $n \to \infty$, $\alpha=0$ which clearly proves that $A^n$ has to be zero matrix because determinant of that matrix is zero, otherwise determinat of that matrix will never be zero.
